Rest and Motion Kinematics

Question 1:

Question 2:

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.

Question 3:

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

Question 4:

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.

Question 5:

Question 6:

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 west and acceleration in the vertically downward direction. 

Question 7:

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

Question 8:

From the velocity–time graph, we have:
Distance covered by A = ($\frac{1}{2}\times \frac{t}{2}\times \frac{at}{2}+\frac{1}{2}\times \frac{t}{2}\times \frac{at}{2}+\frac{t}{2}\times \frac{at}{2})=\frac{a{t}^2}{2}$
Distance covered by B = $(\frac{1}{2}\times at\times \frac{t}{2}+at\times \frac{t}{2}+\frac{1}{2}\times \frac{at}{2}\times \frac{t}{2})=\frac{7}{8}a{t}^2$
∴ Distance covered by A < Distance covered by B

Question 9:

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

Question 10:

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.

Question 11:

(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.

Question 12:

Slope of the x–t graph gives the velocity and change in the slope gives the acceleration.
At t = t₁,
Slope = Positive ⇒ Velocity = Positive
Slope is increasing ⇒ Acceleration = Positive

At t = t₂,
Slope = Zero ⇒ Velocity = Zero
Slope is increasing ⇒ Acceleration = Negative

At t = t₃,
Slope = Negative ⇒ Velocity = Negative
Slope is increasing ⇒ Acceleration = Positive

Question 13:

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

Question 14:

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.

Question 1:

The speed of rain is less, so as we run, the direction of the relative velocity of the rain changes. But as the speed of light is very high, there is a measurable change in the relative velocity of light w.r.t. a person due to the relative motion between the sunrays and the person.

Question 2:

(b) 70 km/h towards south-west

Final velocity, $\overrightarrow{V_f}=-50 \hat{i} \mathrm{km}/\mathrm{h}$
Initial velocity, $\overrightarrow{V_i}=50 \hat{j} \mathrm{km}/\mathrm{h}$
Change in velocity, $∆\overrightarrow{V}=\overrightarrow{V_f}-\overrightarrow{V_i}$
$\left|∆V\right|=\sqrt{{50}^2+{50}^2+2\times 50\times 50\cos \left(90°\right)}=70\mathrm{km}/\mathrm{h}$
It is towards southwest, as shown in the figure.

Question 3:

(d) The particle moves at a constant velocity up to a time t₀ and then stops.

The slope of the x–t graph gives the velocity. In the graph, the slope is constant from t = 0 to t = t₀, so the velocity is constant. After t = t₀, the displacement is zero; i.e., the particle stops.

Question 4:

(d) The information is insufficient to decide the relation of x_A with x_B.

As velocity and acceleration are in opposite directions, velocity will become zero after some time (t) and the particle will return.

 $$\begin{gathered} \therefore 0=u-at \\ \Rightarrow t=\frac{u}{a} \end{gathered}$$
Because the value of acceleration is not given, we cannot say that the particle will return after/before 10 seconds.

Question 5:

(a) $v=\frac{v_1+v_2}{2}$

Velocity is uniform in both cases; that is, acceleration is zero.
We have:
$d_1=v_{1}t$ and $d_2=v_{2}t$
Total displacement, $d=d_1+d_2$
Total time, $t=t+t=2t$
$\therefore$ Average velocity, $v=\frac{d_1+d_2}{2t}=\frac{v_1+v_2}{2}$

Question 6:

(c) $\frac{2}{v}=\frac{1}{v_1}+\frac{1}{v_2}$

Velocity is uniform in both cases; that is, acceleration is zero.
$x=v_1{t}_1\Rightarrow t_1=\frac{x}{v_1}$
$x=v_2{t}_2\Rightarrow t_2=\frac{x}{v_2}$
Total displacement, $x\text{'}=2x$
Total time, $t=t_1+t_2$
$\therefore$ Average velocity, $v=\frac{x\text{'}}{t}=\frac{2v_1{v}_2}{v_1+v_2}$
$\Rightarrow \frac{2}{v}=\frac{1}{v_1}+\frac{1}{v_2}$

Question 7:

(d) g downward

Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction.

Question 8:

(c) v_A = v_B


Total energy of any particle = $\frac{1}{2}m{v}^2+mgh$
Both the particles were at the same height and thrown with equal initial velocities, so their initial total energies are equal. By the law of conservation of energy, their final energies are equal.
At the ground, they are at the same height. So, their P.E. are also equal; this implies that their K.E. should also be equal. In other words, their final velocities are equal.

