Sample Exercises

Hint

Let the man follow the path as follows(fig-a):
The distance is the length of the path taken by the man to reach the field from his home. The length of the path is
\(
\begin{aligned}
& l=50+40+20 \\
& \therefore l=110 m
\end{aligned}
\)
The total distance that the man travels is \(110 \mathrm{~m}\).
Displacement is the shortest distance between two points. It has magnitude as well as direction. While displacement is a vector, distance is a scalar.
The displacement between the man’s home and field will be (fig-b)
In triangle HAF,
\(
\begin{aligned}
& A H=40 m \\
& A F=50-20=30 m
\end{aligned}
\)
Applying Pythagoras theorem in \(\triangle H A F\)
\(
\begin{aligned}
& H F^2=A H^2+A F^2 \\
& \Rightarrow H F^2=(40)^2+(30)^2 \\
& \Rightarrow H F^2=2500 \\
& \therefore H F=50 m
\end{aligned}
\)
Therefore, the displacement between the man’s house and field is \(50 \mathrm{~m}\).
Therefore, the distance travelled by the man is \(110 \mathrm{~m}\), while the displacement between two points is \(50 \mathrm{~m}\).

Hint

Initial velocity \(\mathrm{v_{0}}=0(\therefore\) starts from rest \()\)
Final velocity \(=18 \mathrm{~km} / \mathrm{hr}=5 \mathrm{m/sec}\)
(i.e. max velocity)
Time interval \(t=2 \mathrm{sec}\)
\(
\therefore \text { Acceleration }=\mathrm{a}_{\text {ave }}=\frac{\mathrm{v}-\mathrm{v_{0}}}{\mathrm{t}}=\frac{5}{2}=2.5 \mathrm{~m} / \mathrm{s}^{2} \text {. }
\)

Hint

From the figure we can see, in the interval \(8 \mathrm{sec}\) the velocity changes from 0 to \(20 \mathrm{~m} / \mathrm{s}\).
Average acceleration \(=20 / 8=2.5 \mathrm{~m} / \mathrm{s}^{2}\left(\frac{\text { change in velocity }}{\text { time }}\right)\)
Distance travelled \(S=v_{0} t+1 / 2\) at \(^{2}\)
\(
\Rightarrow 0+1 / 2(2.5) 8^{2}=80 \mathrm{~m} \text {. }
\)

Hint

\(\text { In } 1^{\text {st }} 10 \mathrm{sec}, \mathrm{S}_{1}= \mathrm{v_{0}t}+1 / 2 \mathrm{at}^{2} \Rightarrow 0+\left(1 / 2 \times 5 \times 10^{2}\right)=250 \mathrm{ft}\)

At \(10 \mathrm{sec}, \mathrm{v}=\mathrm{v_{0}}+\) at \(=0+5 \times 10=50 \mathrm{ft} / \mathrm{sec}\)
\(\therefore\) From 10 to \(20 \sec (\Delta t=20-10=10 \mathrm{sec})\) it moves with uniform velocity \(50 \mathrm{ft} / \mathrm{sec}\),
Distance \(S_{2}=50 \times 10=500 \mathrm{ft}\)
From the figure, Between \(20 \mathrm{sec}\) to \(30 \mathrm{sec}\) acceleration is constant i.e. \(-5 \mathrm{ft} / \mathrm{s}^{2}\). At \(20 \mathrm{sec}\) velocity is \(50 \mathrm{ft} / \mathrm{sec}\).
\(
t=30-20=10 \mathrm{~s}
\)
\(
\begin{aligned}
&S_{3}=v_{0} t+1 / 2 \text { at }^{2} \\
&=50 \times 10+(1 / 2)(-5)(10)^{2}=250 \mathrm{~m}
\end{aligned}
\)
Total distance travelled is \(30 \mathrm{sec}=\mathrm{S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3}=250+500+250=1000 \mathrm{ft}\)

Hint

a) Initial velocity \(v_{0}=2 \mathrm{~m} / \mathrm{s}\).
final velocity \(v=8 \mathrm{~m} / \mathrm{s}\)
time \(=10 \mathrm{sec}\),
acceleration \(=\frac{\mathrm{v}-\mathrm{v_{0}}}{\mathrm{ta}}=\frac{8-2}{10}=0.6 \mathrm{~m} / \mathrm{s}^{2}\)
b) \(v^{2}-v_{0}^{2}=2 a S\)
\(\Rightarrow\) Distance \(\mathrm{S}=\frac{\mathrm{v}^{2}-\mathrm{v_{0}}^{2}}{2 \mathrm{a}}=\frac{8^{2}-2^{2}}{2 \times 0.6}=50 \mathrm{~m}\).
c) Displacement is the same as distance travelled.
Displacement \(=50 \mathrm{~m}\).

