Hc Verma I Solutions for Class 11 Science Physics Chapter 3 Rest And Motion: Kinematics are provided here with simple step-by-step explanations. These solutions for Rest And Motion: Kinematics are extremely popular among Class 11 Science students for Physics Rest And Motion: Kinematics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma I Book of Class 11 Science Physics Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma I Solutions. All Hc Verma I Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

#### Page No 48:

#### 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.

#### Page No 48:

#### Answer:

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.

#### Page No 48:

#### 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.

#### Page No 48:

#### Answer:

#### Page No 48:

#### 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 west and acceleration in the vertically downward direction.

#### Page No 48:

#### 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.

#### Page No 48:

#### Answer:

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

#### Page No 48:

#### Answer:

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

#### Page No 48:

#### 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.

#### Page No 48:

#### 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.

#### Page No 48:

#### Answer:

Slope of the *x*–*t* graph gives the velocity and change in the slope gives the acceleration.

At *t = **t*_{1},

Slope = Positive ⇒ Velocity = Positive

Slope is increasing ⇒ Acceleration = Positive

At *t = **t*_{2},

Slope = Zero ⇒ Velocity = Zero

Slope is increasing ⇒ Acceleration = Negative

At *t = **t*_{3},

Slope = Negative ⇒ Velocity = Negative

Slope is increasing ⇒ Acceleration = Positive

#### Page No 49:

#### Answer:

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

#### Page No 49:

#### 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.

#### Page No 49:

#### Answer:

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.

#### Page No 49:

#### Answer:

(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, $\u2206\overrightarrow{V}=\overrightarrow{{V}_{f}}-\overrightarrow{{V}_{i}}$

$\left|\u2206V\right|=\sqrt{{50}^{2}+{50}^{2}+2\times 50\times 50\mathrm{cos}\left(90\xb0\right)}=70\mathrm{km}/\mathrm{h}$

It is towards southwest, as shown in the figure.

#### Page No 49:

#### Answer:

(d) The particle moves at a constant velocity up to a time *t*_{0} 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 = **t*_{0}, so the velocity is constant. After *t = **t*_{0}, the displacement is zero; i.e., the particle stops.

#### Page No 49:

#### Answer:

(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.

$\therefore 0=u-at\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{u}{a}$

Because the value of acceleration is not given, we cannot say that the particle will return after/before 10 seconds.

#### Page No 49:

#### Answer:

(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, $v=\frac{{d}_{1}+{d}_{2}}{2t}=\frac{{v}_{1}+{v}_{2}}{2}$

#### Page No 49:

#### Answer:

(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, $v=\frac{x\text{'}}{t}=\frac{2{v}_{1}{v}_{2}}{{v}_{1}+{v}_{2}}$

$\Rightarrow \frac{2}{v}=\frac{1}{{v}_{1}}+\frac{1}{{v}_{2}}$

#### Page No 49:

#### Answer:

(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.

#### Page No 49:

#### Answer:

(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.

#### Page No 49:

#### Answer:

(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.

#### Page No 49:

#### Answer:

(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 =$\sqrt{\frac{2h}{g}}$

#### Page No 49:

#### Answer:

(d) 100 m

For the same *u *range, $R\propto \mathrm{sin}\left(2\theta \right)$.

So,

$\frac{{R}_{1}}{{R}_{2}}=\frac{\mathrm{sin}\left(2{\theta}_{1}\right)}{\mathrm{sin}\left(2{\theta}_{2}\right)}$

$\Rightarrow {R}_{2}=50\times \frac{\mathrm{sin}\left(90\right)}{\mathrm{sin}\left(30\right)}=100\mathrm{m}$

#### Page No 49:

#### Answer:

(d) The information is insufficient to decide the relation of R_{A} with R_{B}.

Horizontal range for the projectile, $R=\frac{{u}^{2}\mathrm{sin}\left(2\alpha \right)}{g}$

Information of the initial velocity is not given in the question.

#### Page No 49:

#### Answer:

(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.

#### Page No 50:

#### Answer:

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

Along the string, the velocity of each object is the same.

$2v\mathrm{cos}\left(\theta \right)=2u\phantom{\rule{0ex}{0ex}}v=\frac{u}{\mathrm{cos}\left(\theta \right)}$

#### Page No 50:

#### Answer:

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

#### Page No 50:

#### Answer:

(c) the acceleration of the particle is 2*a*

(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)

#### Page No 50:

#### Answer:

(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.

#### Page No 50:

#### Answer:

(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).

#### Page No 50:

#### Answer:

(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

$-a=\frac{dv}{dt}\phantom{\rule{0ex}{0ex}}-a<0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dv}{dt}<0\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{\mathrm{final}}<{V}_{\mathrm{initial}}$

(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.

#### Page No 50:

#### Answer:

(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.

#### Page No 50:

#### Answer:

(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.

#### Page No 50:

#### Answer:

(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.