Question 9:

(c) is perpendicular to the acceleration for one instant only

In projectile motion, velocity is perpendicular to acceleration only at the highest point. Here, velocity is along the horizontal direction and acceleration is along the vertically downward direction.

Question 10:

(c) Both will reach simultaneously.

Because the downward acceleration and the initial velocity in downward direction of the two bullets are the same, they will take the same time to hit the ground and for a half projectile.
Time of flight = T =$\sqrt{\frac{2h}{g}}$

Question 11:

(d) 100 m

For the same u range, $R\propto \sin \left(2\theta \right)$.
So,
$\frac{R_1}{R_2}=\frac{\sin \left(2{\theta }_1\right)}{\sin \left(2{\theta }_2\right)}$
$\Rightarrow R_2=50\times \frac{\sin \left(90\right)}{\sin \left(30\right)}=100 \mathrm{m}$

Question 12:

(d) The information is insufficient to decide the relation of R_A with R_B.

Horizontal range for the projectile, $R=\frac{u^2\sin \left(2\alpha \right)}{g}$
Information of the initial velocity is not given in the question.

Question 13:

(a) due north

If the man swims at any angle east to the north direction, although his relative speed will increase, he will have to travel a larger distance. So, he will take more time.
If the man swims at any angle west to the north direction, his relative speed will decrease. So, he will take more time.

Question 1:

(b) $v=\frac{u}{\cos \left(\theta \right)}$



Along the string, the velocity of each object is the same.
 $$\begin{gathered} 2v\cos \left(\theta \right)=2u \\ v=\frac{u}{\cos \left(\theta \right)} \end{gathered}$$

Question 2:

(a) the displacement is zero
(d) the average velocity is zero

Displacement is zero because the initial and final positions are the same.

Average velocity = $\frac{\mathrm{Displacement} }{\mathrm{Total} \mathrm{time}}=0$

Distance covered = $2\mathrm{\pi }r$ ≠ 0

Average speed =$\frac{\mathrm{Distance} \mathrm{travelled}}{\mathrm{Total} \mathrm{time} \mathrm{taken}}\ne 0$

Question 3:

(c) the acceleration of the particle is 2a
(d) at t = 2 s, the particle is at the origin

Initial velocity = ${\left|\frac{dx}{dt}\right|}_{t=0}$
$\frac{dx}{dt}=u+2a(t-2s)$
${\left|\frac{dx}{dt}\right|}_{t=0}=u-4as\ne u$

Acceleration =$\frac{d^{2}x}{d{t}^2}=2a$
At t = 2 s,
x = u(2 s − 2 s) + a (2 s − 2 s)² = 0 (origin)

Question 4:

(a) Average speed of a particle in a given time is never less than the magnitude of the average velocity.
(b) It is possible to have a situation in which $\left|\frac{d\overrightarrow{v}}{dt}\right|\ne 0 \mathrm{but} \frac{d}{dt}\left|\overrightarrow{v}\right|=0$.
(c) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval, for example, the motion of a particle on a circular track with a constant speed.
Average velocity = $\frac{\mathrm{Displacement}}{\mathrm{Total} \mathrm{time}}$
Displacement ≤ Distance
$\therefore$ Average velocity ≤ Average speed
 In uniform circular motion, speed is constant but velocity is not.
i.e., $\left|\frac{d\overrightarrow{v}}{dt}\right|\ne 0 \mathrm{but} \frac{d}{dt}=\left|\overrightarrow{v}\right|=0$ which proves case (b)

(d) In one complete circle of uniform motion, average velocity is zero. Instantaneous velocity is never zero in the interval.

Question 5:

(b) varying velocity without having varying speed
(d) nonzero acceleration without having varying speed

Velocity and acceleration are vector quantities that can be changed by changing direction only (keeping magnitude constant).

Question 6:

(a) If the velocity and acceleration have opposite signs, the object slows down.
(b) If the position and velocity have opposite signs, the particle moves towards the origin.
(d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

(a) Acceleration is given by
$$\begin{gathered} -a=\frac{dv}{dt} \\ -a<0 \\ \Rightarrow \frac{dv}{dt}<0 \\ \Rightarrow V_{\mathrm{final}}<V_{\mathrm{initial}} \end{gathered}$$

(b) If the position and velocity have opposite signs, the particle moves towards the origin. It can be explained by following figure:

(c) If the velocity is zero in a certain time interval, then the change in the velocity in that time interval will also be zero. As acceleration is rate of change of velocity, it will also be zero at an instant in that time interval.