Hint

a) Displacement in 0 to \(10 \mathrm{sec}\) is \(1000 \mathrm{~m}\).
time \(=10 \mathrm{sec}\).
\(
V_{\text {ave }}=\mathrm{s} / \mathrm{t}=100 / 10=10 \mathrm{~m} / \mathrm{s} \text {. }
\)
b) At \(2 \mathrm{sec}\) it is moving with uniform velocity \(50 / 2.5=20 \mathrm{~m} / \mathrm{s}\). at \(2 \mathrm{sec} . V_{\text {inst }}=20 \mathrm{~m} / \mathrm{s}\).
At \(5 \mathrm{sec}\) it is at rest.
\(
\mathrm{V}_{\text {inst }}=\text { zero. }
\)
At \(8 \mathrm{sec}\) it is moving with uniform velocity \(20 \mathrm{~m} / \mathrm{s}\)
\(
\mathrm{V}_{\text {inst }}=20 \mathrm{~m} / \mathrm{s}
\)
At \(12 \mathrm{sec}\) velocity is negative as it move towards initial position. \(V_{\text {inst }}=-20 \mathrm{~m} / \mathrm{s}\).

Hint

Distance in first \(40 \mathrm{sec}\) is, \(\triangle O A B+\triangle B C D\) \(=\frac{1}{2} \times 5 \times 20+\frac{1}{2} \times 5 \times 20=100 \mathrm{~m}\).
The average velocity is zero as the displacement is zero.

Hint

Consider the point \(B\), at \(t=12 \mathrm{sec}\)
At \(t=0 ; s=20 \mathrm{~m}\)
and \(t=12 \sec s=20 \mathrm{~m}\)
So for time interval 0 to \(12 \mathrm{sec}\)
Change in displacement is zero.
So, average velocity \(=\) displacement/ time \(=0\)
\(\therefore\) The time is \(12 \mathrm{sec}\).

Hint

\(\begin{aligned}
&\mathrm{v_{0}}=4 \mathrm{~m} / \mathrm{s}, \mathrm{a}=1.2 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{t}=5 \mathrm{sec} \\
&\begin{array}{l}
\text { Distance }=\mathrm{s}=\mathrm{v_{0}t}+\frac{1}{2} \mathrm{at}^{2} \\
\qquad=4(5)+1 / 2(1.2) 5^{2}=35 \mathrm{~m}
\end{array}
\end{aligned}\)

Hint

Initial velocity \(v_{0}=43.2 \mathrm{~km} / \mathrm{hr}=12 \mathrm{~m} / \mathrm{s}\)
\(
\mathrm{v_{0}}=12 \mathrm{~m} / \mathrm{s}, \mathrm{v}=0
\)
\(a=-6 \mathrm{~m} / \mathrm{s}^{2}\) (deceleration)
Distance \(S=\frac{v^{2}-v_{0}^{2}}{2(-6)}=12 \mathrm{~m}\)

Hint

\(\begin{aligned}
&\mathrm{v_{0}}=16 \mathrm{~m} / \mathrm{s} \text { (initial), } \mathrm{v}=0, \mathrm{~s}=0.4 \mathrm{~m} . \\
&\text { Deceleration } \mathrm{a}=\frac{\mathrm{v}^{2}-\mathrm{v_{0}}^{2}}{2 \mathrm{~s}}=-320 \mathrm{~m} / \mathrm{s}^{2} \\
&\text { Time }=\mathrm{t}=\frac{\mathrm{v}-\mathrm{v_{0}}}{\mathrm{a}}=\frac{0-16}{-320}=0.05 \mathrm{sec} .
\end{aligned}\)

Hint

\(
\mathrm{v_{0}}=350 \mathrm{~m} / \mathrm{s}, \mathrm{s}=5 \mathrm{~cm}=0.05 \mathrm{~m}, \mathrm{v}=0
\)
Deceleration \(=a=\frac{v^{2}-v_{0}^{2}}{2 s}=\frac{0-(350)^{2}}{2 \times 0.05}=-12.25 \times 10^{5} \mathrm{~m} / \mathrm{s}^{2}\).
Deceleration is \(12.25 \times 10^{5} \mathrm{~m} / \mathrm{s}^{2}\).

Hint

\(
\begin{aligned}
&v_{0}=0, v=18 \mathrm{~km} / \mathrm{hr}=5 \mathrm{~m} / \mathrm{s}, \mathrm{t}=5 \mathrm{sec} \\
&a=\frac{v-v_{0}}{t}=\frac{5-0}{5}=1 \mathrm{~m} / \mathrm{s}^{2} \\
&\mathrm{~s}=v_{0} \mathrm{t}+\frac{1}{2} \mathrm{at}^{2}=12.5 \mathrm{~m}
\end{aligned}
\)
a) Average velocity \(V_{\text {ave }}=(12.5) / 5=2.5 \mathrm{~m} / \mathrm{s}\).
b) Distance travelled is \(12.5 \mathrm{~m}\).