#### Page No 50:

#### Answer:

(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 = $\frac{\mathrm{Total}\mathrm{displacement}}{\mathrm{Total}\mathrm{time}\mathrm{taken}}=\frac{{x}_{\mathrm{final}}-{x}_{\mathrm{initial}}}{t}\phantom{\rule{0ex}{0ex}}$

For the shown time (*t* = 6 s), the displacement of the particle is positive. Therefore, the average velocity is positive.

#### Page No 51:

#### Answer:

(d) The acceleration of S_{2} with respect to S_{1} may be anything between zero and 8 m/s^{2}.

Explanation:

${\overrightarrow{a}}_{{s}_{2}{s}_{1}}={\overrightarrow{a}}_{{s}_{2}\mathrm{p}}+{\overrightarrow{a}}_{\mathrm{p}{s}_{1}}\phantom{\rule{0ex}{0ex}}\left|{\overrightarrow{a}}_{{s}_{2}{s}_{1}}\right|=\sqrt{{4}^{2}+{4}^{2}+32\mathrm{cos}\left(\theta \right)}$

$-1<\mathrm{cos}\left(\theta \right)<1\phantom{\rule{0ex}{0ex}}$

$0<{\overrightarrow{a}}_{{s}_{2}{s}_{1}}<8$

#### Page No 51:

#### Answer:

(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

$\therefore \mathrm{AD}=\sqrt{{\mathrm{AF}}^{2}+{\mathrm{DF}}^{2}}=\sqrt{{30}^{2}+{40}^{2}}\phantom{\rule{0ex}{0ex}}=50\mathrm{m}$

In ∆AED,

$\mathrm{tan}\mathrm{\theta}=\frac{\mathrm{DE}}{\mathrm{AE}}=\frac{30}{40}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\theta}={\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)$

Displacement from the house to the field = 50 m in the direction ${\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)$ north to east.

#### Page No 51:

#### Answer:

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)

#### Page No 51:

#### Answer:

(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)

#### Page No 51:

#### Answer:

(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.

#### Page No 51:

#### Answer:

Initial velocity, *u* = 0

Final velocity, *v* = 18 km/h = 5 m/s

Time interval, *t* = 2 s

Using $v=u+at$, we get:

$\mathrm{Acceleration}={a}_{av}=\frac{\left(v-u\right)}{2}=\frac{5}{2}=2.5\mathrm{m}/{\mathrm{s}}^{2}$

#### Page No 51:

#### Answer:

Slope of the *v–t *graph gives the acceleration.

Slope = $\mathrm{tan}\left(\theta \right)=\frac{20}{8}=2.5\mathrm{m}/{\mathrm{s}}^{2}$

Area in the *v–t *graph gives the distance travelled.

Distance travelled = Area of ∆OAB

= $\frac{1}{2}\times 20\times 8=80\mathrm{m}$

#### Page No 51:

#### Answer:

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:

S_{2} = 50 × 10 = 500 ft

Between 20 s to 30 s, acceleration is constant, i.e., −5 ft/s^{2}.

At 20 s, velocity is 50 ft/s.

*t* = 30 − 20 = 10 s

${\mathrm{S}}_{3}=ut+\frac{1}{2}a{t}^{2}=50\times 10+\frac{1}{2}\left(-5\right){10}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{S}}_{3}=500-250=250\mathrm{ft}$

Total distance travelled is 30 s:

S_{1} + S_{2} + S_{3}

= 250 + 500 + 250

= 1000 ft

The position*–*time graph:

#### Page No 51:

#### Answer:

(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

#### Page No 51:

#### Answer:

(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.

#### Page No 51:

#### Answer:

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.

#### Page No 52:

#### Answer:

Consider point B_{1} 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.

#### Page No 52:

#### Answer:

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).

#### Page No 52:

#### Answer:

Given:

Velocity, *u* = 4.0 m/s

Acceleration, *a *= 1.2 m/s^{2}

Time, *t = *5.0 s

Distance travelled:

$s=ut+\frac{1}{2}a{t}^{2}$

$=4.0\times 5+\frac{1}{2}\times 1.2\times {5}^{2}\phantom{\rule{0ex}{0ex}}=20.0+\frac{1}{2}\times 1.2\times 25\phantom{\rule{0ex}{0ex}}=35\mathrm{m}$

#### Page No 52:

#### Answer:

Initial velocity, *u* = 43.2 km/h = 12 m/s

Final velocity, *v* = 0

Acceleration, *a* = −6 m/s^{2}

From ${v}^{2}={u}^{2}+2as$, we get:

Distance, $s=\frac{{v}^{2}-{u}^{2}}{2\left(a\right)}$

$\Rightarrow s=\frac{0-{\left(12\right)}^{2}}{2\left(-6\right)}=\frac{{\left(12\right)}^{2}}{12}=12\mathrm{m}$

#### Page No 52:

#### Answer:

Initial velocity, *u* = 0

Acceleration, *a* = 2 m/s^{2}

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:

$a\text{'}=\frac{\left(v-u\right)}{t}=\frac{\left(0-60\right)}{60}=-1\mathrm{m}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}{s}_{2}=\frac{{v}^{2}-{u}^{2}}{2a\text{'}}=\frac{{0}^{2}-{60}^{2}}{2\left(-1\right)}=1800\mathrm{m}$