Question 7:

(b) The acceleration at t = 0 may be zero.
(c) If the acceleration is zero from t = 0 s to t = 10 s, the speed is also zero in this interval.
(d) If the speed is zero from t = 0 s to t = 10 s, the acceleration is also zero in this interval.

(b) Acceleration will be zero only when the change in velocity is zero.

(c) Since the acceleration is zero from t = 0 s to t = 10 s, change in velocity is 0.
Velocity in this interval = Initial velocity = 0
Also,
Speed in this interval = Initial speed = 0

(d) From t = 0 s to t = 10 s, speed is zero.
Here, velocity is zero and initial velocity is zero.
So, the change in velocity is zero; i.e., acceleration is zero.

Question 8:

(a) The magnitude of the velocity of a particle is equal to its speed.

(a) Velocity being a vector quantity has magnitude as well as direction, and magnitude of velocity is called speed.

(b) Average velocity = $\frac{\mathrm{Total} \mathrm{displacement}}{\mathrm{Total} \mathrm{time} \mathrm{taken}}$

Average speed = $\frac{\mathrm{Total} \mathrm{distance} \mathrm{travelled}}{\mathrm{Total} \mathrm{time} \mathrm{taken}}$
Distance$\ge$Displacement
∴ Average speed$\ge$Average velocity
The magnitude of average velocity in an interval is not always equal to its average speed in that interval.

(c) If speed is always zero, then the distance travelled is always zero. Hence, the total distance travelled and the average speed will be zero.

(d) If the speed of a particle is never zero, the distance travelled by the particle is never zero. Hence, the average speed will not be zero.

Question 9:

(a) The particle has a constant acceleration.
(d) The average speed in the interval 0 s to 10 s is the same as the average speed in the interval 10 s to 20 s.

Explanation:
(a) The slope of the v–t graph gives the acceleration. For the given graph, the slope is constant. So, acceleration is constant.

(b) From 0 to 10 seconds, velocity is in positive direction and then in negative direction. This means that the particle turns around at t = 10 s.

(c) Area in the v–t curve gives the distance travelled by the particle.
Distance travelled in positive direction$\ne$Distance travelled in negative direction
∴ Displacement$\ne$Zero

(d) The area of the v–t graph from t = 0 s to t = 10 s is the same as that from t = 10 s to t = 20 s. So, the distance covered is the same. Hence, the average speed is the same.

Question 10:

(a) The particle has come to rest 6 times.

Explanation:
(a) The slope of the x–t graph gives the velocity. Here, 6 times the slope is zero. So, the particle has come to rest 6 times.

(b) As the slope is not maximum at t = 6 s, the maximum speed is not at t = 6 s.

(c) As the slope is not positive from t = 0 s to t = 6s, the velocity does not remain positive.

(d) Average velocity = $\frac{\mathrm{Total} \mathrm{displacement}}{\mathrm{Total} \mathrm{time} \mathrm{taken}}=\frac{x_{\mathrm{final} }-x_{\mathrm{initial}}}{t}$
For the shown time (t = 6 s), the displacement of the particle is positive. Therefore, the average velocity is positive.

Question 1:

(d) The acceleration of S₂ with respect to S₁ may be anything between zero and 8 m/s².

Explanation:
$$\begin{gathered} {\overrightarrow{a}}_{s_2{s}_1}={\overrightarrow{a}}_{s_2\mathrm{p}}+{\overrightarrow{a}}_{\mathrm{p}{s}_1} \\ \left|{\overrightarrow{a}}_{s_2{s}_1}\right|=\sqrt{4^2+4^2+32\cos \left(\theta \right)} \end{gathered}$$
$-1<\cos \left(\theta \right)<1$
$0<{\overrightarrow{a}}_{s_2{s}_1}<8$

Question 2:

 


(a) Distance travelled by the man = AB + BC + CD = 50 + 40 + 20 = 110 m

(b) AF = AB − BF = 50 − 20 = 30 m

Displacement = Final position − Initial position = AD
$$\begin{gathered} \therefore \mathrm{AD}=\sqrt{{\mathrm{AF}}^2+{\mathrm{DF}}^2}=\sqrt{{30}^2+{40}^2} \\ =50 \mathrm{m} \end{gathered}$$

In ∆AED,
$$\begin{gathered} \tan \mathrm{\theta }=\frac{\mathrm{DE}}{\mathrm{AE}}=\frac{30}{40} \\ \Rightarrow \mathrm{\theta }={\tan }^{-1} \left(\frac{3}{4}\right) \end{gathered}$$
Displacement from the house to the field = 50 m in the direction ${\tan }^{-1} \left(\frac{3}{4}\right)$ north to east.

