Kinematics (Entrance Paper) Quiz-L3

Hint

\(
V_{\text {avg }}=\frac{2 v_1 v_2}{v_1+v_2}=\frac{2(v)(2 v)}{v+2 v}=\frac{4 v^2}{3 v}=\frac{4 v}{3}
\)

Hint

The centripetal force will exert itself in the north-east direction in order to maintain the player’s motion in a circular arc towards the east, while simultaneously avoiding contact with the opponent.

\(
\begin{array}{l}
\vec{V}_i=(V) \text { southward } \\
\vec{V}_F=(V) \text { Eastward } \\
\overrightarrow{\Delta V}=\vec{V}_F-\vec{V}_i \\
=\text { Along North – East }
\end{array}
\)

When a football player is going south and then quickly turns east to dodge an opponent, they feel a force called the centripetal force. This force is aimed at the middle of the circle path that the person is following while turning. In this case, the centripetal force works toward the middle of the circular arc made by the player’s movement.

Since the player starts going south and then turns east, the centripetal force will act in the opposite direction of the player’s speed when the player turns. This is based on the basic rule that the centripetal force always points toward the center of curve of the path of motion.

Hint

\(
\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2
\)

\(
\begin{aligned}
s & =u t-\frac{1}{2} g t^2 \\
& =16-\frac{1}{2} \times 10 \times 16 \\
& =-64 \mathrm{~m}
\end{aligned}
\)

Height of bridge above water surface \(=64 \mathrm{~m}\)

Hint

\(
v^2=u^2+2 a s
\)
\(
\begin{array}{c}
\frac{u^2}{9}=u^2+2 a(24) \\
\frac{-8 u^2}{9}=2 a(24) \dots(1)
\end{array}
\)
For the entire path
\(
v^2-u^2=2 a(s) \\
0-u^2=2 a(s) \\
-u^2=2 a(s) \dots(2)
\)
Dividing eqns (1) and (2)
\(
\frac{8}{9}=\frac{24}{s} \Rightarrow S=27 \mathrm{~cm}
\)

Hint

(a) Slope of position-time \((x-t)\) graph represents velocity.
In graph (a), the slope is increasing hence the velocity is increasing hence acceleration is positive.

\(
s=\frac{1}{2} a t^2 \quad \text { (if coefficients of } t^2 \text { is positive }(a>0 \text { ) }
\)

Hint

\(
\begin{array}{l}
S_{n^{t h}}=u+\frac{1}{2} a(2 n-1) \\
S_{1^{s t}}=\frac{1}{2} g(2 \times 1-1)=\frac{g}{2} \\
S_{2^{n d}}=\frac{1}{2} g(2 \times 2-1)=3 \times\left(\frac{1}{2} g\right) \\
S_{3^{\text {rd }}}=\frac{1}{2} g(2 \times 3-1)=5 \times\left(\frac{1}{2} g\right) \\
S_{4^{t h}}=\frac{1}{2} g(2 \times 4-1)=7 \times\left(\frac{1}{2} g\right) \\
S_{1^{\text {st }}}: S_{2^{\text {nd }}}: S_{3^{\text {rd }}}: S_{4^{\text {th }}} \\
=1: 3: 5: 7 \\
\end{array}
\)

Hint

Slope of \(x\)-t curves gives the velocity
\(
\therefore \frac{v_1}{v_2}=\frac{\tan \theta_1}{\tan \theta_2}
\)
\(
=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}=\frac{1}{\frac{\sqrt{3}}{1}}=1: \sqrt{3}
\)

Hint

\(
\frac{S_n}{S_{n+1}}=\frac{\frac{a}{2}(2 n-1)}{\frac{a}{2}(2(n+1)-1)}=\frac{2 n-1}{2 n+2-1}=\frac{2 n-1}{2 n+1}
\)

