3.8 Exercise Problems

Q1: Refer to the graphs in Fig 3.1. Match the following.

\(
\begin{array}{|c|c|}
\hline \text { Graph } & \text { Characteristic } \\
\hline \text { (a) } & \text { (i) has } v>0 \text { and a }<0 \text { throughout. } \\
\hline \text { (b) } & \begin{array}{l}
\text { (ii) has } x>0 \text { throughout and has a point with } v=0 \text { and a point } \\
\text { with } a=0 \text {. }
\end{array} \\
\hline \text { (C) } & \text { (iii) has a point with zero displacement fort }>0 \text {. } \\
\hline \text { (d) } & \text { (iv) has } v<0 \text { and } a>0 \text {. } \\
\hline
\end{array}
\)

Answer: (a) (iii); (b) (ii); (c) (iv); (d) -(i)

In graph (a), There is a point (B) on the curve for which displacement is zero. So curve, (a) matches with (iii).

In graph (b), In this graph, \(x\) is positive \((>0)\) throughout and at point \(B\) the highest point of curve the slope of the curve is zero. It means at this point \(v=d x / d t=0\). Also at point \(C\) the \(d t\) curvature changes, which means at this point the acceleration of the particle should be zero or \(\mathrm{a}=0\), So curve (b) matches with (ii).

In graph (c), In this graph the slope is always negative, hence velocity will be negative or \(v<0\). Also \(x\)-t graph opens up, it represents positive acceleration. So curve (c) matches with (iv).

In graph (d), In this graph the slope is always positive, hence velocity will be positive or \(v>0\). Also \(x\)-t graph opens down, it represents negative acceleration. So curve (d) matches with (i).

Q2: A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with time. (Take acceleration in the backward direction as positive).

Answer: Impulsive Force is generated by the bat: If we ignore the effect of gravity just by analyzing the motion of the ball in the horizontal direction only, then the ball moving uniformly will return back with the same speed when a bat hits it. The acceleration of the ball is zero just before it strikes the bat. When the ball strikes the bat, it gets accelerated due to the applied impulsive force by the bat.

Q3: Give examples of a one-dimensional motion where
(a) the particle moving along positive \(x\)-direction comes to rest periodically and moves forward.
(b) the particle moving along positive \(x\)-direction comes to rest periodically and moves backward.

Answer: The equation which contains sine and cosine functions is periodic in nature.
(a) The particle will be moving along positive \(x\)-direction only if \(t>\sin t\) We have displacement as a function of time, \(x(t)=t-\sin t\) By differentiating this equation w.r.t. time we get velocity of the particle as a function of time.
velocity \(v(t)=\frac{d x(t)}{d t}=1-\cos t\)
If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time.
acceleration \(a(t)=\frac{d v}{d t}=\sin t\)
when \(t=0 ; x(t)=0\)
when \(t=\pi ; x(t)=\pi>0\)
when \(t=0 ; x(t)=2 \pi>0\)
(b) Equation can be represented by
\(
\begin{aligned}
& x(t)=\sin t  \\
& v=\frac{d}{d t} x(t)=\cos t \text { and } a=\frac{d v}{d t}=-\sin t
\end{aligned}
\)
At \(t=0 ; x=0, v=1\) (positive) and \(a=0\)
At \(t=\frac{\pi}{2} ; x=1\) (positive), \(v=0\) and \(a=-1\) (negative)
At \(t=\pi, x=0, v=-1\) (negative) and \(a=0\)
At \(t=\frac{3 \pi}{2} ; x=-1\) (negative), \(v=0\) and \(a=+1\) (positive)
At \(t=2 \pi, x=0, v=1\) (positive) and \(a=0\)
Hence the particle moving along positive \(x\)-direction comes to rest periodically and moves backward.
As displacement and velocity is involving sin \(t\) and \(\cos t\), hence these equations represent periodic nature.

Q4: Give an example of a motion where \(x>0, v<0, a>0\) at a particular instant.

Answer: Let the motion is represented by
\(
x(t)=A+B e^{-\gamma t}
\)
Let \(A>B\) and \(\gamma>0 \dots(i)\)
Now velocity \(x(t)=\frac{d x}{d t}=-B \gamma e^{-\gamma t}\)
Acceleration \(a(t)=\frac{d x}{d t}=B \gamma^2 e^{-\gamma t}\)
Suppose we are considering any instant of time \(t\), then from Eq. (i), we can say that
\(
x(t)>0 ; v(t)<0 \text { and } a>0
\)

Q6: An object falling through a fluid is observed to have acceleration given by \(a=g-b v\) where \(g=\) gravitational acceleration and \(\mathrm{b}\) is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?

Answer: Key concept: If a spherical body of radius \(r\) is dropped in a viscous fluid, it is first accelerated and then its acceleration becomes zero and it attains a constant velocity called terminal velocity.
According to the problem, the acceleration of an object is given by the relation \(a=g-b v\)
When speed becomes constant acceleration \(\mathrm{a}=\mathrm{dv} / \mathrm{dt}=0\) (uniform motion).
where, \(g\) = gravitational acceleration
Clearly, from the above equation as speed increases acceleration will decrease. At a certain speed say \(v_0\), acceleration will be zero and speed will remain constant. Hence, \(a=g\) – \(b v_0=0 \Rightarrow v_0=g / b\)

Q7: A ball is dropped and its displacement vs time graph is as shown in Fig. 3.4 (displacement \(x\) is from the ground and all quantities are +ve upwards).
(a) Plot qualitatively velocity vs time graph.
(b) Plot qualitatively acceleration vs time graph.

Answer:

Key concept: To calculate velocity we will find slope which is calculated by \(\frac{d x}{d t}\) for the displacement-time curve and to find acceleration we will find slope \(\frac{d V}{d t}\) of the velocity-time curve.
Sign convention: We are taking downward as negative and upward as positive.
The ball is bouncing on the ground and it is clear from the graph that displacement \(x\) is positive throughout. Ball is dropped from a height and its velocity increases in a downward direction due to gravity pull. In this condition \(v\) is negative but the acceleration of the ball is equal to the acceleration due to gravity i.e., \(a=-g\). When the ball rebounds in an upward direction its velocity is positive but acceleration is \(a=-g\).
(a) The velocity-time graph of the ball is shown in Fig. (i).

(b). The acceleration-time graph of the ball is shown in Fig. (ii).

Q8: A particle executes the motion described by \(x(t)=x_o\left(1-e^{-\gamma t}\right) ; t \geq 0\), \(x_0>0\)
(a) Where does the particle start and with what velocity?
(b) Find maximum and minimum values of \(x(t), v(t), a(t)\). Show that \(x(t)\) and \(a(t)\) increase with time and \(v(t)\) decreases with time.

Answer: Initially, we have to calculate velocity and acceleration and then we can determine the maximum or minimum value accordingly.
We have displacement as a function of time,
\(
x(t)=x_0\left(1-e^{-\gamma t}\right)
\)
By differentiating this equation w.r.t. time we get the velocity of the particle as a function of time.
\(
v(t)=\frac{d x(t)}{d t}=x_0 \gamma e^{-\gamma t}
\)
If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time.
\(
a(t)=\frac{d v(t)}{d t}=-x_0 \gamma^2 e^{-\gamma t}
\)
\(
\begin{gathered}
\text { (a) When } t=0 ; x(t)=x_0\left(1-e^{-0}\right)=x_0(1-1)=0 \\
x(t=0)=x_0 \gamma e^{-0}=x_0 \gamma(1)=\gamma x_0
\end{gathered}
\)
(b) \(x(t)\) is maximum when \(t=\infty[x(t)]_{\max }=x_0\)
\(x(t)\) is minimum when \(t=0[x(t)]_{\min }=0\)
\(v(t)\) is maximum when \(t=0 ; v(0)=x_0 \gamma\)
\(v(t)\) is minimum when \(t=\infty ; v(\infty)=0\)
\(a(t)\) is maximum when \(t=\infty ; a(\infty)=0\)
\(a(t)\) is minimum when \(t=0 ; a(0)=-x_0 \gamma^2\)

Q9: A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18 \(\mathrm{m} / \mathrm{h}\) while the other has the speed of \(27 \mathrm{~km} / \mathrm{h}\). The bird starts moving from the first car towards the other and is moving with the speed of \(36 \mathrm{~km} / \mathrm{h}\) and when the two cars were separated by \(36 \mathrm{~km}\). What is the total distance covered by the bird? What is the total displacement of the bird?

Answer: Concept of relative velocity (for 1-D): If two objects are moving along the same straight line and we are observing the motion from the frame of one object. Then for the relative velocity, it will be subtracted for velocities in same direction and added for velocities in opposite directions. (Remember: add or subtract them with proper sign conventions).
We can find out the relative speed of the cars by adding the speed of the two cars
\(
\begin{aligned}
& =27+18 \\
& =45 \mathrm{~km} / \mathrm{hr}
\end{aligned}
\)
Time taken to meet \(=\frac{\text { Distance between cars }}{\text { relative speed of the cars }}\)
\(
=\frac{36}{45}=\frac{4}{5}{\text { hours }}
\)
\(
\text { Hence, the distance that the bird will cover in } \frac{4}{5} \text { hours }=36\left(\frac{4}{5}\right)=28.8 \mathrm{~km} \text {. }
\)

Q10: A man runs across the rooftop of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 \(\mathrm{m} / \mathrm{s}\), the (horizontal) distance between the two buildings is \(10 \mathrm{~m}\) and the height difference is \(9 \mathrm{~m}\), will he be able to land on the next building? (take \(g=10 \mathrm{~m} / \mathrm{s}^2\) )

Answer:

Horizontal Projectile: When a body is projected horizontally from a certain height ‘ \(y\) ‘ vertically above the ground with initial velocity \(u\). If friction is considered to be absent, then there is no other horizontal force that can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers an equal distance in the horizontal direction in equal intervals of time.
Time of flight: If a body is projected horizontally from a height \(\mathrm{h}\) with velocity \(\mathrm{u}\) and the time taken by the body to reach the ground is \(T\), then
\(h=0+\frac{1}{2} g T^2 \ldots \ldots . …(For ~vertical ~motion)\)
\(T=\sqrt{\frac{2 h}{g}}\)
Horizontal range: Let \(\mathrm{R}\) be the horizontal distance travelled by the body
\(R=u T+\frac{1}{2} 0 T^2 \ldots \ldots . .(\) For horizontal motion \(\mathrm{a}=0\) )
\(R=u \sqrt{\frac{2 h}{g}}\)
We will apply kinematic one by one downward and horizontal. We first consider motion horizontal and there is no horizontal force which can affect horizontal motion. The horizontal velocity therefore remains constant and so the object covers an equal distance in the horizontal direction in equal intervals of time.
\(
\text { According to the problem, the horizontal speed of the } \operatorname{man}\left(u_x\right)=9 \mathrm{~m} / \mathrm{s} \text { Horizontal distance between the two buildings }=10 \mathrm{~m}
\)
\(
\text { Height difference between the two buildings }=9 \mathrm{~m} \text { and } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2
\)

Let the man jumps from point \(A\) and land on the roof of the next building at point \(B\).
Taking motion in the vertical direction,
\(
\begin{aligned}
& y=u t+\frac{1}{2} a t^2 \\
& 9=0 \times t+\frac{1}{2} \times 10 \times t^2 \\
& 9=5 t^2 \\
& \text { or } \mathrm{t}=\sqrt{\frac{9}{5}}=\frac{3}{\sqrt{5}}
\end{aligned}
\)
\(\therefore\) Horizontal distance travelled \(=u_x \times t=9 \times \frac{3}{\sqrt{5}}=\frac{27}{\sqrt{5}} \mathrm{~m} \approx 12 \mathrm{~m}\)
The horizontal distance travelled by the man is greater than \(10 \mathrm{~m}\), therefore, he will land on the next building.