Question 3:

Let the points be O(0,0), A(20 m, 0) and B(−20 m, 0).
(i) Distance travelled = OA + AB = 20 + 40 = 60 m
(ii) Displacement = OB = 20 m (in the negative direction)

Question 4:

(a) Average speed of the plane, s_avg$=\left(\frac{\mathrm{Total} \mathrm{distance} \mathrm{travelled}}{\mathrm{Total} \mathrm{time} \mathrm{taken}}\right)$$=\frac{260}{0.5}=520 \mathrm{km}/\mathrm{h}$

(b) Average speed of the bus, _Savg$=\frac{320}{8}=40 \mathrm{km}/\mathrm{h}$

(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 km
$\therefore {\overrightarrow{V}}_{avg}=\frac{260}{8}=32.5 \mathrm{km}/\mathrm{h}$ (Patna to Ranchi)

Question 5:

(a) Total distance covered = 12416 − 12352 = 64 km
$\therefore \mathrm{Average} \mathrm{speed}=\frac{64}{2}=32 \mathrm{km}/\mathrm{h}$

(b) Because he returns to his house, the displacement is zero. So, the average velocity is zero.

Question 6:

Initial velocity, u = 0         
Final velocity, v = 18 km/h = 5 m/s
Time interval, t = 2 s
Using $v=u+at$, we get:
$\mathrm{Acceleration}=a_{av}=\frac{\left(v-u\right)}{2}=\frac{5}{2}=2.5 \mathrm{m}/{\mathrm{s}}^2$

Question 7:

Slope of the v–t graph gives the acceleration.
Slope = $\tan \left(\theta \right)=\frac{20}{8}=2.5 \mathrm{m}/{\mathrm{s}}^2$

Area in the v–t graph gives the distance travelled.
Distance travelled = Area of ∆OAB      
                              = $\frac{1}{2}\times 20\times 8=80 \mathrm{m}$

Question 8:

In the first 10 seconds,
${\mathrm{S}}_1=ut+\frac{1}{2}a{t}^2$$=0+\frac{1}{2}5\times {10}^2=250 \mathrm{ft}$

At t = 10 s,
v = u + at = 0 + 5 × 10 = 50 ft/s
∴ From 10 to 20 seconds (∆t = 20 − 10 = 10 s), the particle moves with a uniform velocity of 50 ft/s.
Distance covered from t = 10 s to t = 20 s:
S₂ = 50 × 10 = 500 ft

Between 20 s to 30 s, acceleration is constant, i.e., −5 ft/s².
At 20 s, velocity is 50 ft/s.
t = 30 − 20 = 10 s
$$\begin{gathered} {\mathrm{S}}_3=ut+\frac{1}{2}a{t}^2=50\times 10+\frac{1}{2}\left(-5\right){10}^2 \\ \Rightarrow {\mathrm{S}}_3=500-250 = 250 \mathrm{ft} \end{gathered}$$

Total distance travelled is 30 s:
S₁ + S₂ + S₃
= 250 + 500 + 250
= 1000 ft

The position–time graph:

Question 9:

(a) Slope of the v–t graph gives the acceleration.
Acceleration $=\frac{8-2}{10}=\frac{6}{10}=0.6 \mathrm{m}/{\mathrm{s}}^2$

(b) Area in the v–t graph gives the distance travelled.
Distance travelled = Area of ∆ABC + Area of rectangle OABD
                              =$\frac{1}{2}\times 6\times 10+10\times 2=50 \mathrm{m}$

(c) Displacement is the same as the distance travelled.
Displacement = 50 m

Question 10:

(a) Displacement from t = 0 s to t = 10 s:
x = 100 m
Time = 10 s
Average velocity from 0 to 10 seconds,
$v_{avg}=\frac{x}{t}=\frac{100}{10}=10 \mathrm{m}/\mathrm{s}$

(b) Slope of the x–t graph gives the velocity.
At 2.5 s, slope = $\frac{50-0}{2.5-0}=20$.
⇒ v_inst = 20 m/s
At 5 s, v_inst = 0.
At 8 s, v_inst = 20 m/s.
At 12 s, v_inst = −20 m/s.