*s *= *s*_{1} +* **s*_{2} = 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/s^{2}:

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/s^{2}:

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

#### Page No 52:

#### Answer:

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, $t=\frac{v-u}{a}\phantom{\rule{0ex}{0ex}}$

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

#### Page No 52:

#### Answer:

Initial velocity, *u *= 350 m/s

Final velocity, *v* = 0

Distance travelled by the bullet before coming to rest, *s* = 5 m

$\mathrm{Acceleration},a=\frac{{v}^{2}-{u}^{2}}{2s}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow a=\frac{0-{\left(350\right)}^{2}}{2\times 0.05}=-12.2\times {10}^{5}\mathrm{m}/{\mathrm{s}}^{2}$

∴ Deceleration = 12.2 × 10^{5} m/s^{2}

#### Page No 52:

#### Answer:

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/s^{2}

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

#### Page No 52:

#### Answer:

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, *s*_{1} = 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/s^{2}

Distance, ${s}_{2}=\frac{{v}^{2}-{u}^{2}}{2a}$ = 18.75 m

Total distance, *s* = *s*_{1} +* **s*_{2}

*⇒ s* = 3 + 18.75 = 21.75 m

*⇒ **s*$\approx $22 m

#### Page No 52:

#### 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 m/s^{2}

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/s^{2}

Now, we have:

*e = *15 m

*f* = 18 m

*g = *27 m

*h* = 33 m

Car Model |
Driver XReaction Time = 0.20 s |
Driver YReaction Time = 0.30 s |

A (deceleration on hard braking = 6.0 m/s^{2}) |
Speed = 54 km/h Braking distance , a = 19 mTotal stopping distance, b = 22 m |
Speed = 72 km/h Braking distance, c = 33 mTotal stopping distance, d = 39 m |

B (deceleration on hard braking = 7.5 m/s^{2}) |
Speed = 54 km/h Breaking distance, e = 15 mTotal stopping distance, f = 18 m |
Speed = 72 km/h Braking distance, g = 27 mTotal stopping distance, h = 33 m |

#### Page No 52:

#### Answer:

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 _{c}t* = 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*= 25 × 40

_{p}t⇒

*s*' = 1000 m = 1.0 km

Thus, the jeep will catch up with the bike 1.0 km away from the turning.

#### Page No 52:

#### Answer:

Velocity of the first car, *v*_{1}_{ }= 60 km/h = 16.7 m/s

Velocity of the second car, *v*_{2} = 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

#### Page No 52:

#### Answer:

Given:

Initial speed of the ball*, u* = 50 m/s

Acceleration*, a *= −10 m/s^{2}

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 *v*^{2} − *u*^{2} = 2*as*, we have:

$v=\sqrt{\left({u}^{2}+2as\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{{\left(50\right)}^{2}+2\times \left(-10\right)\times \left(62.5\right)}=\sqrt{\left(2500-1250\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{1250}\approx 35\mathrm{m}/\mathrm{s}$

#### Page No 52:

#### Answer:

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/s^{2}

Using the equation of motion, we have:

$s=ut+\frac{1}{2}a{t}^{2}$

$60=-7t+\frac{1}{2}\times 9.8\times {t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 60=-7t+4.9\times {t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4.9{t}^{2}-7t-60=0\phantom{\rule{0ex}{0ex}}t=\frac{7\pm \sqrt{49-4\times 4.9\times \left(-60\right)}}{2\times 4.9}\phantom{\rule{0ex}{0ex}}t=\frac{7\pm \sqrt{\left(49+1176\right)}}{9.8}\phantom{\rule{0ex}{0ex}}t=\frac{7\pm 35}{9.8}$

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

#### Page No 52:

#### Answer:

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/s^{2} (Acceleration due to gravity)

(a) Maximum height can be found using the equation of motion.

Thus, we have:

*v*^{2} − *u*^{2} = 2*as*

$\mathrm{s}=\frac{{v}^{2}-{u}^{2}}{2a}$

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/s^{2}) remains the same. For any initial velocity more than 28 m/s, only the maximum height increases.

#### Page No 52:

#### Answer:

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/s^{2}

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^{rd}^{ }ball 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)

#### Page No 52:

#### Answer:

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/s^{2}

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:

$s=ut+\frac{1}{2}a{t}^{2}$

$\Rightarrow 10=0+\frac{1}{2}\times 9.8\times {t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}=\frac{10}{4.9}=2.04\phantom{\rule{0ex}{0ex}}\Rightarrow t=1.42\mathrm{s}$

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:

$\frac{\mathrm{s}}{t}=\frac{7}{1.42}=4.9\mathrm{m}/\mathrm{s}$

#### Page No 52:

#### Answer:

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/s^{2}

Using the equation of motion, we can find the time taken *t* by the berry to reach the ground.

Thus, we have:

$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 12.1=0+\frac{1}{2}\times 9.8{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}=\frac{12.1}{4.9}=2.46\phantom{\rule{0ex}{0ex}}\Rightarrow t=1.57\mathrm{s}$

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.