Hint

In reaction time the body moves with the speed \(54 \mathrm{~km} / \mathrm{hr}=15 \mathrm{~m} / \mathrm{sec}\) (constant speed)
Distance travelled in this time is \(S_{1}=15 \times 0.2=3 \mathrm{~m}\).
When brakes are applied,
\(
\mathrm{v_{0}}=15 \mathrm{~m} / \mathrm{s}, \mathrm{v}=0, \mathrm{a}=-6 \mathrm{~m} / \mathrm{s}^{2} \text { (deceleration) }
\)

\(
\mathrm{S}_{2}=\frac{\mathrm{v}^{2}-\mathrm{v_{0}}^{2}}{2 \mathrm{a}}=\frac{0-15^{2}}{2(-6)}=18.75 \mathrm{~m}
\)
Total distance \(s=s_{1}+s_{2}=3+18.75=21.75 \mathrm{~m}\).

Hint

\(
\begin{aligned}
&V_{P}=90 \mathrm{~km} / \mathrm{h}=25 \mathrm{~m} / \mathrm{s} . \\
&V_{C}=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)
In 10 sec culprit reaches at point \(B\) from \(A\).
Distance converted by culprit \(S=v t=20 \times 10=200 \mathrm{~m}\).
At time \(t=10 \mathrm{sec}\) the police jeep is \(200 \mathrm{~m}\) behind the culprit.
Time \(=s / v=200 / 5=40 \mathrm{~s}\). (Relative velocity is considered).
In 40 s the police jeep will move from [/latex]A[/latex] to a distance \(S\), where \(\mathrm{S}=\mathrm{vt}=25 \times 40=1000 \mathrm{~m}=1.0 \mathrm{~km}\) away.
\(\therefore\) The jeep will catch up with the bike, \(1 \mathrm{~km}\) far from the turning.

Hint

Velocity of the first car, \(v_{1}=60 \mathrm{~km} / \mathrm{h}=16.7 \mathrm{~m} / \mathrm{s}\)
Velocity of the second car, \(v_{2}=42 \mathrm{~km} / \mathrm{h}=11.7 \mathrm{~m} / \mathrm{s}\)
Relative velocity between the cars \(=(16.7-11.7)=5 \mathrm{~m} / \mathrm{s}\)
Distance travelled by the first car w.r.t. the second car \(=5+5=10 \mathrm{~m}\)
Time,
\(
t=\frac{s}{v}=\frac{10}{5} s=2 \mathrm{~s}
\)
Distance covered by the first car w.r.t. the ground in \(2 \mathrm{~s}=16.7 \times 2=33.4 \mathrm{~m}\) The first car also covers a distance equal to its own length \(=5 \mathrm{~m}\)
\(\therefore\) Total road distance used for the overtake \(=33.4+5 = 38.4  \mathrm{~m}\) = 38 m (approx)

Hint

\(v_{0}=50 \mathrm{~m} / \mathrm{s}, g=-10 \mathrm{~m} / \mathrm{s}^{2}\) when moving upward, \(v=0\) (at highest point).
a) \(S=\frac{v^{2}-u^{2}}{2 a}=\frac{0-50^{2}}{2(-10)}=125 \mathrm{~m}\)
maximum height reached \(=125 \mathrm{~m}\)
b) \(t=(v-v_{0}) / a=(0-50) /-10=5 \mathrm{sec}\)
c) \(s =125 / 2=62.5 \mathrm{~m}, \mathrm{v_{0}}=50 \mathrm{~m} / \mathrm{s}, \mathrm{a}=-10 \mathrm{~m} / \mathrm{s}^{2}\),

\(\begin{aligned}
& v^{2}-v_{0}^{2}=2 a s \\
\Rightarrow & v=\sqrt{\left(v_{0}^{2}+2 a s\right)}=\sqrt{50^{2}+2(-10)(62.5)}=35 \mathrm{~m} / \mathrm{s}
\end{aligned}\)

Hint

Initially, the ball is going upward
\(
\begin{aligned}
&v_{0}=-7 \mathrm{~m} / \mathrm{s}, \mathrm{s}=60 \mathrm{~m}, \mathrm{a}=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} \\
&\mathrm{~s}=\mathrm{v_{0}t}+\frac{1}{2} \mathrm{at}^{2} \Rightarrow 60=-7 \mathrm{t}+1 / 210 \mathrm{t}^{2} \\
&\Rightarrow 5 \mathrm{t}^{2}-7 \mathrm{t}-60=0 \\
&\mathrm{t}=\frac{7 \pm \sqrt{49-4.5(-60)}}{2 \times 5}=\frac{7 \pm 35.34}{10} \\
&\text { taking positive sign } \mathrm{t}=\frac{7+35.34}{10}=4.2 \mathrm{sec}(\therefore t \neq-v e)
\end{aligned}
\)
Therefore, the ball will take \(4.2 \mathrm{sec}\) to reach the ground.