Question 11:

Area shown in the v–t graph gives the distance travelled.
∴ Distance travelled in the first 40 seconds = Area of ∆OAB + Area of ∆BCD
 $=\frac{1}{2}\times 5\times 20+\frac{1}{2}\times 5\times 20=100 \mathrm{m}$

 

As the displacement is zero, the average velocity is zero.

Question 12:

Consider point B₁ at 12 s.



At t = 0 s, S = 20 m and at t = 12 s, S = 20 m.
For the time interval 0–12, change in the displacement is zero.
$\mathrm{Average} \mathrm{velocity}=\frac{\mathrm{Displacement}}{\mathrm{Time}}=0$
Hence, the time is 12 seconds.

Question 13:

At position B , the instantaneous velocity of the particle has the direction along $\overrightarrow{\mathrm{BC}}$.


Average velocity between A and B, ${\mathrm{V}}_{av}=\frac{\mathrm{Displacement}}{\mathrm{Time}}$$=\left(\frac{\overrightarrow{\mathrm{AB}}}{t}\right)$
We can see that $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{BC}}$ are in the same direction. The point is B (5 m, 3 m).

Question 14:

Given:
Velocity, u = 4.0 m/s
Acceleration, a = 1.2 m/s²
Time, t = 5.0 s
Distance travelled:
$s=ut+\frac{1}{2}a{t}^2$
 $$\begin{gathered} =4.0\times 5+\frac{1}{2}\times 1.2\times 5^2 \\ =20.0+\frac{1}{2}\times 1.2\times 25 \\ =35 \mathrm{m} \end{gathered}$$                          

Question 15:

Initial velocity, u = 43.2 km/h = 12 m/s
Final velocity, v = 0
Acceleration, a = −6 m/s²
From $v^2=u^2+2as$, we get:
Distance, $s=\frac{v^2-u^2}{2\left(a\right)}$
$\Rightarrow s=\frac{0-{\left(12\right)}^2}{2\left(-6\right)}=\frac{{\left(12\right)}^2}{12}=12 \mathrm{m}$

Question 16:

Initial velocity, u = 0
Acceleration, a = 2 m/s²
Let the final velocity be v before the brakes are applied.
Now,           
t = 30 s
v = u + at
v = 0 + 2 × 30
 ⇒ v = 60 m/s

(a) ${\mathrm{s}}_1=ut +\frac{1}{2}a{t}^2$
$\Rightarrow s_1=\frac{1}{2}\times 2\times {\left(30\right)}^2=900 \mathrm{m}$
When the brakes are applied:
u' = 60 m/s
v' = 0
t = 1 min = 60 s
Acceleration:
$$\begin{gathered} a\text{'}=\frac{\left(v-u\right)}{t}=\frac{\left(0-60\right)}{60}=-1 \mathrm{m}/{\mathrm{s}}^2 \\ {s}_2=\frac{v^2-u^2}{2a\text{'}}=\frac{0^2-{60}^2}{2\left(-1\right)}=1800 \mathrm{m} \end{gathered}$$
s = s₁ + s₂ = 1800 + 900 = 2700 m
⇒ s = 2.7 km

(b) Maximum speed attained by the train, v = 60 m/s

(c) Half the maximum speed$=\frac{60}{2}=30 \mathrm{m}/\mathrm{s}$
When the train is accelerating with an acceleration of 2 m/s²:

Distance, $\mathrm{s}=\frac{v^2-u^2}{2a\text{'}}$$=\frac{{30}^2-0^2}{2\times 2}$
⇒ s = 225 m

When the train is decelerating with an acceleration of $-$1 m/s²:
Distance, $\mathrm{s}=\frac{v^2-u^2}{2a\text{'}}$$=\frac{{30}^2-{60}^2}{2\left(-1\right)}$
⇒ s = 1350 m
Position from the starting point = 900 + 1350 = 2250
                                                   = 2.25 km

Question 17:

Initial velocity, u = 16 m/s
Final velocity, v = 0
Distance, s = 0.4 m
Acceleration, $a=\frac{v^2-u^2}{2\mathrm{s}}$
$\Rightarrow a=\frac{0-{16}^2}{2\times 0.4}=-320 \mathrm{m}/{\mathrm{s}}^2$

Time, $t=\frac{v-u}{a}$

$\Rightarrow t=\frac{0-16}{-320}=0.05 \mathrm{s}$

Question 18:

Initial velocity, u = 350 m/s
Final velocity, v = 0
Distance travelled by the bullet before coming to rest, s = 5 m
$\mathrm{Acceleration}, a=\frac{v^2-u^2}{2s}$
$\Rightarrow a=\frac{0-{\left(350\right)}^2}{2\times 0.05}=-12.2\times {10}^5 \mathrm{m}/{\mathrm{s}}^2$
∴ Deceleration = 12.2 × 10⁵ m/s²