#### Page No 52:

#### Answer:

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/s^{2} (Acceleration due to gravity)

Using the equation of motion, we get:

$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}6=u\left(0.2\right)+\frac{1}{2}\times 10\times 0.04\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{5.8}{0.2}=29\mathrm{m}/\mathrm{s}$

Let *h* be the height from which the ball is dropped*.*

We have:

*u* = 0 and *v* = 29 m/s

Now,

$h=\frac{{v}^{2}-{u}^{2}}{2a}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow h=\frac{{29}^{2}-{0}^{2}}{2\times 10}=\frac{29\times 29}{20}=42.05\mathrm{m}$

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

#### Page No 52:

#### Answer:

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/s^{2} (Acceleration due to gravity)

Using the equation of motion, we get:

$\mathrm{s}=ut+\frac{1}{2}a{t}^{2}$

$\Rightarrow 5=0+\frac{1}{2}\left(9.8\right){t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}=\frac{5}{4.9}=1.02\phantom{\rule{0ex}{0ex}}\Rightarrow t=1.01\mathrm{s}$

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 *u*_{2}_{ }should be 9.89 m/s and the final velocity *v*_{2} should be 0.

*s*_{s} = 10 cm = 0.1 m

Again using the equation of motion, we get:

${a}_{\mathrm{s}}=\frac{{v}_{2}^{2}-{u}_{2}^{2}}{2{\mathrm{s}}_{s}}=\frac{0-{\left(9.89\right)}^{2}}{2\times 0.1}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{\mathrm{s}}\approx -490\mathrm{m}/{\mathrm{s}}^{2}$

Hence, the sand offers the retardation of 490 m/s^{2}.

#### Page No 52:

#### Answer:

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:

${s}_{\mathrm{c}}=ut+\frac{1}{2}a\text{'}{t}^{2}\phantom{\rule{0ex}{0ex}}$

Here,

*a*' = *g* − *a* ( *a*' is the acceleration felt by the coin.)

*g* = Acceleration due to gravity

*g* = 9.8 m/s^{2} = 32 ft/s^{2}

On substituting the values, we get:

${s}_{\mathrm{c}}=\frac{1}{2}\left(g-a\right){\left(1\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(g-a\right)\phantom{\rule{0ex}{0ex}}$

Therefore, we can write:

$6=\frac{1}{2}\times \left(32-a\right)\phantom{\rule{0ex}{0ex}}12=32-a\phantom{\rule{0ex}{0ex}}\therefore a=20{\mathrm{fts}}^{-2}$

Hence, the acceleration of the elevator is 20 ft/s^{2}.

#### Page No 52:

#### Answer:

Given:

Speed of the ball, *u*_{x} = 20 m/s

Height from which the ball is dropped,* h* = 100 m

(a) Let *t* be the time taken by the ball to reach the ground.

Using the equation of motion, we have:

$h={u}_{y}t+\frac{1}{2}g{t}^{2}$

Here,

Acceleration of gravity, *g* = 9.8 ms^{−2}

Vertical component of velocity, *u*_{y} _{ }= 0

$\therefore t=\sqrt{\left(\frac{2h}{g}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\sqrt{\frac{2\times 100}{9.8}}=4.5\mathrm{s}$

Therefore, the time required by the ball to reach the ground is 4.5 seconds.

(b) Horizontal distance travelled by the ball:

*x* = *u _{x}t *= 20 × 4.5 = 90 m

(c) We know that horizontal velocity remains constant throughout the motion of the ball.

At A,

*v*

_{x}= 20 m/s.

*v*

_{y}=

*u*+

*gt*= 0 + 9.8 × 4.5

⇒

*v*

_{y}= 44.1 m/s

Resultant velocity:

${v}_{\mathrm{r}}=\sqrt{{\left(44.1\right)}^{2}+{\left(20\right)}^{2}}\approx 49\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\beta =\frac{{v}_{y}}{{v}_{x}}=\frac{44.1}{20}=2.205\phantom{\rule{0ex}{0ex}}\Rightarrow \beta ={\mathrm{tan}}^{-1}\left(2.205\right)=66\xb0$

Therefore, the ball strikes the ground with a magnitude of velocity 49 m/s and the direction at an angle of 66° with the ground.

#### Page No 52:

#### Answer:

Given:

Initial speed of the ball, *u* = 40 m/s

Angle of projection of the ball with the horizontal, α = 60°

Also,

*a* = g = 10 m/s^{2}

(a) Maximum height reached by the ball:

$H=\frac{{u}^{2}{\mathrm{sin}}^{2}\mathrm{\alpha}}{2g}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow H=\frac{{40}^{2}{\left(\mathrm{sin}60\xb0\right)}^{2}}{2\times 10}=60\mathrm{m}$

(b) Horizontal range of the ball:

$R=\frac{{u}^{2}\mathrm{sin}2\alpha}{g}\phantom{\rule{0ex}{0ex}}$

$=\frac{{40}^{2}\mathrm{sin}\left(2\times 60\xb0\right)}{10}=80\sqrt{3}\mathrm{m}$

#### Page No 52:

#### Answer:

Given:

Height of the goalpost = 10 ft

The football is kept at a distance of 40 yards, i.e., 120 ft, from the goalpost.