Hint

(a) \(\mathrm{v_{0}}=28 \mathrm{~m} / \mathrm{s}, \mathrm{v}=0, \mathrm{a}=-\mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^{2}
\)
\(S=\frac{v^{2}-v_{0}^{2}}{2 a}=\frac{0^{2}-28^{2}}{2(9.8)}=40 \mathrm{~m}\)

(b) Initial velocity with which the stone is thrown vertically upwards, \(v_{0}=28 \mathrm{~m} / \mathrm{s}\)
When the stone reaches the ground, its final velocity ( \(v)\) is \(0 .\)
Also,
\(a=g=-9.8 \mathrm{~m} / \mathrm{s}^{2}\) (Acceleration due to gravity)
Total time taken by the stone to reach the maximum height:
\(
\begin{aligned}
&t=\frac{(v-v_{0})}{a} \\
&\Rightarrow t=\frac{(0-28)}{-9.8}=2.85 \mathrm{~s}
\end{aligned}
\)
As per the question, we need to find the velocity of the stone one second before it reaches the maximum height.
\(
\mathrm{t}^{\prime}=2.85-1=1.85 \mathrm{~s}
\)
Again, using the equation of motion, we get:
\(
\begin{aligned}
&v^{\prime}=v_{0}+a t^{\prime}=28-9.8 \times 1.85 \\
&\Rightarrow v^{\prime}=28-18.13=9.87 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Hence, the velocity is \(9.87 \mathrm{~m} / \mathrm{s}\)

(c) No it will not change. As after one second velocity becomes zero for any initial velocity and deceleration is \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) remains same. Fro initial velocity more than \(28 \mathrm{~m} / \mathrm{s}\) max height increases.

Hint

A person is releasing balls from a tall building at regular intervals of one second.
It means for each ball, the initial velocity \(v_{0}\) is 0.
Acceleration due to gravity, \(a=g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)
When the \(6^{\text {th }}\) ball is dropped, the \(5^{\text {th }}\) ball moves for 1 second, the \(4^{\text {th }}\) ball moves for 2 seconds and the \(3^{\text {rd }}\) ball moves for 3 seconds.
Position of the \(3^{\text {rd }}\) ball after \(t=3 \mathrm{~s}\):
Using the equation of motion, we get:
\(
\begin{aligned}
&s_{3}=u t+\frac{1}{2} a t^{2} \\
&\Rightarrow s_{3}=0+\frac{1}{2} \times 9.8 \times 3^{2}=44.1 \mathrm{~m}
\end{aligned}
\)
(from the top of the building)
Position of the \(4^{\text {th }}\) ball after \(t=2 \mathrm{~s}\):

\(
\begin{aligned}
&s_{4}=u t+\frac{1}{2} a t^{2} \\
&\Rightarrow s_{4}=0+\frac{1}{2} \times 9.8 \times 2^{2}=19.6 \mathrm{~m}
\end{aligned}
\)
(from the top of the building)
Position of the \(5^{\text {th }}\) ball after \(t=1 \mathrm{~s}\):
\(
\begin{aligned}
&s_{5}=u t+\frac{1}{2} a t^{2} \\
&\Rightarrow s_{5}=0+\frac{1}{2} \times 9.8 \times 1^{2}=4.9 \mathrm{~m}
\end{aligned}
\)
(from the top of the building)

Hint

Given Height of the building \(=11.8 \mathrm{~m}\)
Distance of the young man from the building \(=7 \mathrm{~m}\)
The kid should be caught over \(1.8 \mathrm{~m}\) from the ground.
As the kid is slipping, his initial velocity \(v_{0}\) is 0 .
Acceleration, \(a=9.8 \mathrm{~m} / \mathrm{s}^{2}\)
Let \(s\) be the distance before which the kid has to be caught \(=11.8-1.8=10 \mathrm{~m}\)
Using the equation of motion, we get:
\(
\begin{aligned}
&s=v_{0} t+\frac{1}{2} a t^{2} \\
&\Rightarrow 10=0+\frac{1}{2} \times 9.8 \times t^{2} \\
&\Rightarrow t^{2}=\frac{10}{4.9}=2.04 \\
&\Rightarrow t=1.42 s
\end{aligned}
\)

This is the time in which the man should reach the bottom of the building to catch the kid. The velocity with which the man should run:
\(
\frac{s}{t}=\frac{7}{1.42}=4.9 \mathrm{~m} / \mathrm{s}
\)

Hint

Given Speed of the NCC cadets \(=6 \mathrm{~km} / \mathrm{h}=1.66 \mathrm{~m} / \mathrm{s}\)
Distance of the bird from the ground, \(s=12.1 \mathrm{~m}\)
Initial velocity of the berry dropped by the bird, \(v_{0}=0\)
Acceleration due to gravity, \(a=g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)
Using the equation of motion, we can find the time taken \(t\) by the berry to reach the ground.
Thus, we have:
\(
\begin{aligned}
&s=v_{0} t+\frac{1}{2} a t^{2} \\
&\Rightarrow 12.1=0+\frac{1}{2} \times 9.8 t^{2} \\
&\Rightarrow t^{2}=\frac{12.1}{4.9}=2.46 \\
&\Rightarrow t=1.57 s
\end{aligned}
\)
Distance moved by the cadets \(=v \times t=1.57 \times 1.66=2.6 \mathrm{~m}\)
Therefore, the cadet who is \(2.6 \mathrm{~m}\) away from the tree will receive the berry on his uniform.