Q11: A ball is dropped from a building of height \(45 \mathrm{~m}\). Simultaneously another ball is thrown up with a speed \(40 \mathrm{~m} / \mathrm{s}\). Calculate the relative speed of the balls as a function of time.

Answer: In motion under gravity, if the ball is released or dropped that means its initial velocity is zero. In this problem as ball is dropped, so its initial velocity will be taken as zero. We will apply kinematic equations.
According to the problem, for the ball dropped from the building, \(u_1=0\), \(u_2=40 \mathrm{~m} / \mathrm{s}\)
The velocity of the ball after time \(t\),
\(
\begin{aligned}
& v_1=u_1-g t \\
& v_1=-g t
\end{aligned}
\)
And for another ball which is thrown upward \(u_2=40 \mathrm{~m} / \mathrm{s}\)

The velocity of the ball after time \(t\),
\(
v_2=u_2-g t=(40-g t)
\)
\(\therefore\) Relative velocity of one ball w.r.t. another ball is
\(
\begin{aligned}
& v_{12}=v_1-v_2=-g t-[-40-g t] \\
& v_{12}=v_1-v_2=-g t+40+g t=40 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Q12: The velocity-displacement graph of a particle is shown in Fig. 3.5.
(a) Write the relation between \(v\) and \(x\).
(b) Obtain the relation between acceleration and displacement and plot it.

Answer:

(a) Consider the point \(P(x, v)\) at any time \(t\) on the graph such that angle \(A B O\) is \(\theta\) such that \(\tan \theta=\frac{A Q}{Q P}=\frac{\left(v_0-v\right)}{x}=\frac{v_0}{x_0}\)

When the velocity decreases from \(v_0\) to zero during the displacement, the acceleration becomes negative.
\(
\begin{aligned}
& v_0-v=\left(\frac{v_0}{x_0}\right) x \\
& v=v_0\left(1-\frac{x}{x_0}\right)
\end{aligned}
\)
is the relation between \(v\) and \(x\).
\(
\begin{aligned}
& \text { b) } a=\frac{d v}{d t}=\left(\frac{d v}{d t}\right)\left(\frac{d x}{d x}\right)=\left(\frac{d v}{d x}\right)\left(\frac{d x}{d t}\right) \\
& a=\frac{-v_0}{x_0} v \\
& a=\left(\frac{v_0^2}{x_0^2}\right) x-\left(\frac{v_0^2}{x_0}\right)
\end{aligned}
\)

\(A t x=0\)
\(a=\frac{-v_0^2}{x^0}\)
At \(a=0\)
\(
x=x_0
\)
The points are
\(
\left(0, \frac{-v_0^2}{x_0}\right) \text { and } B\left(x_0, 0\right)
\)

Q13: It is a common observation that rain clouds can be at about a kilometer altitude above the ground.
(a) If a raindrop falls from such a height freely under gravity, what will be its speed? Also calculate in \(\mathrm{km} / \mathrm{h} .\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
(b) A typical raindrop is about \(4 \mathrm{~mm}\) diameter. Momentum is mass \(\mathrm{x}\) speed in magnitude. Estimate its momentum when it hits the ground.
(c) Estimate the time required to flatten the drop.
(d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
(e) Estimate the order of magnitude force on the umbrella. Typical lateral separation between two raindrops is \(5 \mathrm{~cm}\).
(Assume that the umbrella is circular and has a diameter of \(1 \mathrm{~m}\) and the cloth is not pierced through !!)

Answer: Key concept: This problem can be solved by kinematic equations of motion and Newton’s second law that \(F_{\text {ext }}=\mathrm{dp} / \mathrm{dt}\) will be used, where dp is change in momentum over time \(\mathrm{dt}\).
(a) According to the problem \((\mathrm{h})=1 \mathrm{~km}=1000 \mathrm{~m}\) and we know that the initial velocity of the ball is zero. And displacement covered by rain drop in downward direction, so we will taking \(\mathrm{h}\) as negative. (We are neglecting the air resistance.)
\(
\begin{aligned}
& \begin{array}{l}
h=1 \mathrm{~km}=1000 \mathrm{~m} \\
g=10 \mathrm{~m} / \mathrm{s}^2
\end{array} \\
& \begin{array}{l}
d=4 \mathrm{~mm} \text { and } u=0 \mathrm{~m} / \mathrm{s} \\
\qquad r=\frac{4}{2} \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} \\
\end{array} \\
\end{aligned}
\)
(a) Let’s find out the velocity of raindrops on the ground’ \(v^2=u^2-2 a s\)
Velocity attained by the raindrop in freely falling through a height \(h\) is
\(
v^2=u^2-2 g(-h)
\)
As \(u=0\)
So,
\(
v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 1000}=100 \sqrt{2} \mathrm{~m} / \mathrm{s}
\)
\(
=100 \sqrt{2} \times \frac{60 \times 60}{1000} \mathrm{~km} / \mathrm{h}=360 \sqrt{2} \mathrm{~km} / \mathrm{h} \approx 510 \mathrm{~km} / \mathrm{h}
\)
(b) Diameter of the drop \((d)=2 r=4 \mathrm{~mm}\)
Radius of the drop \((r)=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Mass of a raindrop \((m)=V \times \rho\)
\(
=\frac{4}{3} \pi r^3 \rho=\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3
\)
\(\left(\because\right.\) Density of water \(\left.=10^3 \mathrm{~kg} / \mathrm{m}^3\right)\)
\(
\Rightarrow m=3.4 \times 10^{-5} \mathrm{~kg}
\)
Momentum of the rain \(\operatorname{drop}(p)=m v\)
\(
\begin{aligned}
& =3.4 \times 10^{-5} \times 100 \sqrt{2} \\
\Rightarrow \quad p & =4.7 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \approx 5 \times 10^{-3} \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}
\)
(c) Time required to flatten the drop = Time taken by the drop to travel the distance equal to the diameter of the drop near the ground
\(
\begin{aligned}
t=\frac{d}{v}=\frac{4 \times 10^{-3}}{100 \sqrt{2}} & =0.028 \times 10^{-3} \mathrm{~s} \\
& =2.8 \times 10^{-5} \mathrm{~s} \approx 30 \mathrm{~ms}
\end{aligned}
\)
(d) Force exerted by a raindrop is
\(
\begin{aligned}
F & =\frac{\text { Change in momentum }}{\text { Time }}=\frac{p-0}{t} \\
& =\frac{4.7 \times 10^{-3}}{2.8 \times 10^{-5}} \approx 168 \mathrm{~N}
\end{aligned}
\)
(e) Radius of the umbrella \((R)=\frac{1}{2} \mathrm{~m}\)
\(\therefore\) Area of the umbrella
\(
\begin{aligned}
(A) & =\pi R^2=\frac{22}{7} \times\left(\frac{1}{2}\right)^2 \\
& =\frac{22}{28}=\frac{11}{14} \approx 0.8 \mathrm{~m}^2
\end{aligned}
\)
Number of drops striking the umbrella simultaneously with an average separation of \(5 \mathrm{~cm}\left(5 \times 10^{-2} \mathrm{~m}\right)\)
\(
n=\frac{0.8}{\left(5 \times 10^{-2}\right)^2}=320
\)
\(\therefore\) Net force exerted on umbrella \(=320 \times 168=53760 \mathrm{~N} \approx 54000 \mathrm{~N}\)

Q14: A motor car moving at a speed of \(72 \mathrm{~km} / \mathrm{h}\) can not come to a stop in less than \(3.0 \mathrm{~s}\) while for a truck this time interval is \(5.0 \mathrm{~s}\). On a highway the car is behind the truck both moving at \(72 \mathrm{~km} / \mathrm{h}\). The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck? Human response time is \(0.5 \mathrm{~s}\).

Answer: 

\(
\text { According to the problem, speed of car as well as truck }=72 \mathrm{~km} / \mathrm{h}
\)
Given : (for truck)
\(
\begin{aligned}
& U=72 \mathrm{~km} / \mathrm{hr}=2 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} \\
& =20 \mathrm{~m} / \mathrm{s} \\
& V=0, t=5 \mathrm{~s} \\
& a=?,
\end{aligned}
\)
We know that,
\(
V=u+a t
\)
i.e., \(0=20+a(5)\)
thus, \(a=\frac{-20}{5}=\frac{-4 m}{s}\)
Given : (for car)
\(
U=20 \mathrm{~m} / \mathrm{s}
\)
\(
\begin{aligned}
& V=0, t=3 s \\
& a=a_c
\end{aligned}
\)
Again, \(V=u+a t\)
\(
0=20+a_c{ }^3
\)
\(
\text { or } a_c=-\frac{20}{3} \mathrm{~m} / \mathrm{s}^2
\)
Let the car be at a distance \(s\) from the truck, when the truck gives the signal and \(t\) be the time taken to cover this distance.