Hint

(c) Given, \(u=20 \mathrm{~m} / \mathrm{s}, v=80 \mathrm{~m} / \mathrm{s}\) and \(h=\) ?
From kinematic equation of motion, \(v^2=u^2+2 g h\)
\(
\begin{aligned}
\Rightarrow h & =\frac{v^2-u^2}{2 g}=\frac{(80)^2-(20)^2}{2 \times 10} \quad\left(\because \text { Given, } g=10 \mathrm{~m} / \mathrm{s}^2\right) \\
& =300 \mathrm{~m}
\end{aligned}
\)

Hint

(b) According to question, time taken by the ball to cross the window,
\(
t=0.1 \mathrm{~s}, h=1.5 \mathrm{~m}
\)
If \(u\) be the velocity at the top most point of the window, then from equation of motion,
\(
\begin{aligned}
& h=u t+\frac{1}{2} g t^2 \\
\Rightarrow & 1.5=u \times 0.1+\frac{1}{2} \times 10 \times(0.1)^2 \\
\Rightarrow & 1.5=0.1 u+0.05 \\
\Rightarrow & u=\frac{1.5-0.05}{0.1}=\frac{1.45}{0.1}=14.5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(b) For distance \(x\), the person moves with constant velocity \(v_1\) and for another \(x\) distance, he moves with constant velocity of \(v_2\), then
Total distance travelled by the person,
\(
D=x+x=2 x
\)
Total time taken to cover that distance,
\(
T=t_1+t_2=\frac{x}{v_1}+\frac{x}{v_2} \quad\left(\because t=\frac{\text { Distance }}{\text { Velocity }}\right)
\)
Average velocity,
\(
\begin{aligned}
v_{\mathrm{av}} & =\frac{\text { Total distance }}{\text { Total time }}=\frac{D}{T} \\
v & =\frac{2 x}{\frac{x}{v_1}+\frac{x}{v_2}}=\frac{2}{\frac{1}{v_1}+\frac{1}{v_2}} \\
\Rightarrow \quad \frac{2}{v} & =\frac{1}{v_1}+\frac{1}{v_2}
\end{aligned} \quad\left(\because v_{\mathrm{av}}=v\right)
\)

Hint

Speed of swimmer w.r.t river, \(V_{S R}=20 \mathrm{~m} / \mathrm{s}\)
Speed of river flow, \(V_{R G}=10 \mathrm{~m} / \mathrm{s}\)
For the shortest path to cross the river, he should swim at an angle \(\left(90^{\circ}+\theta\right)\) with the stream flow. From the above figure,

\(
\vec{V}_{S G}=\vec{V}_{S R}+\vec{V}_{R G}
\)
So,
\(
\begin{array}{l}
\sin \theta=\frac{\left|\overrightarrow{\mathrm{V}}_{\mathrm{RG}}\right|}{\left|\overrightarrow{\mathrm{V}}_{\mathrm{SR}}\right|} \\
\sin \theta=\frac{10}{20}=\frac{1}{2} \\
\therefore \theta=30^{\circ} \text { west }
\end{array}
\)

As the river is flowing in east direction, so he should swim towards west.

Hint

(d) When a particle completes one revolution in circular motion, then average displacement travelled by particle is zero.
Hence, average velocity
\(
=\frac{\text { average displacement }}{\Delta t}=\frac{0}{\Delta t}=0
\)

Hint

(b) Time interval between 8th and 3rd second,
\(
\Delta t=8-3=5 \mathrm{~s} \text {, i.e. } \Delta t=5 \mathrm{~s}
\)
Change in velocity,
\(
\Delta v=20-0=20 \mathrm{~ms}^{-1}
\)
\(\therefore\) Average acceleration \(=\frac{\Delta v}{\Delta t}=\frac{20}{5}=4 \mathrm{~ms}^{-2}\)

Hint

Given condition can be represented through graph also as shown below.