Hint

Given the distance travelled by the ball in \(0.200\) seconds \(=6 \mathrm{~m}\)
Time, \(t=0.200 \mathrm{~s}\)
Distance, \(\mathrm{s}=6 \mathrm{~m}\)
\(\mathrm{a}=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) (Acceleration due to gravity)
Using the equation of motion, we get:
\(s=v_{0} t+\frac{1}{2} a t^{2}\)
\(6=v_{0}(0.2)+\frac{1}{2} \times 10 \times 0.04\)
\(\Rightarrow v_{0}=\frac{5.8}{0.2}=29 \mathrm{~m} / \mathrm{s}\)
Let \(h\) be the height from which the ball is dropped.
We have \(v_{0}=0\) and \(v=29 \mathrm{~m} / \mathrm{s}\)

Now,
\(
\begin{aligned}
&h=\frac{v^{2}-v_{0}^{2}}{2 a} \\
&\Rightarrow h=\frac{29^{2}-0^{2}}{2 \times 10}=\frac{29 \times 29}{20}=42.05 \mathrm{~m}
\end{aligned}
\)
\(\therefore\) Total height \(=42.05+6=48.05 \mathrm{~m} \)

Hint

A ball is dropped from a height of \(5 \mathrm{~m}\) (s) above the sand level. The same ball penetrates the sand up to \(10 \mathrm{~cm}\) before coming to rest.
Initial velocity of the ball, \(v_{0}=0\) and,
\(a=g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) (Acceleration due to gravity)
Using the equation of motion, we get :
\(
\begin{aligned}
&s=v_{0} t+\frac{1}{2} a t^{2} \\
&\Rightarrow 5=0+\frac{1}{2}(9.8) t^{2} \\
&\Rightarrow t^{2}=\frac{5}{4.9}=1.02 \\
&\Rightarrow t=1.01 s
\end{aligned}
\)
Thus, the time taken by the ball to cover the distance of \(5 \mathrm{~m}\) is \(1.01\) seconds. The velocity of the ball after \(1.01 \mathrm{~s}\):

\(
\begin{aligned}
&v=v_{0}+a t \\
&\Rightarrow v=9.8 \times 1.01=9.89 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Hence, for the motion of the ball in the sand, the initial velocity \(u_{2}\) should be \(9.89 \mathrm{~m} / \mathrm{s}\) and the final velocity \(v_{2}\) should be \(0 .\)
\(
s =10 \mathrm{~cm}=0.1 \mathrm{~m}
\)
Again using the equation of motion, we get:
\(
\begin{aligned}
&a =\frac{v_{2}^{2}-u_{2}^{2}}{2 s}=\frac{0-(9.89)^{2}}{2 \times 0.1} \\
&\Rightarrow a = 489 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}\)
Hence, the sand offers the retardation of \(489 \mathrm{~m} / \mathrm{s}^{2}\).

Hint

Given, the distance between the coin and the floor of the elevator before the coin is dropped \(=6 \mathrm{ft}\)
Let \(a\) be the acceleration of the elevator.
It is given that the coin reaches the floor in 1 second. This means that the coin travels \(6 \mathrm{ft}\) distance.
The initial velocity is \(v_{0}\) for the coin and zero for the elevator.
Using the equation of motion, we get the equation for the coin:
\(
s_{c}=v_{0} t+\frac{1}{2} a^{\prime} t^{2}
\)
Here,
\(a^{\prime}=g-a\left(a^{\prime}\right.\) is the acceleration felt by the coin.)
\(g=\) Acceleration due to gravity
\(
g=9.8 \mathrm{~m} / \mathrm{s}^{2}=32 \mathrm{ft} / \mathrm{s}^{2}
\)

On substituting the values, we get:
\(
\begin{aligned}
&s_{c}=\frac{1}{2}(g-a)(1)^{2} \\
&=\frac{1}{2}(g-a)
\end{aligned}
\)
Therefore, we can write:
\(
\begin{aligned}
&6=\frac{1}{2} \times(32-a) \\
&12=32-a \\
&\therefore a=20 \mathrm{fts}^{-2}
\end{aligned}
\)
Hence, the acceleration of the elevator is \(20 \mathrm{ft} / \mathrm{s}^{2}\).

Hint

Relative velocity
\(
\mathrm{V}_{\mathrm{AB}}=\mathrm{v}_{\mathrm{A}}-\mathrm{v}_{\mathrm{B}}
\)
given
\(
\begin{aligned}
&\mathrm{V}_{\mathrm{AB}}=1500 \mathrm{Km} / \mathrm{h} \\
&\mathrm{v}_{\mathrm{A}}-\mathrm{v}_{\mathrm{B}}=1500 \\
&\mathrm{v}_{\mathrm{A}}-(-500)=1500
\end{aligned}
\)
Since the velocity of combustion products and plane are in opposite directions
\(
\mathrm{V}_{\mathrm{A}}=1000 \mathrm{~km} / \mathrm{h}
\)