As human response time is \(0.5 \mathrm{~s}\), in this time car will cover some distance with uniform velocity. Therefore, the time of retarded motion of the car is \((t-0.5) \mathrm{s}\). The velocity of car after time \(t\),
\(
v_c=u-a t=20-\left(\frac{20}{3}\right)(t-0.5)
\)
The velocity of the truck after time \(t\),
\(
v_t=20-4 t
\)
To avoid the car bumping onto the truck
\(
\begin{aligned}
& 20-\frac{20}{3}(t-0.5)=20-4 t \\
& 4 t=\frac{20}{3}(t-0.5) \\
& \Rightarrow \quad t=\frac{2.5}{2}=\frac{5}{4} \mathrm{~s} \\
&
\end{aligned}
\)
Distance travelled by truck in time \(t\),
\(
\begin{aligned}
s_t & =u_t t+\frac{1}{2} a_t t^2 \\
\Rightarrow \quad s_t & =20 \times \frac{5}{4}+\frac{1}{2} \times(-4) \times\left(\frac{5}{4}\right)^2=21.875 \mathrm{~m}
\end{aligned}
\)
Distance travelled by the truck in time \(t\),
\(
\begin{aligned}
& s_t=u_t t+\frac{1}{2} a_t t^2 \\
\Rightarrow & s_t=20 \times \frac{5}{4}+\frac{1}{2} \times(-4) \times\left(\frac{5}{4}\right)^2=21.875 \mathrm{~m}
\end{aligned}
\)
Distance travelled by car in time \(t=\) Distance travelled by car in \(0.5 \mathrm{~s}\) (without retardation) + Distance travelled by car in \((t-0.5) \mathrm{s}\) (with retardation)
\(
\begin{aligned}
& s_c=(20 \times 0.5)+20\left(\frac{5}{4}-0.5\right)-\frac{1}{2}\left(\frac{20}{3}\right)\left(\frac{5}{4}-0.5\right)^2=23.125 \mathrm{~m} \\
& \therefore \quad s_c-s_t=23.125-21.875=1.250 \mathrm{~m} \\
&
\end{aligned}
\)
Therefore, to avoid the collision with the truck, the car must maintain a distance from the truck more than \(1.250 \mathrm{~m}\).

Q15: A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time \(t\) is given by \(v(t)=2 t(3-t) ; 0<t<3\) and \(v(\mathrm{t})=-(t-3)(6-t)\) for \(3<\mathrm{t}<6 \mathrm{~s}\) in \(\mathrm{m} / \mathrm{s}\). It repeats this cycle till it reaches the height of \(20 \mathrm{~m}\).
(a) At what time is its velocity maximum?
(b) At what time is its average velocity maximum?
(c) At what time is its acceleration maximum in magnitude?
(d) How many cycles (counting fractions) are required to reach the top?

Answer: Given the velocity
\(
v(t)=2 t(3-t)=6 t-2 t^2 \dots(i)
\)
(a) For maximum velocity \(\frac{d v(t)}{d t}=0\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{d}{d t}\left(6 t-2 t^2\right)=0 \\
& \Rightarrow \quad 6-4 t=0 \\
& \Rightarrow \quad t=\frac{6}{4}=\frac{3}{2} \mathrm{~s}=1.5 \mathrm{~s}
\end{aligned}
\)
(b) From Eq. (i) \(v=6 t-2 t^2\)
\(
\begin{aligned}
& \Rightarrow \frac{d s}{d t}=6 t-2 t^2 \\
& \Rightarrow d s=\left(6 t-2 t^2\right) d t
\end{aligned}
\)
where, \(s\) is displacement
\(\therefore\) Distance travelled in time interval 0 to \(3 \mathrm{~s}\),
\(
\begin{aligned}
s & =\int_0^3\left(6 t-2 t^2\right) d t \\
& =\left[\frac{6 t^2}{2}-\frac{2 t^3}{3}\right]_0^3=\left[3 t^2-\frac{2}{3} t^3\right]_0^3
\end{aligned}
\)
\(
=\left[3 \times 9-\frac{2}{3} \times 2\right]=27-18
\)
Thus, \(s=9 m\)
Now average velocity \(
\left(V_{a v}\right)=\frac{9}{3}=\frac{3 m}{s}
\)
\(
\begin{aligned}
& V(t)=6 t-2 t^2 \ldots \ldots \ldots \text { since } 0<t<3 \\
& 3=6 t-2 t^2 \ldots \ldots \ldots . \operatorname{since} V_{a v}=3 \\
& 2 t^2-6 t+3=0 \quad \ldots \ldots \ldots \ldots . \ldots \operatorname{since}(a=2, b=-6 \text { and } c=3) \\
& t=2.36 s
\end{aligned}
\)
Thus, the average velocity is maximum at 2.36 seconds.
(c) In a periodic motion when velocity is zero acceleration will be maximum putting \(v=0\) in Eq. (i).
\(
\begin{aligned}
0 & =6 t-2 t^2 \Rightarrow 0=t(6-2 t) \\
& =t \times 2(3-t)=0 \Rightarrow t=0 \text { or } 3 \mathrm{~s}
\end{aligned}
\)
(d) Distance covered in 0 to \(3 \mathrm{~s}=9 \mathrm{~m}\) Distance covered in 3 to \(6 \mathrm{~s}\)
\(
\begin{aligned}
& =\int_3^6\left(18-9 t+t^2\right) d t=\left(18 t-\frac{9 t^2}{2}+\frac{t^3}{3}\right)_3^6 \\
& =18 \times 6-\frac{9}{2} \times 6^2+\frac{6^3}{3}-\left(18 \times 3-\frac{9 \times 3^2}{2}+\frac{3^3}{3}\right) \\
& =108-9 \times 18+\frac{6^3}{3}-18 \times 3+\frac{9}{2} \times 9-\frac{27}{3} \\
& =-4.5 \mathrm{~m}
\end{aligned}
\)
\(\therefore\) Total distance travelled in one cycle
\(
=s_1+s_2=9-4.5=4.5 \mathrm{~m}
\)
Number of cycles to be covered in the total distance \(=\frac{20}{4.5} \approx 4.44 \approx 5\)

Q16: A man is standing on top of a building \(100 \mathrm{~m}\) high. He throws two balls vertically, one at \(t=0\) and the other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between the first and second ball is \(+15 \mathrm{~m}\) at \(t=2 \mathrm{~s}\). The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Answer:  We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take We solve this problem by using kinematic equations with proper sign convention and to calculate time interval we will take the difference of displacements.

Let the speeds of the two balls ( 1 and 2) be \(v_1\) and \(v_2\) where:
if \(v_1=2 v, v_2=v\)
if \(y_1\) and \(y_2\) and the displacement covered by the balls 1 and 2, respectively, before coming to rest, then
\(
y_1=\frac{v_1^2}{2 g}=\frac{4 v^2}{2 g} \text { and } y_2=\frac{v_2^2}{2 g}=\frac{v^2}{2 g}
\)
Since \(y_1-y_2=15 \mathrm{~m}, \frac{4 v^2}{2 g}-\frac{v^2}{2 g}\)
\(
=15 \mathrm{~m} \text { or } \frac{3 v^2}{2 g}=15 \mathrm{~m}
\)
or \(v^2=\sqrt{5 \mathrm{~m} \times(2 \times 10)} \mathrm{m} / \mathrm{s}^2\)
or \(v=10 \mathrm{~m} / \mathrm{s}\)
Clearly, \(v_1=20 \mathrm{~m} / \mathrm{s}\) and \(v_2=10 \mathrm{~m} / \mathrm{s}\)
as \(y_1=\frac{v_1^2}{2 g}=\frac{(20 \mathrm{~m})^2}{2 \times 10 \mathrm{~m}}=20 \mathrm{~m}\)
\(
y_2=y_1-15 \mathrm{~m}=5 \mathrm{~m}
\)
If \(t_2\) is the time taken by the ball 2 to cover a displacement of \(5 \mathrm{~m}\), then from
\(
\begin{aligned}
& y_2=v_2 t-\frac{1}{2} g t_2^2 \\
& 5=10 t_2-5 t_2^2 \text { or } t_2^2-2 t_2+1=0
\end{aligned}
\)
where \(t_2=1 \mathrm{~s}\)
Since \(t_1\) (time taken by ball 1 to cover the distance of \(20 \mathrm{~m}\) ) is \(2 \mathrm{~s}\), the time interval between the two throws
\(
=t_1-t_2=2 s-1 s=1 s
\)

Q17: In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.

Answer: (a) & (b)

(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point-sized object.
(b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point-sized object on the track.
(c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
(d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.

Q18: The position-time \((x-t)\) graphs for two children A and B returning from their school \(\mathrm{O}\) to their homes \(\mathrm{P}\) and \(\mathrm{Q}\) respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below ;
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) \((\mathrm{A} / \mathrm{B})\) walks faster than \((\mathrm{B} / \mathrm{A})\)
(d) \(A\) and \(B\) reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).

Answer: (a) A lives closer to school than \(B\), because \(B\) has to cover higher distances \([O P<O Q]\),
(b) A starts earlier for school than \(B\), because \(t=0\) for \(A\) but for \(B\), \(t\) has some finite time.
(c) As slope of \(B\) is greater than that of \(A\), thus \(B\) walks faster than \(A\).
(d) \(A\) and \(B\) reach home at the same time.
(e) At the point of intersection (i.e., X), B overtakes \(A\) on the roads once.

Q19: A woman starts from her home at \(9.00 \mathrm{am}\), walks with a speed of \(5 \mathrm{~km} \mathrm{~h}^{-1}\) on a straight road up to her office \(2.5 \mathrm{~km}\) away, stays at the office up to \(5.00 \mathrm{pm}\), and returns home by an auto with a speed of \(25 \mathrm{~km} \mathrm{~h}^{-1}\). Choose suitable scales and plot the \(x\)-t graph of her motion.

Answer: Speed of the woman \(=5 \mathrm{~km} / \mathrm{h}\)
Distance between her office and home \(=2.5 \mathrm{~km}\)
\(
\begin{aligned}
& \text { Time taken }=\frac{\text { Distance }}{\text { Speed }} \\
& =2.5 / 5=0.5 \mathrm{~h}=30 \mathrm{~min}
\end{aligned}
\)
It is given that she covers the same distance in the evening by auto.
Now, speed of the auto \(=25 \mathrm{~km} / \mathrm{h}\)
\(
\begin{aligned}
& \text { Time taken }=\frac{\text { Distance }}{\text { Speed }} \\
& =2.5 / 25=1 / 10=0.1 \mathrm{~h}=6 \mathrm{~min}
\end{aligned}
\)
The suitable \(x\) – \(t\) graph of the motion of the woman is shown in the given figure.

Q20: A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 \(\mathrm{m}\) long and requires \(1 \mathrm{~s}\). Plot the \(x\) – \(t\) graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit \(13 \mathrm{~m}\) away from the start.

Answer: Distance covered with 1 step \(=1 \mathrm{~m}\)
Time taken \(=1 \mathrm{~s}\)
Time taken to move first \(5 \mathrm{~m}\) forward \(=5 \mathrm{~s}\)
Time taken to move \(3 \mathrm{~m}\) backward \(=3 \mathrm{~s}\)
Net distance covered \(=5-3=2 \mathrm{~m}\)
Net time taken to cover \(2 \mathrm{~m}=8 \mathrm{~s}\)
Drunkard covers \(2 \mathrm{~m}\) in \(8 \mathrm{~s}\).
Drunkard covered \(4 \mathrm{~m}\) in \(16 \mathrm{~s}\).
Drunkard covered \(6 \mathrm{~m}\) in \(24 \mathrm{~s}\).
Drunkard covered \(8 \mathrm{~m}\) in \(32 \mathrm{~s}\).
In the next \(5 \mathrm{~s}\), the drunkard will cover a distance of \(5 \mathrm{~m}\) and a total distance of \(13 \mathrm{~m}\) and falls into the pit.
Net time taken by the drunkard to cover \(13 \mathrm{~m}=32+5=37 \mathrm{~s}\)
The \(x-t\) graph of the drunkard’s motion can be shown as:

Q21: Read each statement below carefully and state with reasons and examples, if it is true or false ;
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.