Initial speed *u* with which the ball is hit = 64 ft/s

Acceleration due to gravity, *a* = *g* = 9.8 m/s^{2} = 32.2 ft/s^{2}

For the given question, 40 yards is the horizontal range (*R*).

Angle of projection, *α* = 45°

We know that the horizontal range is given by

*R* = *u*cos*α*(*t*)

$\Rightarrow t=\frac{R}{u\mathrm{cos}\alpha}\phantom{\rule{0ex}{0ex}}=\frac{120}{64\mathrm{cos}45\xb0}=2.65\mathrm{s}$

Vertical distance covered by the football:

$y=u\mathrm{sin\alpha}\left(t\right)-\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}=64\times \frac{1}{\sqrt{2}}\times 2.65-\frac{1}{2}\times 32.2\times {\left(2.65\right)}^{2}$

= 6.86 ft < Height of the goalpost

Yes, the football will reach the goalpost.

#### Page No 52:

#### Answer:

Given:

Distance between the golis of the first and second players = 2.0 m *= R* = Horizontal range

Height *h* from which the goli is projected by the second player = 19.6 cm = 0.196 m

We know that the goli moves in projectile motion.

Acceleration due to gravity *a* = *g* = 9.8 m/s^{2}

The time in which the goli will reach the ground is given by the equation of motion.

As the initial velocity *u*_{y} in vertical direction is zero, we have:

$s={u}_{\mathrm{y}}t+\frac{1}{2}g{t}^{2}$

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 0.196}{9.8}}\phantom{\rule{0ex}{0ex}}=0.4=0.2\mathrm{s}$

Let us assume that the goli is projected with horizontal velocity *u*_{x} m/s.

The horizontal range is given by

*R* = *u*_{x}*t*

$\Rightarrow u=\frac{R}{t}=\frac{2}{0.2}=10\mathrm{m}/\mathrm{s}$

Hence, if the second player projects the goli with a speed of 10 m/s, then his goli will hit the goli of the first player.

#### Page No 52:

#### Answer:

Given:

Width of the ditch = 11.7 ft

Length of the bike = 5 ft

The approach road makes an angle of 15˚ (α) with the horizontal.

Total horizontal range that should be covered by the biker to cross the ditch safely, *R* = 11.7 + 5 = 16.7 ft

Acceleration due to gravity, *a = g* = 9.8 m/s = 32.2 ft/s^{2}

We know that the horizontal range is given by

$R=\frac{{u}^{2}\mathrm{sin}2\alpha}{g}\phantom{\rule{0ex}{0ex}}$

By putting respective values, we get:

${u}^{2}=\frac{Rg}{\mathrm{sin}2\alpha}=\frac{16.7\times 32.2}{\mathrm{sin}30\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow u\approx 32\mathrm{ft}/\mathrm{s}$

Therefore, the minimum speed with which the motorbike should be moving is 32 ft/s.

#### Page No 52:

#### Answer:

Given:

Height (*h*) of the cliff = 171 ft

Horizontal distance from the bottom of the cliff = 228 ft

As per the question, the person throws the packet directly aiming to his friend at the initial speed (*u*) of 15.0 ft/s.

From the diagram, we can write:

$\mathrm{tan}\mathrm{\theta}=\frac{\mathrm{P}}{\mathrm{B}}=\frac{171}{228}$

$\Rightarrow \mathrm{\theta}={\mathrm{tan}}^{-1}\left(\frac{171}{228}\right)$

∴ θ = 37°

When the person throws the packet from the top of the cliff, it moves in projectile motion.

Let us take the reference axis at point A.

*u* is below the *x*-axis.

*a* = *g* = 32.2 ft/s^{2} (Acceleration due to gravity)

Using the second equation of motion, we get:

$y=u\mathrm{sin}\left(\theta \right)T+\frac{1}{2}g{T}^{2}\phantom{\rule{0ex}{0ex}}y=171\mathrm{ft}\phantom{\rule{0ex}{0ex}}\theta =37\xb0\phantom{\rule{0ex}{0ex}}g=32\mathrm{ft}/{\mathrm{s}}^{2}\phantom{\rule{0ex}{0ex}}T=\mathrm{Time}\mathrm{of}\mathrm{flight}\phantom{\rule{0ex}{0ex}}171=15\mathrm{sin}\left(37\right)T+\frac{1}{2}\times 32\times {T}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{solving}\mathrm{this}\mathrm{quadratic}\mathrm{equation}\mathrm{in}T,\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}T=2.99\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Range}=15\mathrm{cos}\left(37\right)\times 2.99=35.81\mathrm{ft}\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{by}\mathrm{which}\mathrm{the}\mathrm{packet}\mathrm{will}\mathrm{fall}\mathrm{short}=228-35.81=192.19\mathrm{ft}$