\(\therefore\) Displacement in three seconds
\(=\) Area under the graph
\(
=\frac{1}{2} \times 1 \times 6+\frac{1}{2} \times 1 \times 6-\frac{1}{2} \times 6 \times 1=3 \mathrm{~m}
\)
\(
\text { Average velocity }=\frac{\text { net displacement }}{\text { total time }}=\frac{3 \mathrm{~m}}{3 \mathrm{~s}}=1 \mathrm{~m} / \mathrm{s}
\)
Total distance travelled, \(d=9 \mathrm{~m}\)
Hence, average speed
\(
=\frac{\text { total distance }}{\text { total time }}=\frac{9}{3}=3 \mathrm{~ms}^{-1}
\)

Hint

(c) When a particle is released from rest position under gravity, then \(v=0\), but \(a \neq 0\).
Also, a body is momentarily at rest at the instant, if it reverse the direction.
Thus, Assertion is correct but Reason in incorrect.

Hint

(a) To find value of time at which velocity is maximum, taking differentiation of \(v\) with respect to time
Given,
\(
\frac{d v}{d t}=0
\)
\(
\text { Given, } \begin{aligned}
v & =4 t(1-2 t) \\
v & =4 t-8 t^2 \Rightarrow \frac{d}{d t}\left(4 t-8 t^2\right)=0 \\
\Rightarrow \quad 4-16 t & =0 \Rightarrow t=\frac{1}{4} \mathrm{~s}=0.25 \mathrm{~s}
\end{aligned}
\)
Again, taking differentiation, we get
\(
\Rightarrow \quad \frac{d^2 v}{d t^2}=-16<0. \operatorname{maxima} \text { at } \mathrm{t}=1 / 4
\)

So, at \(t=0.25 \mathrm{~s}\), velocity is maximum.

Hint

(a) Runner starts from \(O\) and goes to \(O\) following the path \(O Q R O\), so net displacement is zero.
\(
\begin{aligned}
\because \text { Average speed } & =\frac{\text { Total distance }}{\text { Total time }}=\frac{O Q+Q R+R O}{\text { Total time }} \\
& =\frac{1 \mathrm{~km}+(2 \pi r)(1 / 4)+1 \mathrm{~km}}{1 \mathrm{~h}} \\
& =2+\frac{\pi}{2}=3.57 \mathrm{~kmh}^{-1}
\end{aligned}
\)

Hint

(c) Time taken to reach the highest point from the height \(h\) is obtained from equation, \(v=u-g t\)
\(
\therefore \quad 0=u-g t \text { or } t=\frac{u}{g}
\)
( \(\because\) At highest point, final velocity of ball \(=0\) )
Height attained above \(h\) is obtained from \(v^2-u^2=2(-g) h_1\)
or \(\quad 0-u^2=2(-g) h_1\) or \(h_1=\frac{u^2}{2 g}\)
Total height, \(h_2=\left(h_1+h=\frac{u^2}{2 g}+h\right)\)
Time taken to hit the ground is obtained from
\(
h_2=u t+\frac{1}{2} a t^2 \text { or } \frac{u^2}{2 g}+h=0+\frac{1}{2} g t^2
\)
\(\therefore\) Total time taken, \(t=\sqrt{\frac{\left(u^2+2 g h\right)}{g}}\)

Hint

(c) Speed of walking, \(v_1=\frac{h}{t_1}\) Speed of escalator, \(v_2=\frac{h}{t_2}\)
Time taken when she walks over moving escalator,
\(
\begin{array}{ll}
& t=\frac{h}{v_1+v_2} \\
\Rightarrow & \frac{1}{t}=\frac{v_1}{h}+\frac{v_2}{h}=\frac{1}{t_1}+\frac{1}{t_2} \\
\Rightarrow \quad & t=\frac{t_1 \times t_2}{t_1+t_2}
\end{array}
\)