Answer:

(a) True
When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity \((\mathrm{g})\) that acts in the downward direction at that point.
(b) False
Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.
(c) True
A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, the acceleration of the car is also zero.
(d) This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is a slowing down of the particle. Such a case happens when a particle is projected upwards.

This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

Q22: Explain clearly, with examples, the distinction between :
(a) the magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) the magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].

Answer: (a) Suppose a particle goes from point A to B along a straight path and returns to A along the same path. The magnitude of the displacement of the particle is zero, because the particle has returned to its initial position. The total length of path covered by the particle is \(A B+B A=A B+A B=2 A B\). Thus, the second quantity is greater than the first.
(b) Suppose, in the above example, the particle takes time \(t\) to cover the whole journey. Then, the magnitude of the average velocity of the particle over time-interval \(\mathrm{t}\) is = Magnitude of displacement \(/\) Time-interval \(=0 / \mathrm{t}=0\)
While the average speed of the particle over the same time-interval is =Total path length \(/\) Time-interval \(=2 \mathrm{AB} / \mathrm{t}\)
Again, the second quantity (average speed) is greater than the first (magnitude of average velocity).

Q23: The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Answer: Instantaneous velocity is given by the first derivative of distance with respect to time i.e.,
\(
v_{\mathrm{inst}}=\frac{d x}{d t}
\)
Here, the time interval \(d t\) is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time.
Therefore, instantaneous speed is always equal to instantaneous velocity.

Note: Because, for an arbitrarily small interval of time, the magnitude of displacement is equal to the length of the path.

Q24: Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer: All four graphs are impossible. (a) a particle cannot have two different positions at the same time; (b) a particle cannot have velocity in opposite directions at the same time; (c) speed is always non-negative; (d) the total path length of a particle can never decrease with time. (Note, the arrows on the graphs are meaningless).

Q25: Figure 3.21 shows the \(x\)-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for \(t<0\) and on a parabolic path for \(t>0\)? If not, suggest a suitable physical context for this graph.

Answer: No, wrong. \(x\) – \(t\) plot does not show the trajectory of a particle. Context: A body is dropped from a tower \((x=0)\) at \(t=0\).

It is not correct to say that the particle moves in a straight line for \(t<0\) (i.e., -ve) and on a parabolic path for \(t>0\) (i.e., + ve) because the \(x-t\) graph can not show the path of the particle.
For the graph, a suitable physical context can be the particle thrown from the top of a tower at the instant \(t=0\).

Q26: Suggest a suitable physical situation for each of the following graphs (Fig 3.22):

Answer: (a) A ball at rest on a smooth floor is kicked, it rebounds from a wall with reduced speed and moves to the opposite wall which stops it; (b) A ball thrown up with some initial velocity rebounding from the floor with reduced speed after each hit; (c) A uniformly moving cricket ball turned back by hitting it with a bat for a very short time interval.

Q27: Figure 3.23 gives the \(x\)-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at \(t=0.3 \mathrm{~s}, 1.2 \mathrm{~s},-1.2 \mathrm{~s}\)

Answer:

\(
\text { In S.H.M., acceleration, } a=-k x \text {. }
\)
(i) At \(t=0.3 \mathrm{~s}, \mathrm{x}<0\) i.e., \(x\) is in -ve direction. Moreover, as \(x\) is becoming more negative with time, it shows that \(v\) is also – ve (i.e., \(v<0)\). However, \(a=-k x\) will be + ve \((a>0)\).
(ii) At \(t=1.2 \mathrm{~s}, x>0, v>0\) and \(a<0\).
(iii) At \(t=-1.2 \mathrm{~s}, x<0\), but here on increasing the time \(t\), value of \(x\) becomes less negative.
It means that \(v\) is \(+v e\) (i.e., \(v>0\) ). Again \(a=-k x\) will be positive (i.e., \(a>0\) ).

Q28: Figure 3.24 gives the \(x\)-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Answer: Interval 3 (Greatest), Interval 2 (/Least)
Positive (Intervals \(1 \& 2\) ), Negative (Interval 3)
The average speed of a particle shown in the \(x\) – \(t\) graph is obtained from the slope of the graph in a particular interval of time.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

\(
\text { Greater in } 3 \text {, least in } 2 ; v>0 \text { in } 1 \text { and } 2, v<0 \text { in interval } 3 \text {. }
\)

Q29: Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, gives the signs of \(v\) and \(a\) in the three intervals. What are the accelerations at points A, B, C, and D?

Answer: The acceleration is greatest in magnitude in interval 2 as the change in speed in the same time is maximum in this interval. The average speed is greatest in interval 3 (peak D is at maximum on speed axis).
Acceleration magnitude greatest in \(2 ;\) speed greatest in \(3 ; v>0\) in 1,2 and \(3 ; a>0\) in 1 and \(3, a<0\) in \(2 ; a=0\) at A, B, C, D.

Q30: A three-wheeler starts from rest, and accelerates uniformly with \(1 \mathrm{~m} \mathrm{~s}^{-2}\) on a straight road for \(10 \mathrm{~s}\), and then moves with uniform velocity. Plot the distance covered by the vehicle during the \(n^{\text {th }}\) second \((n=1,2,3 \ldots\) ) versus \(n\). What do you expect this plot to be during accelerated motion: a straight line or a parabola?

Answer: Since \(S_{n^{t h}}=u+\frac{1}{2} a(2 n-1)\)
when \(\mathrm{u}=0, \mathrm{a}=1 \mathrm{~ms}^{-2}\)
\(\therefore S_{n^{t h}}=0+\frac{1}{2}(2 n-1)=\frac{1}{2}(2 n-1)\)
\(
\begin{aligned}
& S_1=\frac{1}{2}(2 \times 1-1)=0.5 m \\
& S_2=\frac{1}{2}(2 \times 2-1)=1.5 m \\
& S_3=\frac{1}{2}(2 \times 3-1)=2.5 m \\
& S_4=\frac{1}{2}(2 \times 4-1)=3.5 m \\
& S_5=\frac{1}{2}(2 \times 5-1)=4.5 m \\
& S_6=\frac{1}{2}(2 \times 6-1)=5.5 m \\
& S_7=\frac{1}{2}(2 \times 7-1)=6.5 m \\
& S_8=\frac{1}{2}(2 \times 8-1)=7.5 m \\
& S_9=\frac{1}{2}(2 \times 9-1)=8.5 \mathrm{~m} \\
& S_{10}=\frac{1}{2}(2 \times 10-1)=9.5 \mathrm{~m}
\end{aligned}
\)
Hence, the plot during accelerated motion is a straight line.

Note: A straight line inclined with the time-axis for uniformly accelerated motion; parallel to the time-axis for uniform motion.

Q31: On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed \(9 \mathrm{~km} \mathrm{~h}^{-1}\) (with respect to the belt) between his father and mother located \(50 \mathrm{~m}\) apart on the moving belt. The belt moves with a speed of \(4 \mathrm{~km} \mathrm{~h}^{-1}\). For an observer on a stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt?
(b) speed of the child running opposite to the direction of motion of the belt?
(c) time taken by the child in (a) and (b)?
Which of the answers alter if motion is viewed by one of the parents?

Answer: (a) Speed of the belt, \(v_B=4 \mathrm{~km} / \mathrm{h}\)
Speed of the boy, \(v_{\mathrm{b}}=9 \mathrm{~km} / \mathrm{h}\)
Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
\(
v_{\mathrm{bB}}=v_{\mathrm{b}}+v_{\mathrm{B}}=9+4=13 \mathrm{~km} / \mathrm{h}
\)
(b) Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:
\(
v_{\mathrm{bB}}=v_{\mathrm{b}}+\left(-v_{\mathrm{B}}\right)=9-4=5 \mathrm{~km} / \mathrm{h}
\)
(c) Distance between the child’s parents \(=50 \mathrm{~m}\)
As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., \(9 \mathrm{~km} / \mathrm{h}=2.5 \mathrm{~m} / \mathrm{s}\).
Hence, the time taken by the child to move towards one of his parents is \(\frac{50}{2.5}=20 \mathrm{~s}\)

If the motion is viewed by any one of the parents, answers obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., \(9 \mathrm{~km} / \mathrm{h}\).
For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.

Q32: Two stones are thrown up simultaneously from the edge of a cliff \(200 \mathrm{~m}\) high with initial speeds of \(15 \mathrm{~m} \mathrm{~s}^{-1}\) and \(30 \mathrm{~m} \mathrm{~s}^{-1}\). Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\). Give the equations for the linear and curved parts of the plot.

Answer:  For the first stone:
Initial velocity, \(u_1=15 \mathrm{~m} / \mathrm{s}\)
Acceleration, \(\mathrm{a}=-\mathrm{g}=-10 \mathrm{~m} / \mathrm{s}^2\)
Using the relation,
\(
x_1=x_0+u_1 t+\frac{1}{2} a t^2
\)
Where, height of the cliff, \(x_0=200 \mathrm{~m}\)
\(
x_1=200+15 t-5 t^2 \dots(i)
\)
When this stone hits the ground, \(\mathrm{x}_1=0\)
\(
\therefore-5 t^2+15 t+200=0
\)
On solving we can get:
\(
\mathrm{t}=8 \mathrm{~s} \text { or } \mathrm{t}=-5 \mathrm{~s}
\)
Since the stone was projected at time \(t=0\), the negative sign before time is meaningless.
\(
\therefore \mathrm{t}=8 \mathrm{~s}
\)
For the second stone:
Initial velocity, \(\mathrm{u}_2=30 \mathrm{~m} / \mathrm{s}\)
Acceleration, \(\mathrm{a}=-\mathrm{g}=-10 \mathrm{~m} / \mathrm{s}^2\)
Using the relation,
\(
\begin{aligned}
& x_2=x_0+u_2 t+\frac{1}{2} a t^2 \\
& =200+30 t-5 t^2 \ldots \ldots(ii)
\end{aligned}
\)
At the moment when this stone hits the ground; \(x_2=0\)
\(
\begin{aligned}
& -5 t^2+30 t+200=0 \\
& t=10 \mathrm{~s} \text { or } t=-4 \mathrm{~s}
\end{aligned}
\)
Here again, the negative sign is meaningless.
\(
\therefore \mathrm{t}=10 \mathrm{~s}
\)
Subtracting equations (i) and (ii), we get
\(
\begin{aligned}
& x_2-x_1=\left(200+30 t-5 t^2\right)-\left(200+15 t-5 t^2\right) \\
& x_2-x_1=15 t \ldots \ldots . . \text { (iii) }
\end{aligned}
\)
Equation (iii) represents the linear path of both stones. Due to this linear relation between \(\left(\mathrm{x}_2-\mathrm{x}_1\right)\) and \(\mathrm{t}\), the path remains a straight line till \(8 \mathrm{~s}\). Maximum separation between the two stones is at \(\mathrm{t}=8 \mathrm{~s}\) \(\left(\mathrm{x}_2-\mathrm{x}_1\right)_{\max }=15 \times 8=120 \mathrm{~m}\)
This is in accordance with the given graph.
After \(8 \mathrm{~s}\), only the second stone is in motion whose variation with time is given by the quadratic equation:
\(
\mathrm{x}_2-\mathrm{x}_1=200+30 \mathrm{t}-5 \mathrm{t}^2
\)
Hence, the equation of linear and curved path is given by
\(
\begin{aligned}
& x_2-x_1=15 t \text { (Linear path) } \\
& x_2-x_1=200+30 t-5 t^2 \text { (Curved path) }
\end{aligned}
\)