#### Page No 53:

#### Answer:

Given:

Initial speed of the ball, *u* = 15 m/s

Angle of projection with horizontal, *α* = 60°

Distance of the wall from the point of projection = 5 m

*a* = *g* = 9.8 m/s^{2} (Acceleration due to gravity)

We know that the horizontal range for a projectile is given by

$R=\frac{{u}^{2}\mathrm{sin}2\alpha}{g}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow R=\frac{{15}^{2}\times \mathrm{sin}\left(2\times 60\xb0\right)}{9.8}=19.88\mathrm{m}$

As the horizontal range of the projectile is 19.88 m, the ball will hit the wall 5 m away from the point of projection. If the wall is 22 m away from the point of projection, the ball will hit the wall because it is not in its horizontal range.

#### Page No 53:

#### Answer:

Given:

Initial velocity of the projectile = *u*

Angle of projection = *θ*

To find: Average velocity of the projectile

Average velocity$=\frac{\mathrm{Change}\mathrm{in}\mathrm{displacement}}{\mathrm{Time}}$

Consider the projectile motion in the figure given below.

By the symmetry of figure, it can be said that the line joining points A and B is horizontal.

So, there will be no effect of the vertical component of velocity of the projectile during displacement AB.

We know that the projectile moves at a constant velocity *u*cos*θ* in horizontal direction.

Hence, the average velocity of the projectile is *u*cos*θ**.*

#### Page No 53:

#### Answer:

The plane is flying horizontally with a uniform speed. Therefore, the bomb also has the same speed.

Let the speed of the plane be represented by *u*.

Now, let *t *be the time taken by the bomb to reach the ground.

Distance travelled by the bomb in horizontal direction = *ut*

Both the plane and bomb are travelling in the same direction.

Distance travelled by the plane in the same time = *ut*

Hence, the bomb will explode vertically below the plane.

(i)

(ii)

When the plane is flying with a uniform speed but not horizontally:

Let us consider it will make an angle of projection *θ *along the horizontal direction.

So, both the plane and the bomb will be flying with the same angle of projection.

Therefore, both will have the same horizontal speed *u*cos*θ**,* where *u *is the initial speed of the plane and the bomb.

When the bomb is released, the time taken by the bomb to reach the ground is *t.*

The distance travelled by the bomb and the plane will be *u*cos*θt**.*

Hence, again the bomb will explode vertically below the plane.

(i) During the motion of bomb, its horizontal velocity *u* remains constant and is the same as that of the plane at every point of its path.

Let the bomb reach the ground in time *t.*

Distance travelled in horizontal direction by the bomb = *ut*

Distance travelled in horizontal direction by the bomb is the same as that travelled by the plane.

So, the bomb will explode vertically below the plane.

(ii) Let the plane move making an angle α with the horizontal.

Horizontal distance for both the bomb and the plane = *u*cosα*t*'

*t*' = Time taken by the bomb to reach the ground

So, in this case also, the bomb will explode vertically below the plane.

#### Page No 53:

#### Answer:

Given:

Acceleration of the car = 1 m/s^{2}

Projection velocity of the ball (considered as a projectile) in the vertical direction = 9.8 m/s

Angle of projection, *α *= 90˚

Let *u* be the initial velocity of the car when the ball is thrown.

Both the car and the ball have the same horizontal velocity.

We know that the distance travelled by the ball in horizontal direction is given by*
s*

_{ }=

*ut*

Here,

*t*is the time.

Also, the distance travelled by the car in horizontal direction is given by

$s\text{'}=ut+\frac{1}{2}a{t}^{2}$

Time of flight of the projectile:

$t=\frac{2u\mathrm{sin}\alpha}{g}$

*g*= 9.8 m/s

^{2}

$\Rightarrow t=\frac{2\times 9.8}{9.8}=2\mathrm{s}$

Distance between the accelerated car and the projectile:

$s\text{'}-s=\frac{1}{2}a{t}^{2}=\frac{1}{2}\times 1\times {2}^{2}=2\mathrm{m}$

Therefore, the ball drops 2 m behind the boy.

#### Page No 53:

#### Answer:

Given:

Height of one step = 10 cm

Width of one step = 20 cm

Total height of the staircase = *y* = 30 cm

Total width of the staircase = *x* = 40 cm

To directly hit the lowest plane, the ball should just touch point E.

Let point A be the origin of reference coordinate.

Let *u* be the minimum speed of the ball.

We have:

*x* = 40 cm

*y* = −20 cm

*θ* = 0°

*g* = 10 m/s^{2} = 1000 cm/s^{2}

$\therefore y=x\mathrm{tan}\theta -g\frac{{x}^{2}{\mathrm{sec}}^{2}\theta}{2{u}^{2}}$

$\Rightarrow -20=-\frac{800000}{2{u}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow u=200\mathrm{cm}/\mathrm{s}=2\mathrm{m}/\mathrm{s}$

Thus, the minimum horizontal velocity of the ball is 2 m/s.