Hint

(c) For the given \(v-x\) graph,
Slope \(=-v_0 / x_0\)
and intercept \(=v_0\)
From the general equation of straight line, i.e.
\(
y=m x+c
\)
where, \(m\) is slope and \(c\) is the intercept.
The equation of motion of the \(v-x\) graph can be given as
\(
v=-\frac{v_0 x}{x_0}+v_0 \dots(i)
\)
(-ve sign signifies that the slope is decreasing)
On differentiating Eq. (i) with respect to \(t\), we get
\(
\frac{d v}{d t}=-\frac{v_0}{x_0} \frac{d x}{d t}+0 \text { or } a=-\frac{v_0}{x_0} v \dots(ii)
\)
Putting the value of \(v\) from Eq. (i) in Eq. (ii), we get
\(
a=-\frac{v_0}{x_0}\left[-\frac{v_0}{x_0} x+v_0\right]=\frac{v_0^2}{x_0^2} x-\frac{v_0^2}{x_0}
\)
This equation shows that at \(x=0\), \(a=-\frac{v_0^2}{x_0}=\) constant and at \(a=0, x=x_0\).
So, option (c) is correct.

Hint

(b) For a car in motion, if we describe this event w.r.t. a frame of reference attached to the person sitting inside the car, the car will appear to be at rest as the person inside the car (i.e. observer) is also moving with same velocity and in the same direction as car.

Hint

(b) Velocity of the particle is given as \(v=A t+B t^2\) where, \(A\) and \(B\) are constants.
\(
\begin{aligned}
\Rightarrow & \frac{d x}{d t} =A t+B t^2 & \left(\because v=\frac{d x}{d t}\right) \\
\Rightarrow & d x =\left(A t+B t^2\right) d t &
\end{aligned}
\)
Integrating both sides, we get
\(
\begin{aligned}
\int_{x_1}^{x_2} d x & =\int_1^2\left(A t+B t^2\right) d t \\
\Rightarrow \quad \Delta x & =x_2-x_1=A \int_1^2 t d t+B \int_1^2 t^2 d t \\
& =A\left[\frac{t^2}{2}\right]_1^2+B\left[\frac{t^3}{3}\right]_1^2 \\
& =\frac{A}{2}\left(2^2-1^2\right)+\frac{B}{3}\left(2^3-1^3\right)
\end{aligned}
\)
\(\therefore\) Distance travelled by the particle between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
\(
\Delta x=\frac{A}{2} \times(3)+\frac{B}{3}(7)=\frac{3 A}{2}+\frac{7 B}{3}
\)

Hint

(b)

\(
\text { (b) Given, } \begin{aligned}
v & =\beta x^{-2 n} \\
a & =\frac{d v}{d t}=\frac{d x}{d t} \cdot \frac{d v}{d x}
\end{aligned}
\)
\(
\begin{array}{l}
a=v \frac{d v}{d x}=\left(\beta x^{-2 n}\right)\left(-2 n \beta x^{-2 n-1}\right) \\
a=-2 n \beta^2 x^{-4 n-1}
\end{array}
\)

Hint

(c) Let the ball hit water in \(t\) second.

For first ball, \(\quad s=u t+\frac{1}{2} a t^2\)
\(
\begin{aligned}
122.5 & =0+\frac{1}{2} \times 9.8 \times t^2=4.9 t^2 \\
\Rightarrow \quad t & =\sqrt{\frac{122.5}{4.9}}=\sqrt{25}=5 \mathrm{~s}
\end{aligned}
\)
\(
\begin{array}{l}
\text { For second ball, } \begin{aligned}
122.5 & =u(5-2)+\frac{1}{2} \times 9.8 \times(5-2)^2 \\
& =3 u+44.1 \\
3 u & =122.5-44.1 \\
3 u & =78.4 \Rightarrow u=26.1 \mathrm{~ms}^{-1}
\end{aligned}
\end{array}
\)

Hint

(c) Given, upward velocity of ball \(=25.2 \mathrm{~ms}^{-1}\)

Height attained by the ball,
\(
H=\frac{u^2}{2 g}=\frac{25.2 \times 25.2}{2 \times 9.8}=32.4 \mathrm{~m}
\)