Q33: The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval \(t_1\) to \(t_2\) :
(a) \(x\left(t_2\right)=x\left(t_1\right)+v\left(t_1\right)\left(t_2-t_1\right)+(1 / 2) a\left(t_2-t_1\right)^2\)
(b) \(v\left(t_2\right)=v\left(t_1\right)+a\left(t_2-t_1\right)\)
(c) \(v_{\text {average }}=\left(x\left(t_2\right)-x\left(t_1\right)\right) /\left(t_2-t_1\right)\)
(d) \(a_{\text {arerage }}=\left(v\left(t_2\right)-v\left(t_1\right)\right) /\left(t_2-t_1\right)\)
(e) \(x\left(t_2\right)=x\left(t_1\right)+v_{\text {average }}\left(t_2-t_1\right)+(1 / 2) a_{\text {average }}\left(t_2-t_1\right)^2\)
(f) \(x\left(t_2\right)-x\left(t_1\right)=\) area under the \(v\) – \(t\) curve bounded by the \(t\)-axis and the dotted line shown.

Answer:

The equations given in options (a), (b) and (e), are of straight lines.
From the given graph, it is observed that the graph between \(\mathrm{t_1}\) and \(\mathrm{t_2}\) has a non-uniform slope. The motion in this interval cannot be a straight line. Therefore, the equations given in options (a), (b) and (e) do not define the motion of the particle over the time interval \(\mathrm{t_1}\) and \(\mathrm{t_2}\).

The average velocity is the ratio of the difference in the displacement at two positions to the difference in the time interval.
The average velocity between time \(t_2\) and \(t_1\) is given as, \(
v_{\text {average }}=\left(x\left(t_2\right)-x\left(t_1\right)\right) /\left(t_2-t_1\right)
\)
The average acceleration is the ratio of the difference in the velocity at two positions to the difference in the time interval.
The average acceleration between time \(t_2\) and \(t_1\) is given as,
\(
a_{\text {average }}=\left(v\left(t_2\right)-v\left(t_1\right)\right) /\left(t_2-t_1\right)
\)
The area under the velocity-time graph represents the displacement of the body between two-time intervals. This is because displacement is the product of the velocity and the respective time.
The displacement of the body between time \(t_2\) and \(t_1\) is given as, \(x(t_2)-x(t_1)=\) area under the \(v-t\) curve bounded by the \(t\)-axis and the dotted line Thus, the correct equations are (c), (d) and ( f ).

Q34: A man walks at a speed of \(6 \mathrm{~km} / \mathrm{hr}\) for \(1 \mathrm{~km}\) and \(8 \mathrm{~km} / \mathrm{hr}\) for the next \(1 \mathrm{~km}\). What is his average speed for the walk of \(2 \mathrm{~km}\)?

Answer: Distance travelled is \(2 \mathrm{~km}\).
\(
\begin{aligned}
\text { Time taken } & =\frac{1 \mathrm{~km}}{6 \mathrm{~km} / \mathrm{hr}}+\frac{1 \mathrm{~km}}{8 \mathrm{~km} / \mathrm{hr}} \\
& =\left(\frac{1}{6}+\frac{1}{8}\right) \mathrm{hr}=\frac{7}{24} \mathrm{hr} . \\
\text { Average speed } & =\frac{2 \mathrm{~km} \times 24}{7 \mathrm{hr}}=\frac{48}{7} \mathrm{~km} / \mathrm{hr} \\
& \approx 7 \mathrm{~km} / \mathrm{hr} .
\end{aligned}
\)

Q35: The I.Sc. lecture theatre of a college is \(40 \mathrm{ft}\) wide and has a door at a corner. A teacher enters at 12.00 noon through the door and makes 10 rounds along the \(40 \mathrm{ft}\) wall back and forth during the period and finally leaves the classroom at 12.50 p.m. through the same door. Compute his average speed and average velocity.

Answer: Total distance travelled in 50 minutes \(=800 \mathrm{ft}\). Average speed \(=\frac{800}{50} \mathrm{ft} / \mathrm{min}=16 \mathrm{ft} / \mathrm{min}\).
At 12.00 noon he is at the door and at \(12.50 \mathrm{pm}\) he is again at the same door.
The displacement during the \(50 \mathrm{~min}\) interval is zero.
Average velocity \(=\) zero.

Q36: The position of a particle moving on \(X\)-axis is given by \(x=A t^2+B t^2+C t+D\).
The numerical values of \(A, B, C, D\) are \(1,4,-2\) and 5 respectively, and SI units are used. Find (a) the dimensions of \(A, B, C\) and \(D\), (b) the velocity of the particle at \(t=4 \mathrm{~s}\), (c) the acceleration of the particle at \(t=4 \mathrm{~s}\), (d) the average velocity during the interval \(t=0\) to \(t=4 \mathrm{~s}\), (e) the average acceleration during the interval \(t=0\) to \(t=4 \mathrm{~s}\).

Answer: (a) Dimensions of \(x, A t^3, B t^2, C t\) and \(D\) must be identical and in this case each is length. Thus,
\(
\begin{array}{rlrl}
{\left[A t^3\right]} & =\mathrm{L},  \text { or, } {[A]=\mathrm{LT}^{-3}} \\
{\left[B t^2\right]} & =\mathrm{L},  \text { or, } {[B]=\mathrm{LT}^{-2}} \\
{[C t]} & =\mathrm{L},  \text { or, }[C]=\mathrm{LT}^{-1} \\
{[D]} & =\mathrm{L} & &
\end{array}
\)
(b) \(x=A t^2+B t^2+C t+D\)
or, \(v=\frac{d x}{d t}=3 A t^2+2 B t+C\).
Thus, at \(t=4 \mathrm{~s}\), the velocity
\(
\begin{aligned}
& =3\left(1 \mathrm{~m} / \mathrm{s}^3\right)\left(16 \mathrm{~s}^2\right)+2\left(4 \mathrm{~m} / \mathrm{s}^2\right)(4 \mathrm{~s})+(-2 \mathrm{~m} / \mathrm{s}) \\
& =(48+32-2) \mathrm{m} / \mathrm{s}=78 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)
(c) \(v=3 A t^2+2 B t+C\)
or, \(a=\frac{d v}{d t}=6 A t+2 B\).
At \(t=4 \mathrm{~s}, a=6\left(1 \mathrm{~m} / \mathrm{s}^3\right)(4 \mathrm{~s})+2\left(4 \mathrm{~m} / \mathrm{s}^2\right)=32 \mathrm{~m} / \mathrm{s}^2\).
(d) \(x=A t^3+B t^2+C t+D\).
Position at \(t=0\) is \(x=D=5 \mathrm{~m}\).
Position at \(t=4 \mathrm{~s}\) is
\(
\begin{aligned}
& \left(1 \mathrm{~m} / \mathrm{s}^2\right)\left(64 \mathrm{~s}^2\right)+\left(4 \mathrm{~m} / \mathrm{s}^2\right)\left(16 \mathrm{~s}^2\right)-(2 \mathrm{~m} / \mathrm{s})(4 \mathrm{~s})+5 \mathrm{~m} \\
& \quad=(64+64-8+5) \mathrm{m}=125 \mathrm{~m}
\end{aligned}
\)
Thus, the displacement during 0 to \(4 \mathrm{~s}\) is \(125 \mathrm{~m}-5 \mathrm{~m}=120 \mathrm{~m}\).
Average velocity \(=\frac{120 \mathrm{~m}}{4 \mathrm{~s}}=30 \mathrm{~m} / \mathrm{s}\).
(e) \(v=3 A t^2+2 B t+C\).
Velocity at \(t=0\) is \(C=-2 \mathrm{~m} / \mathrm{s}\).
Velocity at \(t=4 \mathrm{~s}\) is \(=78 \mathrm{~m} / \mathrm{s}\).
Average acceleration \(=\frac{v_2-v_1}{t_2-t_1}=20 \mathrm{~m} / \mathrm{s}^2\).

Q37: From the velocity-time graph of a particle given in figure (3-W1), describe the motion of the particle qualitatively in the interval 0 to \(4 \mathrm{~s}\). Find (a) the distance travelled during first two seconds, (b) during the time \(2 \mathrm{~s}\) to \(4 \mathrm{~s}\), (c) during the time 0 to \(4 \mathrm{~s}\), (d) displacement during 0 to \(4 \mathrm{~s}\), \((e)\) acceleration at \(t=1 / 2 \mathrm{~s}\) and \((f)\) acceleration at \(t=2 \mathrm{~s}\).

Answer: At \(t=0\), the particle is at rest, say at the origin. After that the velocity is positive, so that the particle moves in the positive \(x\) direction. Its speed increases till 1 second when it starts decreasing. The particle continues to move further in positive \(x\) direction. At \(t=2 \mathrm{~s}\), its velocity is reduced to zero, it has moved through a maximum positive \(x\) distance. Then it changes its direction, velocity being negative, but increasing in magnitude. At \(t=3 \mathrm{~s}\) velocity is maximum in the negative \(x\) direction and then the magnitude starts decreasing. It comes to rest at \(t=4 \mathrm{~s}\).
(a) Distance during 0 to \(2 \mathrm{~s}=\) Area of \(O A B\)
\(
=\frac{1}{2} \times 2 \mathrm{~s} \times 10 \mathrm{~m} / \mathrm{s}=10 \mathrm{~m}
\)
(b) Distance during 2 to \(4 \mathrm{~s}=\) Area of \(B C D=10 \mathrm{~m}\). The particle has moved in negative \(x\) direction during this period.
(c) The distance travelled during 0 to \(4 \mathrm{~s}=10 \mathrm{~m}+10 \mathrm{~m}\) \(=20 \mathrm{~m}\)
(d) displacement during 0 to \(4 \mathrm{~s}=10 \mathrm{~m}+(-10 \mathrm{~m})=0\).
(e) at \(t=1 / 2 \mathrm{~s}\) acceleration \(=\) slope of line \(O A=10 \mathrm{~m} / \mathrm{s}^2\).
(f) at \(t=2 \mathrm{~s}\) acceleration = slope of line \(A B C=-10 \mathrm{~m} / \mathrm{s}^2\).