Hint

Given, Velocity of car \(\mathrm{v}=126 \mathrm{~km} / \mathrm{h}=35 \mathrm{~m} / \mathrm{s}\)
Displacement \(\mathrm{s}=200 \mathrm{~m}\)
Final velocity \(\mathrm{v}=0 \mathrm{~m} / \mathrm{s}\)
Apply second kinematic equation to calculate retardation a
\(
\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}
\)
\(\Rightarrow \mathrm{a}=\frac{\mathrm{v}^{2}-\mathrm{u}^{2}}{2 \mathrm{~s}}=\frac{0-35^{2}}{2 \times 200}=-3.0625 \mathrm{~m} / \mathrm{s}^{2}\)
Apply first kinematic equation to calculate time t \(\mathrm{v}=\mathrm{u}+\mathrm{at}\)
\(\Rightarrow \mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}=\frac{0-35}{3.0625}=11.43 \mathrm{sec}\)
Hence, retardation is \(3.06 \mathrm{~m} / \mathrm{s}^{2}\) and time take to stop is \(11.4 \mathrm{sec}\)

Hint

Given that,
\(
\mathrm{u}_{\mathrm{A}}=\mathrm{u}_{\mathrm{B}}=72 \mathrm{kmh}^{-1}=72 \times \frac{5}{18}=20 \mathrm{~ms}^{-1}
\)
Using the relations, \(s=u t+\frac{1}{2}\) at \(^{2}\), we get
\(
\begin{aligned}
&\mathrm{S}_{\mathrm{B}}=\mathrm{u}_{\mathrm{B}} \mathrm{t}+\frac{1}{2} a \mathrm{t}^{2}=20 \times 50+\frac{1}{2} \times 1 \times(50)^{2} \\
&\mathrm{~S}_{\mathrm{B}}=1000+1250=2250 \mathrm{~m}
\end{aligned}
\)
Also, let \(\mathrm{S}_{\mathrm{A}}\) be the distance covered by the train \(\mathrm{A}\), then \(\mathrm{S}_{\mathrm{A}}=\mathrm{u}_{\mathrm{A}} \times \mathrm{t}\) \(=20 \times 50=1000 \mathrm{~m}\)
Original distance between the two trains
\(
\begin{aligned}
&=\mathrm{S}_{\mathrm{B}}-\mathrm{S}_{\mathrm{A}} \\
&=2250-1000=1250 \mathrm{~m}
\end{aligned}
\)

Hint

Speed of \(\mathrm{A}=36 \mathrm{~km} / \mathrm{hr}=36 \times \frac{\mathrm{5}}{18} \mathrm{~m} / \mathrm{s}=10 \mathrm{~m} / \mathrm{s}\)
Speed of \(B=\) Speed of \(C=54 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=15 \mathrm{~m} / \mathrm{s}\)
Relative speed of A w.r.t \(C=10+15=25 \mathrm{~m} / \mathrm{s}\)
Time taken by C to overtake \(A=\frac{1000}{25}\)
\(=40 \mathrm{sec}\).
Distance travelled by A all this time \(=10 \times 40=400 \mathrm{~m}\)
So, B have to cover distance of \((1000 \mathrm{~m}+400 \mathrm{~m})\)
i.e. \(1400 \mathrm{~m}\) to take over a \(\mathrm{A}\) before \(\mathrm{C}\) does. in \(40 \mathrm{sec}\).
Now putting in the formula
\(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\)
\(
1400=15 \times 40+\frac{1}{2} \times \mathrm{a} \times(40)^{2}; a = 1 \mathrm{~m} \mathrm{~s}^{-2}\)

Hint

As both the bus and cyclist are moving in the same direction their relative velocity will be
Relative velocity of bus \(=\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}} \quad-(+20)\)
Distance covered \(=\mathrm{V}_{\mathrm{T}}\)
Speed \(\times\) Time \(=\) Distance
\(
\left(\mathrm{V}_{\mathrm{B}}-20\right) \times 18=\mathrm{V}_{\mathrm{BT}}
\)
Case II – Bus moving from \(\mathrm{B}\) to \(\mathrm{A}\)
As the bus and cyclist are moving in the opposite direction their relative velocity will be –
Relative velocity of bus \(=\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}} \quad-(-20)\)
Distance covered \(=\mathrm{V}_{\mathrm{T}}\)

Speed \(\times\) Time \(=\) Distance
\(
\left(V_{B}+20\right) \times 6=V_{B T}
\)
Solving the above equations
\(
\begin{aligned}
&\left(\mathrm{V}_{\mathrm{B}}-20\right) \times 18 \\
&\mathrm{~V}_{\mathrm{B}}=40 \mathrm{~km} / \mathrm{hr}
\end{aligned}
\)
Substituting value of \(\mathrm{V}_{\mathrm{B}}\) in equation
\(
(40+20) \times 6=40 \mathrm{~T}
\)
\(\mathrm{T}=9\) minutes
The speed of the bus is \(40 \mathrm{~km} / \mathrm{h}\) and the time interval after which each bus leaves is 9 minutes.