Now, time taken by the ball to attain \(32.4 \mathrm{~m}\) is
\(
t=\frac{u}{g}=\frac{25.2}{9.8}=2.57 \mathrm{~s}
\)

Hint

(c) According to question, the given condition can be depicted as

For motion from \(A\) to \(C\),
\(
\frac{s}{2}=2 u t+\frac{1}{2} a \times(2 t)^2=2 u t+2 a t^2 \dots(i)
\)
Also, for motion from \(C\) to \(B\),
\(
\frac{s}{2}=v^{\prime} t+\frac{1}{2} a t^2=(u+2 a t) t+\frac{1}{2} a t^2
\)
(Let initial velocity be \(v^{\prime}=u+2 a t\) )
\(
=u t+2 a t^2+\frac{1}{2} a t^2
\)
Now, from Eq. (i), we get
\(
2 u t+2 a t^2=u t+2 a t^2+\frac{1}{2} a t^2 \Rightarrow u=\frac{a t}{2}
\)
For overall motion,
\(
v=u+3 a t=\frac{a t}{2}+3 a t=\frac{7 a t}{2}=7 u \quad\left(\because u=\frac{a t}{2}\right)
\)

Hint

(d) At first, the slope is decreasing, therefore the motion is retarded. Finally, the displacement becomes constant, thus the particle stops.

Hint

(d) Given, \(u=0, v=180 \mathrm{kmh}^{-1}=50 \mathrm{~ms}^{-1}\)

Time taken, \(t=10 \mathrm{~s}, a=\frac{v-u}{t}=\frac{50}{10}=5 \mathrm{~ms}^{-2}\)
Distance covered by the car,
\(
\begin{aligned}
s & =u t+\frac{1}{2} a t^2=0+\frac{1}{2} \times 5 \times(10)^2 \\
& =\frac{500}{2}=250 \mathrm{~m}
\end{aligned}
\)

Hint

(d) The slope of velocity-time graph gives acceleration, i.e. \(\quad \frac{a_A}{a_B}=\frac{\tan \theta_1}{\tan \theta_2}\)

Here, \(\theta_1\) is the angle made by line \(A\) with time axis and \(\theta_2\) is the angle made by line \(B\) with time axis
\(
\Rightarrow \quad \frac{a_A}{a_B}=\frac{\tan 25^{\circ}}{\tan 50^{\circ}}
\)

Hint

(a) The position of an object is always expressed w.r.t. some reference point. If the initial position of an object w.r.t. a reference points is \(s_1\) and after sometime, it changes to \(s_2\), then the magnitude of the displacement of the object is \(s_2-s_1\). Since, the given ball returns to its starting point in \(10 \mathrm{~s}\). So, in \(5 \mathrm{~s}\), it reaches the highest point which is at \(s_2\).
Given, \(s_2\) is at a height \(h\) w.r.t. \(s_1\). Thus, the displacement after 5 s will be \(s_2-s_1=h\).

Hint

It is given that the velocity of the police jeep is \(45 \mathrm{~km} / \mathrm{h}\).
First, let us convert it to \(\mathrm{m} / \mathrm{s}\) as the options are in this unit.
\(
\begin{array}{l}
45 \mathrm{~km} / \mathrm{h}=45 \times \frac{1000}{3600} \mathrm{~m} / \mathrm{s} \\
=45 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} =12.5 \mathrm{~m} / \mathrm{s}
\end{array}
\)
The muzzle velocity of the bullet is \(180 \mathrm{~m} / \mathrm{s}\). As the muzzle of the gun is also moving at a velocity of \(12.5 \mathrm{~m} / \mathrm{s}\), a ground observer will see the velocity of the bullet to be,
Observed bullet velocity = Muzzle velocity of the bullet + Muzzle’s velocity (Velocity of the police jeep).
\(
=(180+12.5)=192.5 \mathrm{~m} / \mathrm{s}
\)
Hence, the bullet velocity with respect to a ground observer is \(=192.5 \mathrm{~m} / \mathrm{s}\)
Now, let us convert the velocity of the thief’s jeep into \(\mathrm{m} / \mathrm{s}\),
\(
\begin{array}{l}
153 \mathrm{~km} / \mathrm{h}=153 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} =42.5 \mathrm{~m} / \mathrm{s}
\end{array}
\)
Hence the relative velocity of the bullet with respect to the thief’s jeep is given by,
Relative velocity \(=(\) Velocity of bullet – Velocity of the thief’s jeep \()\).
Both these velocities are taken with respect to a ground observer.
So, the bullet will hit the thief’s jeep at a speed of,
\(
=(192.5-42.5)=150 \mathrm{~m} / \mathrm{s}
\)