Q38: A particle starts from rest with a constant acceleration. At a time \(t\) second, the speed is found to be \(100 \mathrm{~m} / \mathrm{s}\) and one second later the speed becomes \(150 \mathrm{~m} / \mathrm{s}\). Find (a) the acceleration and (b) the distance travelled during the \((t+1)^{t h}\) second.

Answer: (a) Velocity at time \(t\) second is
\(100 \mathrm{~m} / \mathrm{s}=a .(t\) second \() \dots(1)\)
and velocity at time \((t+1)\) second is
\(150 \mathrm{~m} / \mathrm{s}=a \cdot(t+1)\) second ……(2)
Subtracting (1) from (2), \(a=50 \mathrm{~m} / \mathrm{s}^2\)
(b) Consider the interval \(t\) second to \((t+1)\) second, time elapsed \(=1 \mathrm{~s}\) |nitial velocity \(=100 \mathrm{~m} / \mathrm{s}\) final velocity \(=150 \mathrm{~m} / \mathrm{s}\).
Thus, \((150 \mathrm{~m} / \mathrm{s})^2=(100 \mathrm{~m} / \mathrm{s})^2+2\left(50 \mathrm{~m} / \mathrm{s}^2\right) x\) or, \(x=125 \mathrm{~m}\).

Q39: A boy stretches a stone against the rubber tape of a catapult or ‘gulel’ (a device used to detach mangoes from the tree by boys in Indian villages) through a distance of \(24 \mathrm{~cm}\) before leaving it. The tape returns to its normal position accelerating the stone over the stretched length. The stone leaves the gulel with a speed \(2 \cdot 2 \mathrm{~m} / \mathrm{s}\). Assuming that the acceleration is constant while the stone was being pushed by the tape, find its magnitude.

Answer: Consider the accelerated \(24 \mathrm{~cm}\) motion of the stone.
Initial velocity \(=0\)
Final velocity \(=2 \cdot 2 \mathrm{~m} / \mathrm{s}\)
Distance travelled \(=24 \mathrm{~cm}=0.24 \mathrm{~m}\)
Using \(v^2=u^2+2 a x\),
\(
a=\frac{4 \cdot 84 \mathrm{~m}^2 / \mathrm{s}^2}{2 \times 0.24 \mathrm{~m}}=10 \cdot 1 \mathrm{~m} / \mathrm{s}^2
\)

Q40: A police inspector in a jeep is chasing a pickpocket on a straight road. The jeep is going at its maximum speed \(v\) (assumed uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance \(d\) away, and the motorcycle starts with a constant acceleration \(a\). Show that the pickpocket will be caught if \(v \geq \sqrt{2 a d}\).

Answer: Suppose the pickpocket is caught at a time \(t\) after the motorcycle starts. The distance travelled by the motorcycle during this interval is
\(
s=\frac{1}{2} a t^2 \dots(i)
\)
During this interval, the jeep travels a distance
\(
s+d=v t \text {. } \dots(ii)
\)
By (i) and (ii),
\(
\begin{aligned}
\frac{1}{2} a t^2-v t+d & =0 \\
t & =\frac{v \pm \sqrt{v^2-2 a d}}{a} .
\end{aligned}
\)
The pickpocket will be caught if \(t\) is real and positive. This will be possible if
\(
v^2 \geq 2 a d \text { or, } v \geq \sqrt{2 a d} .
\)

Q41: A car is moving at a constant speed of \(40 \mathrm{~km} / \mathrm{h}\) along a straight road which heads towards a large vertical wall and makes a sharp \(90^{\circ}\) turn by the side of the wall. A fly flying at a constant speed of \(100 \mathrm{~km} / \mathrm{h}\), starts from the wall towards the car at an instant when the car is \(20 \mathrm{~km}\) away, flies until it reaches the glass pane of the car and returns to the wall at the same speed. It continues to fly between the car and the wall till the car makes the \(90^{\circ}\) turn. (a) What is the total distance the fly has travelled during this period? (b) How many trips has it made between the car and the wall?

Answer: (a) The time taken by the car to cover \(20 \mathrm{~km}\) before the turn is \(\frac{20 \mathrm{~km}}{40 \mathrm{~km} / \mathrm{h}}=\frac{1}{2} \mathrm{~h}\). The fly moves at a constant speed of \(100 \mathrm{~km} / \mathrm{h}\) during this time. Hence the total distance coverd by it is \(100 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1}{2} \mathrm{~h}=50 \mathrm{~km}\).
(b) Suppose the car is at a distance \(x\) away (at A) when the fly is at the wall (at \(O\) ). The fly hits the glass pane at \(B\), taking a time \(t\). Then \(
\begin{aligned}
& A B=(40 \mathrm{~km} / \mathrm{h}) t \text {, } \\
& \text { and } O B=(100 \mathrm{~km} / \mathrm{h}) t \text {. } \\
& \text { Thus, } \quad x=A B+O B \\
& =(140 \mathrm{~km} / \mathrm{h}) t \\
& \text { or, } \quad t=\frac{x}{140 \mathrm{~km} / \mathrm{h}} \text {, or } O B=\frac{5}{7} x \text {. } \\
&
\end{aligned}
\)

The fly returns to the wall and during this period the car moves the distance \(B C\). The time taken by the fly in this return path is
\(
\left(\frac{5 x / 7}{100 \mathrm{~km} / \mathrm{h}}\right)=\frac{x}{140 \mathrm{~km} / \mathrm{h}}
\)
Thus,
\(
\begin{aligned}
& B C=\frac{40 x}{140}=\frac{2}{7} x \\
& O C=O B-B C=\frac{3}{7} x .
\end{aligned}
\)
If at the beginning of the round trip (wall to the car and back), the car is at a distance \(x\) away, it is \(\frac{3}{7} x\) away when the next trip again starts.

Distance of the car at the beginning of the 1 st trip \(=20 \mathrm{~km}\).
Distance of the car at the beginning of the 2nd trip
\(
=\frac{3}{7} \times 20 \mathrm{~km}
\)
Distance of the car at the beginning of the 3rd trip
\(
=\left(\frac{3}{7}\right)^2 \times 20 \mathrm{~km} .
\)
Distance of the car at the beginning of the 4 th trip
\(
=\left(\frac{3}{7}\right)^3 \times 20 \mathrm{~km} .
\)
Distance of the car at the beginning of the \(n\)th trip
\(
=\left(\frac{3}{7}\right)^{n-1} \times 20 \mathrm{~km} \text {. }
\)
Trips will go on till the car reaches the turn that is the distance reduces to zero. This will be the case when \(n\) becomes infinity. Hence the fly makes an infinite number of trips before the car takes the turn.

Q42: A ball is dropped from a height of \(19.6 \mathrm{~m}\) above the ground. It rebounds from the ground and raises itself up to the same height. Take the starting point as the origin and vertically downward as the positive \(X\)-axis. Draw approximate plots of \(x\) versus \(t\), versus \(t\) and a versus t. Neglect the small interval during which the ball was in contact with the ground.

Answer: Since the acceleration of the ball during the contact is different from ‘ \(g\) ‘, we have to treat the downward motion and the upward motion separately. For the downward motion : \(a=g=9.8 \mathrm{~m} / \mathrm{s}^2\),
\(
x=u t+\frac{1}{2} a t^2=\left(4 \cdot 9 \mathrm{~m} / \mathrm{s}^2\right) t^2 \text {. }
\)
The ball reaches the ground when \(x=19 \cdot 6 \mathrm{~m}\). This gives \(t=2 \mathrm{~s}\). After that it moves up, \(x\) decreases and at \(t=4 \mathrm{~s}, x\) becomes zero, the ball reaching the initial point.
We have at
\(
\begin{array}{ll}
t=0, & x=0 \\
t=1 \mathrm{~s}, & x=4.9 \mathrm{~m} \\
t=2 \mathrm{~s}, & x=19.6 \mathrm{~m} \\
t=3 \mathrm{~s}, & x=4.9 \mathrm{~m} \\
t=4 \mathrm{~s}, & x=0 .
\end{array}
\)

Velocity: During the first two seconds,
\(
v=u+a t=\left(9 \cdot 8 \mathrm{~m} / \mathrm{s}^2\right) t
\)
at \(t=0 \quad v=0\)
at \(t=1 \mathrm{~s}, \quad v=9.8 \mathrm{~m} / \mathrm{s}\)
at \(t=2 \mathrm{~s}, \quad v=19.6 \mathrm{~m} / \mathrm{s}\).
During the next two seconds, the ball goes upward, velocity is negative, magnitude decreasing and at \(t=4 \mathrm{~s}, v=0\). Thus,
\(
\begin{array}{ll}
\text { at } t=2 \mathrm{~s}, & v=-19.6 \mathrm{~m} / \mathrm{s} \\
\text { at } t=3 \mathrm{~s}, & v=-9.8 \mathrm{~m} / \mathrm{s} \\
\text { at } t=4 \mathrm{~s}, & v=0 .
\end{array}
\)

At \(t=2 \mathrm{~s}\) there is an abrupt change in velocity from \(19.6 \mathrm{~m} / \mathrm{s}\) to \(-19.6 \mathrm{~m} / \mathrm{s}\). In fact, this change in velocity takes place over a small interval during which the ball remains in contact with the ground.
Acceleration: The acceleration is constant \(9.8 \mathrm{~m} / \mathrm{s}^2\) throughout the motion (except at \(t=2 \mathrm{~s}\) ).

Q43: A stone is dropped from a balloon going up with a uniform velocity of \(5.0 \mathrm{~m} / \mathrm{s}\). If the balloon was \(50 \mathrm{~m}\) high when the stone was dropped, find its height when the stone hits the ground. Take \(g=10 \mathrm{~m} / \mathrm{s}^2\).

Answer: At \(t=0\), the stone was going up with a velocity of \(5^{\circ} 0 \mathrm{~m} / \mathrm{s}\). After that it moved as a freely falling particle with downward acceleration \(g\). Take vertically upward as the positive \(X\)-axis. If it reaches the ground at time \(t\)
\(
x=-50 \mathrm{~m}, \quad u=5 \mathrm{~m} / \mathrm{s}, \quad a=-10 \mathrm{~m} / \mathrm{s}^2 .
\)
We have
\(
x=u t+\frac{1}{2} a t^2
\)
or,
\(
-50 \mathrm{~m}=(5 \mathrm{~m} / \mathrm{s}) \cdot t+\frac{1}{2} \times\left(-10 \mathrm{~m} / \mathrm{s}^2\right) t^2
\)
or
\(
t=\frac{1 \pm \sqrt{41}}{2} \mathrm{~s} \text {. }
\)
or,
\(
t=-2 \cdot 7 \mathrm{~s} \quad \text { or, } 3 \cdot 7 \mathrm{~s} \text {. }
\)
Negative \(t\) has no significance in this problem. The stone reaches the ground at \(t=3 \cdot 7 \mathrm{~s}\). During this time the balloon has moved uniformly up. The distance covered by it is
\(
5 \mathrm{~m} / \mathrm{s} \times 3.7 \mathrm{~s}=18.5 \mathrm{~m}
\)
Hence, the height of the balloon when the stone reaches the ground is \(50 \mathrm{~m}+18.5 \mathrm{~m}=68.5 \mathrm{~m}\).