Hint

(a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.
(b) At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).
(c) During upward motion, the sign of position is positive, the sign of velocity is negative, and the sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.
(d) Initial velocity of the ball, \(u=29.4 \mathrm{~m} / \mathrm{s}\)
The final velocity of the ball, \(v=0\) (At maximum height, the velocity of the ball becomes zero)
Acceleration, \(a=-\mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^{2}\)
From the third equation of motion, height (s) can be calculated as:

\(
\begin{aligned}
&v^{2}-u^{2}=2 g s \\
&s=\frac{v^{2}-u^{2}}{2} g \\
&=\frac{(0)^{2}-(29.4)^{2}}{2 \times(-9.8)}=44.1 m
\end{aligned}
\)
From the first equation of motion, the time of ascent \((t)\) is given as:
\(
\begin{aligned}
&v=\mathrm{u}+\mathrm{at} \\
&t=\frac{v-u}{a}=\frac{-29.4}{-9.8}=3 \mathrm{~s}
\end{aligned}
\)
Time of ascent \(=\) Time of descent
Hence, the total time taken by the ball to return to the player’s hands \(=3+3=6 \mathrm{~s}\)

Hint

Given, height from which the ball is dropped \(\mathrm{s}=90 \mathrm{~m}\)
The initial velocity of the ball, \(\mathrm{u}=0\)
Let us consider that the final velocity of the ball is \(\mathrm{V}\)
The time taken by the ball to reach the ground is given by:
\(
\mathrm{s}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^{2}
\)
\(
90=0+\frac{1}{2} \times 9.8 \mathrm{t}^{2}
\)
\(
\mathrm{t}=\sqrt{18.38}=4.29 \mathrm{~s}
\)
The final velocity of the ball as it reaches the ground is given by:
\(
\mathrm{V}=\mathrm{u}+\text { at }
\)
\(
\Rightarrow \mathrm{V}=0+9.8 \times 4.29=42.04 \mathrm{~m} / \mathrm{s}
\)
The ball losses it’s one-tenth of the velocity at collision. So, the rebound velocity of the ball,
\(
\mathrm{u}_{\mathrm{r}}=\frac{9 \mathrm{v}}{10}
\)

\(
=\frac{9 \times 42.04}{10}=37.84 \mathrm{~m} / \mathrm{s}
\)
Now, the ball rebounds and reaches a new maximum height from the ground. The time taken by the ball is given by:
\(
\begin{aligned}
&\mathrm{v}=\mathrm{u}_{\mathrm{r}}+\mathrm{at}^{\prime} \\
&0=37.84+(-9.8) \mathrm{t}^{\prime} \\
&\mathrm{t}^{\prime}=-\frac{37.84}{-9.8}=3.86 \mathrm{~s}
\end{aligned}
\)
Total time the ball takes to reach the maximum height is :
\(
\begin{aligned}
&\mathrm{T}=\mathrm{t}+\mathrm{t}^{\prime} \\
&=4.29+3.86=8.15 \mathrm{~s}
\end{aligned}
\)
Now the ball travels back to the ground in the same time as it takes to reach the maximum height i.e. \(3.86 \mathrm{~s}\)

The final velocity of the ball with which it reaches the ground will be the same as the speed with which it goes up. So, the velocity of the ball after rebound:
\(
\mathrm{v}^{\prime}=\frac{9 \times 37.84}{10}=34.05 \mathrm{~m} / \mathrm{s}
\)
Total time taken by the ball to rebound for the second time:
\(
\mathrm{T}_{\mathrm{t}}=8.15+3.86=12.01 \mathrm{~s}
\)

Hint

Distance to market \(\mathrm{s}=2.5 \mathrm{~km}=2.5 \times 10^{3}=2500 \mathrm{~m}\)
Speed with which he goes to market \(=5 \mathrm{~km} / \mathrm{h}=5 \frac{10^{3}}{3600}=\frac{25}{18} \mathrm{~m} / \mathrm{s}\)
Speed with which he comes back \(=7.5 \mathrm{~km} / \mathrm{h}=7.5 \times \frac{10^{3}}{3600}=\frac{75}{36} \mathrm{~m} / \mathrm{s}\)
(a)Average velocity is zero since his displacement is zero.
(b)
(i)Since the initial speed is \(5 \mathrm{~km} / \mathrm{s}\) and the market is \(2.5 \mathrm{~km}\) away,time taken to reach market: \(\frac{2.5}{5}=1 / 2 \mathrm{~h}=30\) minutes.
Average speed over this interval \(=5 \mathrm{~km} / \mathrm{h}\)
(ii)After 30 minutes, the man is travelling with \(7.5 \mathrm{~km} / \mathrm{h}\) speed for \(50-30=20\) minutes.The distance he covers in 20 minutes : \(7.5 \times \frac{1}{3}=2.5 \mathrm{~km}\). His average speed in o to 50 minutes: \(\mathrm{V}_{\text {avg }}=\frac{\text { distance traveled }}{\text { time }}\) \(=\frac{2.5+2.5}{(50 / 60)}=6 \mathrm{~km} / \mathrm{h}\)

(iii)In 40-30=10 minutes he travels a distance of : \(7.5 \times \frac{1}{6}=1.25 \mathrm{~km}\)
\(
\mathrm{V}_{\mathrm{avg}}=\frac{2.5+1.25}{(40 / 60)}=5.625 \mathrm{~km} / \mathrm{h}
\)