Hint

\(
\begin{array}{c}
\text { (b) We know that, } s_{n \text {th }}=u+\frac{1}{2} a(2 n-1) \\
s_{3 \text { rd }}=0+\frac{1}{2} a(2 \times 3-1)=\frac{5}{2} a \text { (For } n=3 \mathrm{~s} \text { ) }
\end{array}
\)
\(
\left.s_{\text {th }}=0+\frac{1}{2} a(2 \times 4-1)=\frac{7}{2} a \quad \text { (For } n=4 \mathrm{~s}\right)
\)
\(
\begin{array}{l}
\text { So, the percentage increase }=\frac{s_{4 \text { th }}-s_{3 \text { rd }}}{s_{3 \text { rd }}} \times 100 \\
=\frac{\frac{7}{2} a-\frac{5}{2} a}{\frac{5}{2} a} \times 100=\frac{\frac{2 a}{2}}{\frac{5}{2} a} \times 100=2 \times 20=40 \%
\end{array}
\)

Hint

Average speed is \(\mathrm{V}=\frac{\text { distance }}{\text { time }}\)
if a car travels first half with \(V_1\) and second half of distance with \(V_2\) then formula of average velocity is given by
\(
\mathrm{V}=\frac{2 \mathrm{~V}_2 \mathrm{~V}_1}{\mathrm{~V}_1+\mathrm{V}_2}=\frac{2 \times 60 \times 40}{60+40}=48 \mathrm{~km} / \mathrm{h}
\)

Hint

(a) The distance covered in \(2 \mathrm{~s}, \quad x=\frac{1}{2} a(2)^2 \quad(\because u=0)\)
\(
=2 a
\)

The distance covered in next \(2 \mathrm{~s}\),
\(
\begin{aligned}
y & =\frac{1}{2} a(4)^2-\frac{1}{2} a(2)^2=6 a \\
x / y & =\frac{2 a}{6 a}=\frac{1}{3} \Rightarrow y=3 x
\end{aligned}
\)

Hint

(c) \(A \rightarrow v_A\) (constant velocity)
\(B \rightarrow a\) (constant acceleration)
\(C\) be the fixed point at which both \(A\) and \(B\) meet in time \(t\).
Now, \(s=v_A t\) and \(v_B=a t\)
\(
\begin{array}{l}
s=\frac{1}{2} a t^2=v_A t \\
a=\frac{2 v_A}{t} \dots(i)
\end{array}
\)
\(
\begin{array}{rlr}
v_B & =u+a t & (\because v=0) \\
v_B & =a t=2 v_A & \text { [From Eq. (i)] } \\
v_{B A} & =\text { velocity of } B \text { w.r.t. } A & \\
v_{B A} & =v_B-v_A=2 v_A-v_A=v_A & \\
v_{B A} & =v & \left(\because v_A=v\right)
\end{array}
\)

Hint

\(
v_{\mathrm{av}}=\frac{2 v_{A B} v_{B A}}{v_{A B}+v_{B A}}=24 \mathrm{kmh}^{-1}
\)

Hint

(c)