Question 19:

Initial velocity of the particle, u = 0
Final velocity of the particle, v = 18 km/h = 5 m/s
Time, t = 5 s
Acceleration, a = (v − u)/t
⇒ a = (5 − 0)/5 = 1 m/s²
Distance, $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=\frac{1}{2}\times 1\times \left(5\times 5\right)=12.5 \mathrm{m}$
(a) Average velocity, $v_{avg}=\frac{\mathrm{Total} \mathrm{displacement}}{\mathrm{Total} \mathrm{time} \mathrm{taken}}$
$\Rightarrow v_{avg}=\frac{\left(12.5\right)}{5}=2.5 \mathrm{m}/\mathrm{s}$

(b) Distance travelled, s = 12.5 m

Question 20:

In the reaction time, the car moves with a constant speed of 54 km/h, i.e., 15 m/s.
Distance travelled in this time, s₁ = 15 × 0.2 = 3 m
When the brakes are applied:
Initial velocity of the car, u = 15 m/s
Final velocity of the car, v = 0
Acceleration, a = −6 m/s²
Distance, $s_2=\frac{v^2-u^2}{2a}$ = 18.75 m
Total distance, s = s₁ + s₂
⇒ s = 3 + 18.75 = 21.75 m
⇒ s$\approx$22 m

Question 21:

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 m/s²
For driver X:
Initial velocity, u = 54 km/h = 15 m/s
Final velocity, v = 0
Braking distance, $a=\frac{0^2-{15}^2}{2\left(-6\right)}\approx 19 \mathrm{m}$
Distance travelled in the reaction time = 15 × 0.20 = 3 m
Total stopping distance, b = 19 + 3 = 22 m

For driver Y:
Initial velocity, u = 72 km/h = 20 m/s
Final velocity, v = 0
Braking distance, $c=\frac{0^2-{20}^2}{2\left(-6\right)}\approx 33 \mathrm{m}$
Distance travelled in the reaction time = 20 × 0.30 = 6 m
Total stopping distance, d = 33 + 6 = 39 m

Case B:
Deceleration = 6.0 m/s²
Now, we have:
e = 15 m
f = 18 m
g = 27 m
h = 33 m
 

Question 22:

Velocity of the police jeep, v_p = 90 km/h = 25 m/s
Velocity of the culprit riding the motorbike, v_c = 72 km/h = 20 m/s



In 10 seconds, the culprit reaches point B from point A.
Distance covered by the culprit:
s = v_ct = 20 × 10 = 200 m

At time t = 10 s, the police jeep is 200 m behind the culprit.
Relative velocity between the police jeep and the culprit:
25 − 20 = 5 m/s
$\mathrm{Time}=\frac{\mathrm{Distance} \mathrm{to} \mathrm{be} \mathrm{covered}}{\mathrm{Relative} \mathrm{velocity}}=\frac{200}{5}= 40 \mathrm{s}$
In 40 seconds, the police jeep moves from point A to a distance s' to catch the culprit.
Here,
s' = v_pt = 25 × 40
⇒ s' = 1000 m = 1.0 km
Thus, the jeep will catch up with the bike 1.0 km away from the turning.

Question 23:

Velocity of the first car, v₁= 60 km/h = 16.7 m/s
Velocity of the second car, v₂ = 42 km/h = 11.7 m/s
Relative velocity between the cars = (16.7 − 11.7) = 5 m/s
Distance travelled by the first car w.r.t. the second car = 5 + 5 = 10 m
Time, $t=\frac{s}{v}=\frac{10}{5} \mathrm{s}$ = 2 s
Distance covered by the first car w.r.t. the ground in 2 s = 16.7 × 2 = 33.4 m
The first car also covers a distance equal to its own length = 5 m
∴ Total road distance used for the overtake = 33.4 + 5 ≈ 38 m

Question 24:

Given:
Initial speed of the ball, u = 50 m/s
Acceleration, a = −10 m/s²
At the highest point, velocity v of the ball is 0.