Q44: A football is kicked with a velocity of \(20 \mathrm{~m} / \mathrm{s}\) at an angle of \(45^{\circ}\) with the horizontal. (a) Find the time taken by the ball to strike the ground. (b) Find the maximum height it reaches. (c) How far away from the kick does it hit the ground? Take \(g=10 \mathrm{~m} / \mathrm{s}^2\).

Answer: (a) Take the origin at the point where the ball is kicked, vertically upward as the \(Y\)-axis and the horizontal in the plane of motion as the \(X\)-axis. The initial velocity has the components
\(
\begin{aligned}
& u_x=(20 \mathrm{~m} / \mathrm{s}) \cos 45^{\circ}=10 \sqrt{ } 2 \mathrm{~m} / \mathrm{s} \\
& u_y=(20 \mathrm{~m} / \mathrm{s}) \sin 45^{\circ}=10 \sqrt{ } 2 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)
When the ball reaches the ground, \(y=0\).
Using
\(
\begin{aligned}
\text { Using } & y =u_y t-\frac{1}{2} g t^2, \\
& 0=(10 \sqrt{2} \mathrm{~m} / \mathrm{s}) t-\frac{1}{2} \times\left(10 \mathrm{~m} / \mathrm{s}^2\right) \times t^2 \\
\text { or, } & t =2 \sqrt{2} \mathrm{~s}=2.8 \mathrm{~s} .
\end{aligned}
\)
Thus, it takes \(2.8 \mathrm{~s}\) for the football to fall on the ground.
(b) At the highest point \(v_y=0\). Using the equation
\(
\begin{aligned}
v_y^2 & =u_y^2-2 g y, \\
0 & =(10 \sqrt{2} \mathrm{~m} / \mathrm{s})^2-2 \times\left(10 \mathrm{~m} / \mathrm{s}^2\right) H \\
\text { or, } \quad H & =10 \mathrm{~m} .
\end{aligned}
\)
Thus, the maximum height reached is \(10 \mathrm{~m}\).
(c) The horizontal distance travelled before falling to the ground is \(x=u_x t\)
\(
=(10 \sqrt{2} \mathrm{~m} / \mathrm{s})(2 \sqrt{2} \mathrm{~s})=40 \mathrm{~m}
\)

Q45: Galileo was punished by the Church for teaching that the sun is stationary and the earth moves around it. His opponents held the view that the earth is stationary and the sun moves around it. If the absolute motion has no meaning, are the two viewpoints not equally correct or equally wrong?

Answer: The absolute motion has no meaning. In the relative motion view, the two viewpoints are the same. Hence, both viewpoints are equally correct or equally wrong.

Q46: When a particle moves with constant velocity, its average velocity, its instantaneous velocity and its speed are all equal. Comment on this statement.

Answer: Constant velocity means that a particle has the same direction and speed at every point. So, it’s average velocity and instantaneous velocity are equal. Its speed being a scalar quantity is equal in magnitude only.

Q47: A car travels at a speed of \(60 \mathrm{~km} / \mathrm{hr}\) due north and the other at a speed of \(60 \mathrm{~km} / \mathrm{hr}\) due east. Are the velocities equal? If no, which one is greater? If you find any of the questions irrelevant, explain.

Answer: No, as the directions of the cars are different, their velocities are not equal, as velocity is a vector quantity. We cannot compare velocities on the basis of direction only, so the question is irrelevant.

Q48: A ball is thrown vertically upward with a speed of 20 \(\mathrm{m} / \mathrm{s}\). Draw a graph showing the velocity of the ball as a function of time as it goes up and then comes back.

Answer: A graph showing the velocity of the ball as a function of time.

Q49: The velocity of a particle is towards the west at an instant. Its acceleration is not towards the west, not towards the east, not towards the north, and not towards the south. Give an example of this type of motion.

Answer: Motion of a projectile at the highest point of its path: Because at this point, the projectile has only horizontal component of velocity. So, it can have velocity towards the west and acceleration in the vertically downward direction.

Q50: At which point on its path a projectile has the smallest speed?

Answer: A projectile has the lowest speed at the highest point of its path because the vertical component of velocity is zero at this point.

Q51: Two particles \(A\) and \(B\) start from rest and move for equal time on a straight line. The particle \(A\) has an acceleration \(a\) for the first half of the total time and \(2 a\) for the second half. The particle \(B\) has an acceleration \(2 a\) for the first half and \(a\) for the second half. Which particle has covered larger distance?

Answer: 

\(
\text { For particle } A
\):
For first half \(\left(\right.\) time = \(\left.\frac{\mathrm{t}}{2}\right)\)
\(
\mathrm{s}_1=O P=\frac{1}{2} a\left(\frac{t}{2}\right)^2=\frac{a t^2}{8}
\)
Velocity at \(P=a\left(\frac{t}{2}\right)=\frac{a t}{2}\)
For the next half time
\(
\begin{aligned}
& \mathrm{s}_2=\mathrm{PQ}=\left(\frac{\mathrm{at}}{2}\right)\left(\frac{\mathrm{t}}{2}\right)+\frac{1}{2}(2 \mathrm{a})\left(\frac{\mathrm{t}}{2}\right)^2 \\
& =\frac{a t^2}{4}+\frac{a t^2}{4}=\frac{a t^2}{2}
\end{aligned}
\)
Total distance \(\mathrm{s}=\frac{\mathrm{at}^2}{8}+\frac{\mathrm{at}^2}{2}=\frac{5 a \mathrm{t}^2}{8}\)

\(
\text { For particle } B
\):
For first half time = \(\frac{t}{2}\)
\(
O R=\frac{1}{2}(2 a)\left(\frac{t}{2}\right)^2=\frac{a t^2}{4}
\)
For the second half time
Velocity point \(R=(2 a)\left(\frac{t}{2}\right)=\) at
\(
\begin{aligned}
& R T=(a t)\left(\frac{t}{2}\right)+\frac{1}{2}(a)\left(\frac{t}{2}\right)^2 \\
& =\frac{a t^2}{2}+\frac{a t^2}{8} \\
& =\frac{5 a t^2}{8}
\end{aligned}
\)
Total distance \(=O R+R T\)
\(
=\frac{a t^2}{4}+\frac{5 a t^2}{8}=\frac{7 a t^2}{8}
\)

\(
\therefore \text { Distance covered by } \mathrm{A}<\text { Distance covered by } \mathrm{B}
\)

Q52: If a particle is accelerating, it is either speeding up or speeding down. Do you agree with this statement?

Answer: Acceleration does not mean speeding up or speeding down. It means the change of velocity either in direction or in magnitude.

Q53: A food packet is dropped from a plane going at an altitude of \(100 \mathrm{~m}\). What is the path of the packet as seen from the plane? What is the path as seen from the ground? If someone asks “what is the actual path”, what will you answer?

Answer: The path of the packet (as seen from the plane) is a vertically downward straight line, as the horizontal velocity of the packet and the plane is the same.
As seen from the ground, the path of the packet is a parabola. The path is defined with respect to some reference frame. As there is no absolute reference frame, no actual path is defined.

Q54: Give examples where (a) the velocity of a particle is zero but its acceleration is not zero, (b) the velocity is opposite in direction to the acceleration, (c) the velocity is perpendicular to the acceleration.

Answer: (a) At the highest point when a particle is thrown vertically upwards.
(b) While going up when a particle is thrown vertically upwards.
(c) At the highest point of a full projectile.

Q55: Figure (3-Q1) shows the \(x\) coordinate of a particle as a function of time. Find the signs of \(v_x\) and \(a_x\) at \(t=t_1\), \(t=t_2\) and \(t=t_3\).

Answer: The slope of the \(x-t\) graph gives the velocity and the change in the slope gives the acceleration.
At \(t=t_1\),
Slope \(=\) Positive \(\Rightarrow\) Velocity \(=\) Positive
Slope is increasing \(\Rightarrow\) Acceleration \(=\) Positive
At \(\mathrm{t}=\mathrm{t}_2\)
Slope \(=\) Constant \(\Rightarrow\) Velocity \(=\) Zero
Slope is constant \(\Rightarrow\) Acceleration \(=\) Negative
At \(\mathrm{t}=\mathrm{t}_3\),
Slope \(=\) Negative \(\Rightarrow\) Velocity \(=\) Negative
Slope is increasing \(\Rightarrow\) Acceleration \(=\) Negative

Q56: A player hits a baseball at some angle. The ball goes high up in space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?

Answer: The displacement of the ball and the player is the same, as the initial and final points are the same.

Q57: The increase in the speed of a car is proportional to the additional petrol put into the engine. Is it possible to accelerate a car without putting more petrol or less petrol into the engine?

Answer: Yes, it is possible to accelerate a car without putting more petrol or less petrol in the engine. This can be done by driving the car on a circular or curved track at a uniform speed.

Q58: Rain is falling vertically. A man running on the road keeps his umbrella tilted but a man standing on the street keeps his umbrella vertical to protect himself from the rain. But both of them keep their umbrella vertical to avoid the vertical sun-rays. Explain.

Answer: Let us visualize this practical scenario using vector diagrams.
Man running will have greater velocity than the rain falling; both velocities are with respect to the street. Therefore, the resultant velocity is of rain with respect to the running man. The figure depicting this is shown below.

Man standing with umbrella in the vertical falling rain is shown below:

Now, the velocity of man running is much less than the velocity of sun rays falling vertically. Therefore, the resultant velocity is in the same direction of the vertical sun rays. This is depicted in the figure below:

Now, man standing with umbrella in vertical falling sun rays is shown below:

Q59: A particle starts from the origin, goes along the \(X\)-axis to the point \((20 \mathrm{~m}, 0)\) and then returns along the same line to the point \((-20 \mathrm{~m}, 0)\). Find the distance and displacement of the particle during the trip.

Answer: The starting point for the particle is origin as shown in the figure.
The Distance traveled by the particle is:
\(
\begin{aligned}
& \mathrm{D}=20+20+20 \\
& =60 \mathrm{~m}
\end{aligned}
\)
The displacement is the distance between the initial and the final points. Displacement \(\Rightarrow \mathrm{OB}=20 \mathrm{~m}\) in the negative direction
The distance is \(60 \mathrm{~m}\) and displacement is \(20 \mathrm{~m}\) in negative direction.