Let \(A\) be the original point.
From equation of motion,
\(
x=u t+\frac{1}{2} a t^2=\frac{1}{2} a t^2 \quad(\because u=0)
\)
For point \(C, \quad x_1=\frac{1}{2} a(t / 2)^2 \Rightarrow x_1=\frac{1}{8} a t^2\)
Distance travelled by the body in \(t\) second is
\(
\begin{aligned}
x_1+x_2 & =\frac{1}{2} a(t)^2=4 x_1 \\
x_2 & =4 x_1-x_1=3 x_1
\end{aligned}
\)

Hint

\(
a=\frac{d v}{d t}=\text { Slope of } v-t \text { curve }
\)

Hint

(b) The distance \(h_1\) covers by stone,
\(
h_1=\frac{1}{2} g t^2 \times \frac{1}{2} 10 \times(5)^2=125
\)

The distance \(h_2\) covers by stone,
\(
h_2=\frac{1}{2} \times 10 \times(10)^2-\frac{1}{2} \times 10 \times(5)^2=375
\)

The distance \(h_3\) covers by stone,
\(
h_3=\frac{1}{2} \times 10 \times(15)^2-\frac{1}{2} \times 10 \times(10)^2=625
\)

The relation between \(h_1, h_2\) and \(h_3\) is \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\)

Hint

(b) When velocity of a body changes by equal amount in equal intervals of time, then the body is said to have uniform acceleration, this holds true for straight line motion.

Hint

(c) Velocity-time graph gives the instantaneous value of acceleration at any instant. For non-uniformly accelerated motion, \(v-t\) graph is non-linear.

Hint

(b) The initial velocity of the body, \(u^2=2 g h\)
\(
\Rightarrow \quad u=\sqrt{2 g h}=\sqrt{2 \times 10 \times 125}=\sqrt{2500}=50 \mathrm{~ms}^{-1}
\)

The final velocity from equation of motion,
\(
\begin{aligned}
v^2 & =u^2+2 g h \\
v & =\sqrt{(50)^2+2 \times 10 \times 55} \\
v & =\sqrt{2500+1100} \\
& =\sqrt{3600}=60 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

(a) Given, velocity of water, \(v_w=4 \mathrm{~ms}^{-1}\)

Velocity of person, \(v_p=5 \mathrm{~ms}^{-1}\), width of the river \(=846 \mathrm{~m}\)
We know that, time taken, \(t=\frac{s}{\sqrt{v_p^2-v_W^2}}\)
\(
\begin{array}{ll}
\Rightarrow & t=\frac{846}{\sqrt{25-16}} \\
\Rightarrow & t=\frac{84.6}{3} \cong 28.2 \mathrm{~s}
\end{array}
\)

Alternate: Let the angle between the path traversed by person and the direction in which he tries to swim with respect to river be \(\theta\).
Since the net path traveled is a straight line directly towards the opposite end, the component of net velocity does not exist in the direction of river’s flow.
Thus \(\operatorname{vin} \theta=v_{\text {river }}\)
\(
\Longrightarrow \sin \theta=\frac{4}{5}
\)
Thus net velocity of man in the direction of crossing=
\(
v \cos \theta=5 \mathrm{~m} / \mathrm{s} \times \frac{3}{5}=3 \mathrm{~m} / \mathrm{s}
\)
Thus total time taken to cross the river \(=\frac{84.6 \mathrm{~m}}{3 \mathrm{~m} / \mathrm{s}}=28.2 \mathrm{~s}\)

Hint

(c) From graph, velocity of robber’s car \(=10 \mathrm{~ms}^{-1}\)

Let police car crosses it after \(t\) second.
Distance travelled by robber’s car \(=10 t \mathrm{~m}\)
Police car is moving with a constant acceleration of \(1 \mathrm{~ms}^{-2}\) as it attains a velocity of \(10 \mathrm{~ms}^{-1}\) in \(10 \mathrm{~s}\) after starting from rest.
Distance travelled in \(t\) second \(=\frac{1}{2} \cdot a \cdot t^2=\frac{1}{2} t^2\)
When the police car crosses the robber’s car, distance travelled by the both cars should be same from the starting point of chase.
\(
\therefore \quad \frac{1}{2} t^2=10 t \Rightarrow t=20 \mathrm{~s}
\)