(a) $s=\frac{v^2-u^2}{2a}$
   ⇒ $s=\frac{\left(0^2 - {50}^2\right)}{2\times \left(-10\right)}=125 \mathrm{m}$
Maximum height = 125 m

(b) $t=\frac{\left(v-u\right)}{a}=\frac{\left(0-50\right)}{-10}=5 \mathrm{s}$

(c) $s=\frac{125}{2}=62.5 \mathrm{m}$
From v² − u² = 2as, we have:
$$\begin{gathered} v=\sqrt{\left(u^2+2as\right)} \\ \Rightarrow v=\sqrt{{\left(50\right)}^2+2\times \left(-10\right)\times \left(62.5\right)}=\sqrt{\left(2500-1250\right)} \\ \Rightarrow v=\sqrt{1250}\approx 35 \mathrm{m}/\mathrm{s} \end{gathered}$$

Question 25:

Given:
Height of the balloon from the ground, s = 60 m
Balloon is moving upwards with velocity 7 m/s.
The balloon and the ball are moving upwards with the same speed.
When the ball is dropped, its initial velocity (u) is −7 m/s.
Acceleration due to gravity, a = g = 9.8 m/s²
Using the equation of motion, we have:
$s=ut+\frac{1}{2}a{t}^2$
$$\begin{gathered} 60=-7t+\frac{1}{2}\times 9.8\times t^2 \\ \Rightarrow 60=-7t+4.9\times t^2 \\ \Rightarrow 4.9t^2-7t-60=0 \\ t=\frac{7\pm \sqrt{49-4\times 4.9 \times \left(-60\right)}}{2\times 4.9} \\ t=\frac{7\pm \sqrt{\left(49+1176\right)}}{9.8} \\ t=\frac{7\pm 35}{9.8} \end{gathered}$$

We will ignore the −ve sign in the above equation because time can never be negative.

$\therefore t=\frac{7+35}{9.8}=4.28 \mathrm{s}$
Time taken by the ball to reach the ground = 4.3 s

Question 26:

Given:
Initial velocity with which the stone is thrown vertically upwards, u = 28 m/s
When the stone reaches the ground, its final velocity (v) is 0.
Also,
a = g = −9.8 m/s²  (Acceleration due to gravity)
(a) Maximum height can be found using the equation of motion.
Thus, we have:
v² − u² = 2as
$\mathrm{s}=\frac{v^2-u^2}{2a}$
On putting respective values, we get:
$s=\frac{0^2-{28}^2}{2\left(-9.8\right)}=40 \mathrm{m}$

(b) Total time taken by the stone to reach the maximum height:
$t=\frac{\left(v-u\right)}{a}$
$\Rightarrow t=\frac{\left(0-28\right)}{-9.8}=2.85 \mathrm{s}$
As per the question, we need to find the velocity of the stone one second before it reaches the maximum height.
t' = 2.85 − 1 = 1.85 s
Again, using the equation of motion, we get: 
v' = u + at' = 28 − 9.8 × 1.85
⇒ v' = 28 − 18.13 = 9.87 m/s
Hence, the velocity is 9.87 m/s

(c) No answer of part (b) will not change, as after one second, the velocity becomes zero for any initial velocity and acceleration (a = − 9.8 m/s²) remains the same. For any initial velocity more than 28 m/s, only the maximum height increases.

Question 27:

A person is releasing balls from a tall building at regular intervals of one second.
It means for each ball, the initial velocity u is 0.
Acceleration due to gravity, a = g = 9.8 m/s²
When the 6^th ball is dropped, the 5^th ball moves for 1 second, the 4^th ball moves for 2 seconds and the 3^rdball moves for 3 seconds.
Position of the 3^rd ball after t = 3 s:
Using the equation of motion, we get:
${\mathrm{s}}_3=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s_3 =0+\frac{1}{2}\times 9.8\times 3^2=44.1 \mathrm{m}$
(from the top of the building)

Position of the 4^th ball after t = 2 s:
 ${\mathrm{s}}_4=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s_4 =0+\frac{1}{2}\times 9.8\times 2^2=19.6 \mathrm{m}$
(from the top of the building)

Position of the 5^th ball after t = 1 s:
 ${\mathrm{s}}_5=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s_5 =0+\frac{1}{2}\times 9.8\times 1^2=4.9 \mathrm{m}$
(from the top of the building)

Question 28:

Given:
Height of the building = 11.8 m
Distance of the young man from the building = 7 m
The kid should be caught over 1.8 m from ground.
As the kid is slipping, his initial velocity u is 0.
Acceleration, a = 9.8 m/s²
Let s be the distance before which the kid has to be caught = 11.8 − 1.8 = 10 m
Using the equation of motion, we get:
$s=ut+\frac{1}{2}a{t}^2$
$$\begin{gathered} \Rightarrow 10=0+\frac{1}{2}\times 9.8\times t^2 \\ \Rightarrow t^2=\frac{10}{4.9}=2.04 \\ \Rightarrow t=1.42 \mathrm{s} \end{gathered}$$
This is the time in which the man should reach the bottom of the building to catch the kid.
Velocity with which the man should run:
$\frac{\mathrm{s}}{t}=\frac{7}{1.42}=4.9 \mathrm{m}/\mathrm{s}$