Hint

Speed of police van \(=v_{p}=30 k m h^{-1}=30 \times \frac{1000}{3600} m s^{-1}=\frac{25}{3} m s^{-1}\)
Speed of thief’s car \(=v_{t} =192 \mathrm{~km} \mathrm{~h}^{-1}\)
\(=192 \times \frac{5}{18} m s^{-1}=\frac{160}{3} m s^{-1}\)
Speed of bullet \(v_{b}\) = Speed of police van + speed with which bullet is actually fired
\(
\therefore v_{b}=\left(\frac{25}{3}+150\right) m s^{-1}=\frac{475}{3} m s^{-1}
\)
Relative velocity of bullet w.r.t theif’s car
\(
v_{b t}=v_{b}-v_{t}=\left(\frac{475}{3}-\frac{160}{3}\right) m s^{-1}=105 m s^{-1}
\)

Hint

Initial velocity of the ball, \(u=49 \mathrm{~m} / \mathrm{s}\)
Acceleration, \(a=-\mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^{2}\)
Case-1: When the lift was stationary, the boy throws the ball. Taking upward motion of the ball, the Final velocity, \(v\) of the ball becomes zero at the highest point.
From the first equation of motion, the time of ascent (t) is given as:
\(
\begin{aligned}
&v=u+a t \\
&t=\frac{v-u}{a} \\
&=\frac{-49}{-9.8}=5 \mathrm{~s}
\end{aligned}
\)
But, the time of ascent is equal to the time of descent. Hence, the total time taken by the ball to return to the boy’s hand \(=5+5=10 \mathrm{~s}\).
Case-2: The lift was moving up with a uniform velocity of \(5 \mathrm{~m} / \mathrm{s}\). In this case, the relative velocity of the ball with respect to the boy remains the same i.e., \(49 \mathrm{~m} / \mathrm{s}\). Therefore, in this case, also, the ball will return back to the boy’s hand after \(10 \mathrm{~s}\).

Hint

(a) Distance travelled by the particle = Area under the given graph.
\(
\begin{aligned}
&=(1 / 2) \times(10-0) \times(12-0) \\
&=60 \mathrm{~m}
\end{aligned}
\)
Average Speed \(=\frac{\text { Distance }}{\text { Time }}=\frac{60}{10}=6 \mathrm{~m} / \mathrm{s}\)
(b) Let \(s_{1}\) and \(s_{2}\) be the distances covered by the particle between time \(t=2 \mathrm{~s}\) to \(5 \mathrm{~s}\) and \(t=5 \mathrm{~s}\) to \(6 \mathrm{~s}\) respectively.
Total distance \((s)\) covered by the particle in time \(t=2 \mathrm{~s}\) to \(6 \mathrm{~s}\) is,
\(
s=s_{1}+s_{2} \quad \text {… (i) }
\)
For distance \(\mathbf{s}_{1}\),
Let \(\mathrm{u}_{1}\) be the velocity of the particle after \(2 \mathrm{~s}\), and
\(a_{1}\) be the acceleration of the particle in \(t=0\) to \(t=5 \mathrm{~s}\).
The particle undergoes uniform acceleration in the first 5 seconds. Therefore, using the first equation of motion \(v=u+a t\), we have
\(
12=0+a_{1} \times 5
\)
i.e., \(a_{1}=12 / 5=2.4 \mathrm{~m} / \mathrm{s}^{2}\)
Again, using the first equation of motion, we have
\(
\begin{aligned}
v &=u+a t \\
&=0+2.4 \times 2 \\
&=4.8 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Distance travelled by the particle between time \(2 \mathrm{~s}\) and \(5 \mathrm{~s}\) i.e., in \(3 \mathrm{~s}\)
\(
\begin{aligned}
s_{1} &=u_{1} t+(1 / 2) a_{1} t^{2} \\
&=4.8 \times 3+(1 / 2) \times 2.4 \times(3)^{2} \\
&=25.2 \mathrm{~m}
\end{aligned}
\)
For distance \(\mathbf{s}_{\mathbf{2}}\):
Let \(a_{2}\) be the acceleration of the particle between time \(t=5 \mathrm{~s}\) and \(t=10 \mathrm{~s}\).
Using the first equation of motion,
\(
v=u+a t
\)

\(0=12+a_{2} \times 5\)
\(
a_{2}=-12 / 5=-2.4 \mathrm{~ms}^{-2}
\)
Therefore,
Distance travelled by the particle in \(1 \mathrm{~s}\) (i.e., between \(t=5 \mathrm{~s}\) and \(t=6 \mathrm{~s}\) ),
\(
\begin{aligned}
\mathrm{s}_{2} &=u_{2} t+(1 / 2) a_{2} t^{2} \\
&=12 \times 1+(1 / 2)(-2.4) \times(1)^{2} \\
&=12-1.2=10.8 \mathrm{~m}
\end{aligned}
\)
Now solving the above equations, we have
\(
s=25.2+10.8=36 \mathrm{~m}
\)
\(\therefore\) Average speed \(=36 / 4=9 \mathrm{~m} / \mathrm{s}\)