Hint

(b) Given, initial speed, \(u=1 \mathrm{kms}^{-1}=1000 \mathrm{~ms}^{-1}\)
Final speed, \(v=9 \mathrm{kms}^{-1}=9000 \mathrm{~ms}^{-1}\)
By using the relation, \(v^2=u^2+2 a s\)
\(
(9000)^2=(1000)^2+2 \times a \times 4 \Rightarrow a=10^7 \mathrm{~ms}^{-2}
\)
\(\therefore\) The time for which the particle remains in the tube
\(
\begin{aligned}
v & =u+a t \\
\Rightarrow \quad t & =\frac{v-u}{a}=\frac{9000-1000}{10^7}=8 \times 10^{-4} \mathrm{~s}
\end{aligned}
\)

Hint

(d) Given, \(x=8+12 t-t^3\)

We know that , \(v=\frac{d x}{d t}[latex] and [latex]a=\frac{d v}{d t}\)
So,
\(v=12-3 t^2\)
and
\(t=2 \mathrm{~s}\),
\(v=0\)
and
\(a=-12 \mathrm{~ms}^{-2}\)

So, retardation of the particle \(=12 \mathrm{~ms}^{-2}\).

Hint

(d) Distance, \(x=b_0+b_1 t+b_2 t^2\)
Velocity, \(v=\left(\frac{d x}{d t}\right)=b_1+2 b_2 t\)
Acceleration, \(a=\frac{d^2 x}{d t^2}=2 b_2\)

Hint

(c) Let \(t\) second be the time of flight of the first body after meeting, then \((t-4)\) second will be the time of flight of the second body.
As the initial velocity at which the bodies \(A\) and \(B\) projected are same and also the position of meeting will be also same.
So, \(\quad h_x=h_y\)
\(
\therefore \quad 98 t-\frac{1}{2} g t^2=98(t-4)-\frac{1}{2} g(t-4)^2
\)
On solving, \(t=12 \mathrm{~s}\)

Hint

(b) Let after time \(t\), car catches the scooter and the distance travelled by scooter in time \(t\),
\(
x=\frac{1}{2} \times(1) \times t^2=\frac{t^2}{2} \dots(i)
\)
The distance travelled by car in time \(t\)
\(
x+150=\frac{1}{2} \times 2 \times t^2=t^2 \dots(ii)
\)
Solving Eqs. (i) and (ii), we get
\(
t=\sqrt{300} \mathrm{~s}
\)

Hint

\(
\begin{array}{l}
\text { (c) Given, } \mathbf{r}_1(t)=3 t \hat{\mathbf{i}}+4 t^2 \hat{\mathbf{j}} \\
\therefore \quad \frac{d \mathbf{r}_1}{d t}=3 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}
\end{array}
\)
\(
\begin{array}{l}
\text { At } t=1 \mathrm{~s} \text {, } \\
\mathbf{v}_1=\frac{d \mathbf{r}_1}{d t}=3 \hat{\mathbf{i}}+8 \hat{\mathbf{j}} \\
\text { Again, } \quad \mathbf{r}_2(t)=4 t^2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \\
\Rightarrow \quad \frac{d \mathbf{r}_2}{d t}=8 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \\
\end{array}
\)
At
\(
\begin{aligned}
t & =1 \mathrm{~s}, \\
\mathbf{v}_2 & =\frac{d \mathbf{r}_2}{d t}=8 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}
\end{aligned}
\)
\(
\begin{aligned}
\text { Relative speed, } \mathbf{v}_{\text {rel }} & =\mathbf{v}_2-\mathbf{v}_1 \\
& =(8 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})-(3 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}) \\
& =5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}} \\
\therefore \quad\left|\mathbf{v}_{\text {rel }}\right| & =\sqrt{(5)^2+(-5)^2} \\
& =5 \sqrt{2} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)