Q60: It is \(260 \mathrm{~km}\) from Patna to Ranchi by air and \(320 \mathrm{~km}\) by road. An aeroplane takes 30 minutes to go from Patna to Ranchi whereas a delux bus takes 8 hours. (a) Find the average speed of the plane. (b) Find the average speed of the bus. (c) Find the average velocity of the plane. (d) Find the average velocity of the bus.

Answer: (a) Average speed of the plane
\(
\begin{aligned}
& \mathrm{S}_{\mathrm{avg}} =\left(\frac{\text { Total distance travelled }}{\text { Total time taken }}\right) \\
& =\frac{260}{0.5}=520 \mathrm{~km} / \mathrm{h}
\end{aligned}
\)
(b) Average speed of the bus, \(\mathrm{S}_{\mathrm{avg}}\) \(=\frac{320}{8}=40 \mathrm{~km} / \mathrm{h}\)
(c) The plane moves in a straight path.
Average velocity \(=\) Average speed \(=520 \mathrm{Km} / \mathrm{h}\) (Patna to Ranchi)
(d) Straight path distance from Patna to Ranchi = Displacement of the bus \(=260 \mathrm{~km}\) \(\therefore \vec{V}_{a v g}=\frac{260}{8}=32.5 \mathrm{~km} / h\) (Patna to Ranchi)

Q61: When a person leaves his home for sightseeing by his car, the meter reads \(12352 \mathrm{~km}\). When he returns home after two hours the reading is \(12416 \mathrm{~km}\). (a) What is the average speed of the car during this period? (b) What is the average velocity?

Answer: (a) Total distance covered \(=12416-12352=64 \mathrm{~km}\) \(\therefore\) Average speed \(=\frac{64}{2}=32 \mathrm{~km} / h\)
(b) Because he returns to his house, the displacement is zero. So, the average velocity is zero.

Q62: A particle starts from a point \(A\) and travels along the solid curve shown in Figure (3-E7). Find approximately the position \(B\) of the particle such that the average velocity between the positions \(A\) and \(B\) has the same direction as the instantaneous velocity at \(B\).

Answer:
\(
\text { At position } \mathrm{B} \text {, the instantaneous velocity of the particle has the direction along } \overrightarrow{B C} \text {. }
\)

Average velocity between \(A\) and \(B\),
\(
\begin{aligned}
& V_{a v}=\frac{\text { Displacement }}{\text { Time }} \\
& =\left(\frac{\overrightarrow{A B}}{t}\right)
\end{aligned}
\)
We can see that \(\overrightarrow{A B}\) and \(\overrightarrow{B C}\) are in the same direction. The point is \(\mathrm{B}(5 \mathrm{~m}, 3 \mathrm{~m})\).

Q63: A train starts from rest and moves with a constant acceleration of \(2.0 \mathrm{~m} / \mathrm{s}^2\) for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.

Answer: (a) Initial velocity, \(u=0\)
Acceleration, \(a=2 \mathrm{~m} / \mathrm{s}^2\)
Let the final velocity be \(v\) before the brakes are applied.
Now,
\(
\begin{aligned}
& t=30 \mathrm{~s} \\
& v=u+a t \\
& v=0+2 \times 30 \\
& \Rightarrow v=60 \mathrm{~m} / \mathrm{s} \\
& s_1=u t+\frac{1}{2} a t^2 \\
& \Rightarrow s_1=\frac{1}{2} \times 2 \times(30)^2=900 \mathrm{~m}
\end{aligned}
\)
When the brakes are applied:
\(
\begin{aligned}
& u^{\prime}=60 \mathrm{~m} / \mathrm{s} \\
& v^{\prime}=0 \\
& t=1 \mathrm{~min}=60 \mathrm{~s}
\end{aligned}
\)
Acceleration:
\(
\begin{aligned}
& a^{\prime}=\frac{(v-u)}{t}=\frac{(0-60)}{60}=-1 \mathrm{~m} / \mathrm{s}^2 \\
& s_2=\frac{v^2-u^2}{2 a^{\prime}}=\frac{0^2-60^2}{2(-1)}=1800 \mathrm{~m} \\
& s=s_1+s_2=1800+900=2700 \mathrm{~m} \\
& \Rightarrow s=2.7 \mathrm{~km}
\end{aligned}
\)
(b) Initial velocity, \(u=0\)
Acceleration, \(a=2 \mathrm{~m} / \mathrm{s}^2\)
Let the final velocity be \(v\) before the brakes are applied.
Now,
\(
\begin{aligned}
& t=30 \mathrm{~s} \\
& v=u+a t \\
& v=0+2 \times 30 \\
& \Rightarrow v=60 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Maximum speed attained by the train, \(v=60 \mathrm{~m} / \mathrm{s}\)
(c) Initial velocity, \(u=0\)
Acceleration, \(a=2 \mathrm{~m} / \mathrm{s}^2\)
Let the final velocity be \(v\) before the brakes are applied.
Now,
\(
\begin{aligned}
& t=30 \mathrm{~s} \\
& v=u+a t \\
& v=0+2 \times 30 \\
& \Rightarrow v=60 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Half the maximum speed \(=\frac{60}{2}=30 \mathrm{~m} / \mathrm{s}\)
When the train is accelerating with an acceleration of \(2 \mathrm{~m} / \mathrm{s}^2\) :
Distance,
\(
\begin{aligned}
& s=\frac{v^2-u^2}{2 a^{\prime}} \\
& =\frac{30^2-0^2}{2 \times 2} \\
& \Rightarrow s=225 \mathrm{~m}
\end{aligned}
\)
When the train is decelerating with an acceleration of \(-1 \mathrm{~m} / \mathrm{s}^2\) :
Distance,
\(
\begin{aligned}
& s=\frac{v^2-u^2}{2 a^{\prime}} \\
& =\frac{30^2-60^2}{2(-1)} \\
& \Rightarrow s=1350 \mathrm{~m}
\end{aligned}
\)
Position from the starting point \(=900+1350=2250\)
\(
=2.25 \mathrm{~km}
\)

Q64: Complete the following table:

\(
\begin{array}{|c|c|c|}
\hline \text { Car Model } & \begin{array}{c}
\text { Driver } X \\
\text { Reaction time } 0^{\circ} 20 \mathrm{~s}
\end{array} & \begin{array}{c}
\text { Driver } Y \\
\text { Reaction time } 0.30 \mathrm{~s}
\end{array} \\
\hline \begin{array}{l}
\text { A (deceleration } \\
\text { on hard braking } \\
=6.0 \mathrm{~m} / \mathrm{s}^2 \text { ) }
\end{array} & \begin{array}{l}
\text { Speed }=54 \mathrm{~km} / \mathrm{h} \\
\text { Braking distance } \\
a=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \\
\text { Total stopping } \\
\text { distance } \\
b=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\end{array} & \begin{array}{l}
\text { Speed }=72 \mathrm{~km} / \mathrm{h} \\
\text { Braking distance } \\
c=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \\
\text { Total stopping } \\
\text { distance } \\
d=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
.
\end{array} \\
\hline \begin{array}{l}
B \text { (deceleration } \\
\text { on hard braking } \\
=7.5 \mathrm{~m} / \mathrm{s}^2 \text { ) }
\end{array} & \begin{array}{l}
\text { Speed }=54 \mathrm{~km} / \mathrm{h} \\
\text { Braking distance } \\
e=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \\
\text { Total stopping } \\
\text { distance } \\
f=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . .
\end{array} & \begin{array}{l}
\text { Speed }=72 \mathrm{~km} / \mathrm{h} \\
\text { Braking distance } \\
g=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \\
\text { Total stopping } \\
\text { distance } \\
h=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . .
\end{array} \\
\hline
\end{array}
\)

Answer: Braking distance: Distance travelled after the brakes are applied.
Total stopping distance \(=\) Braking distance + Distance travelled in the reaction time
Case A:
Deceleration \(=6.0 \mathrm{~m} / \mathrm{s}^2\)
For driver \(\mathrm{X}\) :
Initial velocity, \(u=54 \mathrm{~km} / \mathrm{h}=15 \mathrm{~m} / \mathrm{s}\)
Final velocity, \(v=0\)
Braking distance,
\(
a=\frac{0^2-15^2}{2(-6)} \approx 19 \mathrm{~m}
\)
Distance travelled in the reaction time \(=15 \times 0.20=3 \mathrm{~m}\)
Total stopping distance, \(b=19+3=22 \mathrm{~m}\)
For driver \(\mathrm{Y}\):
Initial velocity, \(u=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}\)
Final velocity, \(v=0\)
Braking distance,
\(
a=\frac{0^2-20^2}{2(-6)} \approx 33 \mathrm{~m}
\)
Distance travelled in the reaction time \(=20 \times 0.30=6 \mathrm{~m}\)
Total stopping distance, \(d=33+6=39 \mathrm{~m}\)
Case B:
Deceleration \(=6.0 \mathrm{~m} / \mathrm{s}^2\)
Now, we have:
\(
\begin{aligned}
& e=15 \mathrm{~m} \\
& f=18 \mathrm{~m} \\
& g=27 \mathrm{~m} \\
& h=33 \mathrm{~m}
\end{aligned}
\)

\(
\begin{array}{|c|c|c|}
\hline \text { Car Model } & \begin{array}{c}
\text { Driver } X \\
\text { Reaction time } 0^{\circ} 20 \mathrm{~s}
\end{array} & \begin{array}{c}
\text { Driver } Y \\
\text { Reaction time } 0.30 \mathrm{~s}
\end{array} \\
\hline \begin{array}{l}
\text { A (deceleration } \\
\text { on hard braking } \\
=6.0 \mathrm{~m} / \mathrm{s}^2 \text { ) }
\end{array} & \begin{array}{l}
\text { Speed }=54 \mathrm{~km} / \mathrm{h} \\
\text { Braking distance } \\
a=19 m \\
\text { Total stopping } \\
\text { distance } \\
b=22 m
\end{array} & \begin{array}{l}
\text { Speed }=72 \mathrm{~km} / \mathrm{h} \\
\text { Braking distance } \\
c=33 m \\
\text { Total stopping } \\
\text { distance } \\
d=39 m \\
\end{array} \\
\hline \begin{array}{l}
B \text { (deceleration } \\
\text { on hard braking } \\
=7.5 \mathrm{~m} / \mathrm{s}^2 \text { ) }
\end{array} & \begin{array}{l}
\text { Speed }=54 \mathrm{~km} / \mathrm{h} \\
\text { Braking distance } \\
e=15 m\\
\text { Total stopping } \\
\text { distance } \\
f=18 m
\end{array} & \begin{array}{l}
\text { Speed }=72 \mathrm{~km} / \mathrm{h} \\
\text { Braking distance } \\
g=27 m \\
\text { Total stopping } \\
\text { distance } \\
h=33 m
\end{array} \\
\hline
\end{array}
\)