Question 29:

Speed of the NCC cadets = 6 km/h = 1.66 m/s
Distance of the bird from the ground, s = 12.1 m
Initial velocity of the berry dropped by the bird, u = 0
Acceleration due to gravity, a = g = 9.8 m/s²
Using the equation of motion, we can find the time taken t by the berry to reach the ground.
Thus, we have:
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 12.1=0+\frac{1}{2}\times 9.8 t^2 \\ \Rightarrow t^2=\frac{12.1}{4.9}=2.46 \\ \Rightarrow t=1.57 \mathrm{s} \end{gathered}$$

Distance moved by the cadets = v × t = 1.57 × 1.66 = 2.6 m
Therefore, the cadet who is 2.6 m away from tree will receive the berry on his uniform.

Question 30:

Given:
Distance travelled by the ball in 0.200 seconds = 6 m
Let:
Time, t = 0.200 s
Distance, s = 6 m
a = g = 10 m/s² (Acceleration due to gravity)
Using the equation of motion, we get:
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 6=u\left(0.2\right)+\frac{1}{2}\times 10\times 0.04 \\ \Rightarrow u=\frac{5.8}{0.2}=29 \mathrm{m}/\mathrm{s} \end{gathered}$$
Let h be the height from which the ball is dropped.
We have:
u = 0 and v = 29 m/s
Now,
$h=\frac{v^2-u^2}{2a}$
$\Rightarrow h=\frac{{29}^2-0^2}{2\times 10}=\frac{29\times 29}{20}=42.05 \mathrm{m}$

∴ Total height = 42.05 + 6 = 48.05 m ≈ 48 m

Question 31:

A ball is dropped from a height of 5 m (s) above the sand level.
The same ball penetrates the sand up to 10 cm (s_s) before coming to rest.
Initial velocity of the ball, u = 0
And,
a = g = 9.8 m/s² (Acceleration due to gravity)
Using the equation of motion, we get:
$\mathrm{s}=ut+\frac{1}{2}a{t}^2$
$$\begin{gathered} \Rightarrow 5=0+\frac{1}{2}\left(9.8\right)t^2 \\ \Rightarrow t^2=\frac{5}{4.9}=1.02 \\ \Rightarrow t=1.01 \mathrm{s} \end{gathered}$$

Thus, the time taken by the ball to cover the distance of 5 m is 1.01 seconds.
Velocity of the ball after 1.01 s:
 v = u + at
 ⇒ v = 9.8 × 1.01 = 9.89 m/s     
Hence, for the motion of the ball in the sand, the initial velocity u₂should be 9.89 m/s and the final velocity v₂ should be 0.
s_s = 10 cm = 0.1 m
Again using the equation of motion, we get:
$$\begin{gathered} a_{\mathrm{s}}=\frac{v_2^2-u_2^2}{2{\mathrm{s}}_s}=\frac{0-{\left(9.89\right)}^2}{2\times 0.1} \\ \Rightarrow a_{\mathrm{s}}\approx -490 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Hence, the sand offers the retardation of 490 m/s².

Question 32:

Given:
Distance between the coin and the floor of the elevator before the coin is dropped = 6 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 ft distance.
The initial velocity is u for the coin and  zero for the elevator.
Using the equation of motion, we get:
Equation for the coin:
$s_{\mathrm{c}}=ut+\frac{1}{2}a\text{'}{t}^2$
Here,
a' = g − a ( a' is the acceleration felt by the coin.)
g = Acceleration due to gravity
g = 9.8 m/s² = 32 ft/s²
On substituting the values, we get:
$$\begin{gathered} s_{\mathrm{c}}=\frac{1}{2}\left(g-a\right){\left(1\right)}^2 \\ =\frac{1}{2}\left(g-a\right) \end{gathered}$$
Therefore, we can write:
$$\begin{gathered} 6=\frac{1}{2}\times \left(32-a\right) \\ 12=32-a \\ \therefore a=20 {\mathrm{fts}}^{-2} \end{gathered}$$
Hence, the acceleration of the elevator is 20 ft/s².