#### Page No 53:

#### Answer:

Given:

Velocity of the truck = 14.7 m/s

Distance covered by the truck when the ball returns again to the truck = 58.8 m

(a) Therefore, we can say that the time taken by the truck to cover 58.8 m distance is equal to the time of the flight of the truck.

Time in which the truck has moved the distance of 58.8 m:

$\mathrm{Time}\left(T\right)=\frac{s}{v}=\frac{58.8}{14.7}=4\mathrm{s}$

We consider the motion of the ball going upwards.

*T* = 4 s

Time taken to reach the maximum height when the final velocity *v *= 0:

$t=\frac{T}{2}=\frac{4}{2}=2s$

*a* = *g *= −9.8 m/s^{2} (Acceleration due to gravity)

∴ *v* = *u* − *at*

⇒ 0 = *u* + 9.8 × 2

⇒ *u *= 19.6 m/s

19.6 m/s is the initial velocity with which the ball is thrown upwards.

(b) From the road, the motion of ball seems to be a projectile motion.

Total time of flight (*T*) = 4 seconds

Horizontal range covered by the ball in this time, *R* = 58.8 m

We know:

*R* = *u*cos*αt*

Here, *α *is the angle of projection.

Now,

*u*cosα = 14.7 ...(i)

Now, take the vertical component of velocity.

Using the equation of motion, we get:

${v}^{2}-{u}^{2}=2ay$

Here, *v* is the final velocity.

Thus, we get:

$y=\frac{{0}^{2}-{\left(19.6\right)}^{2}}{2\times \left(-9.8\right)}\phantom{\rule{0ex}{0ex}}=19.6\mathrm{m}$

Vertical displacement of the ball:

$y=u\mathrm{sin}\alpha t-\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 19.6=u\mathrm{sin}\alpha \left(2\right)-\frac{1}{2}\times 9.8\times {2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2u\mathrm{sin\alpha}=19.6\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow u\mathrm{sin}\alpha =19.6...\left(\mathrm{ii}\right)$

Dividing (ii) by (i), we get:

$\frac{u\mathrm{sin}\alpha}{u\mathrm{cos}\alpha}=\frac{19.6}{14.7}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\alpha =1.333\phantom{\rule{0ex}{0ex}}\alpha ={\mathrm{tan}}^{-1}(1.333)\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =53\xb0$

From (i), we get:

*u*cosα = 14.7

$\Rightarrow u=\frac{14.7}{\mathrm{cos}53\xb0}=24.42\mathrm{m}/\mathrm{s}\approx 25\mathrm{m}/\mathrm{s}$

Therefore, when seen from the road, the speed of the ball is 25 m/s and the angle of projection is 53° with horizontal.

#### Page No 53:

#### Answer:

Given:

Angle of projection of the ball, α = 53°

Width and height of the bench = 1 m

Initial speed of the ball = 35 m/s

Distance of the first bench from the batsman = 110 m

The batsman strikes the ball 1 m above the ground.

Let the ball land on the n^{th} bench.

*∴ y* = (*n* − 1) ...(i)

And,

$x=110+n-1=110+y$

$\mathrm{Now},\phantom{\rule{0ex}{0ex}}y=x\mathrm{tan}\alpha -\left(\frac{g{x}^{2}{\mathrm{sec}}^{2}\alpha}{2{u}^{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y=\left(110+y\right)\left(\frac{4}{3}\right)-\frac{10\times {\left(110+y\right)}^{2}\left({\mathrm{sec}}^{2}53\xb0\right)}{2\times {\left(35\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{440}{3}+\frac{4}{3}y-\frac{250{\left(110+y\right)}^{2}}{18\times {\left(35\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{440}{3}+\frac{4}{3}y-\frac{250{\left(110+y\right)}^{2}}{18\times {35}^{2}}$

Solving the above equation, we get:

*y* = 5

⇒ *n* − 1 = 5

⇒ *n* = 6

The ball will hit the sixth bench of the gallery.

#### Page No 53:

#### Answer:

Given:

Length of the boat = 1.0 m

Distance between the man and the centre of the boat (*R*) = 5.5 m

Initial speed (*u*) of throwing the apple by the man = 10 m/s

Acceleration due to gravity (*g*) = 10 m/s^{2}

We know that the horizontal range is given by

$R=\frac{{u}^{2}\mathrm{sin}2\mathrm{\alpha}}{g}\phantom{\rule{0ex}{0ex}}\Rightarrow 5=\frac{{\left(10\right)}^{2}\mathrm{sin}2\mathrm{\alpha}}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}2\mathrm{\alpha}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =15\xb0\mathrm{or}75\xb0$

Similarly, for the end point of the boat, i.e., point C, we have:

Horizontal range (*R*) = 6 m

$R=\frac{{u}^{2}\mathrm{sin}2\mathrm{\alpha}}{g}\phantom{\rule{0ex}{0ex}}\Rightarrow 6=\frac{{\left(10\right)}^{2}\mathrm{sin}2\mathrm{\alpha}}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}2\mathrm{\alpha}=\frac{3}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =18\xb0\mathrm{or}71\xb0$

For a successful shot, the angle of projection α with initial speed 10 m/s may vary from 15° to 18° or from 71° to 75°. The minimum angle is 15° and the maximum angle is 75°, but there is an interval of 53° for which the successful shot is not allowed. We can show this by putting the successive value of α from 15° to 75°.

#### Page No 54:

#### Answer:

Given:

Distance between the opposite shore of the river or width of the river = 400 m

Rate of flow of the river = 2.0 m/s

Boat is sailing at the rate of 10 m/s.

The vertical component of velocity 10 m/s takes the boat to the opposite shore. The boat sails at the resultant velocity v_{r}.

Time taken by the boat to reach the opposite shore:

$\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Time}}=\frac{400}{10}=40\mathrm{s}$

From the figure, we have:

$\mathrm{tan}\mathrm{\theta}=\frac{2}{10}=\frac{1}{5}$

The boat will reach point C.

$\mathrm{In}\u2206\mathrm{ABC},\phantom{\rule{0ex}{0ex}}\mathrm{tan}\mathrm{\theta}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{BC}}{400}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=\frac{400}{5}=80\mathrm{m}$

Magnitude of velocity $\left|{v}_{r}\right|=\sqrt{{10}^{2}+{2}^{2}}=10.2\mathrm{m}/\mathrm{s}$

Let α be the angle made by the boat sailing with respect to the direction of flow.

$\mathrm{tan}\left(\alpha \right)=\frac{10}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =78.7\xb0$

Distance the boat need to travel to reach the opposite shore = $\frac{400}{\mathrm{sin}\left(\alpha \right)}=407.9\mathrm{m}$

a) Time=$\frac{\mathrm{Total}\mathrm{distance}}{\mathrm{Total}\mathrm{velocity}}=\frac{407.9}{10.2}=40\mathrm{s}$

b) Using Pythagoras' theorem, we get:

Distance = $\sqrt{407.{9}^{2}-{400}^{2}}=79.9\mathrm{m}\approx 80\mathrm{m}$

#### Page No 54:

#### Answer:

Given:

Width of the river = 500 m

Rate of flow of the river = 5 km/h

Swimmer's speed with respect to water = 3 km/h

As per the question, the swimmer heads in a direction making an angle *θ* with the flow.

We know that the vertical component of velocity 3sin*θ* takes him to the opposite side of the river.

Distance to be travelled = 0.5 km

Vertical component of velocity = 3sin*θ* km/h

Thus, we have:

$\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Velocity}}=\frac{0.5}{3\mathrm{sin}\theta}\mathrm{h}=\frac{500\times 6}{5\mathrm{sin}\theta}=\frac{600}{\mathrm{sin}\theta}\mathrm{s}\phantom{\rule{0ex}{0ex}}$

∴ Required time = $\frac{10\mathrm{minutes}}{\mathrm{sin}\theta}$

(b) Shortest possible time to cover the river:

Take *θ* = 90^{∘}

$\mathrm{Time}=\frac{0.5}{3\mathrm{sin}\theta}\mathrm{h}=\frac{500\times 6}{5\mathrm{sin}\theta}=\frac{600}{\mathrm{sin}90\xb0}\mathrm{s}=600\mathrm{s}\phantom{\rule{0ex}{0ex}}$

Hence, the required time is 10 minutes.

#### Page No 54:

#### Answer:

Given:

Width of the river = 500 m

Rate of flow of the river = 5 km/h

Swimmer's speed with respect to water = 3 km/h

As per the question, the man has to reach the other shore at the point directly opposite to his starting point.

From

Horizontal distance is BD for the resultant velocity *v*_{r}.

*x*-component of the resultant velocity, *R* = 5 - 3 cos *θ*

Vertical component of velocity = 3sin*θ* km/h

$\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Velocity}}=\frac{0.5}{3\mathrm{sin}\theta}\mathrm{h}\phantom{\rule{0ex}{0ex}}$

This is the same as the horizontal component of velocity.

Thus, we have:

$\mathrm{BD}=\left(5-3\mathrm{cos}\mathrm{\theta}\right)\left(\frac{0.5}{3\mathrm{sin}\mathrm{\theta}}\right)\phantom{\rule{0ex}{0ex}}=\frac{5-3\mathrm{cos}\mathrm{\theta}}{6\mathrm{sin}\mathrm{\theta}}$

For *H* (horizontal distance) to be minimum, $\left(\frac{dH}{d\mathrm{\theta}}\right)=0$

$\Rightarrow \frac{d}{d\mathrm{\theta}}\left(\frac{5+3\mathrm{cos}\mathrm{\theta}}{6\mathrm{sin}\mathrm{\theta}}\right)=0$

⇒ 18 (sin^{2} θ + cos^{2} θ) + 30 cos θ = 0

⇒ −30 cos θ = 18

$\Rightarrow \mathrm{cos}\mathrm{\theta}=-\frac{18}{30}=-\frac{3}{5}\phantom{\rule{0ex}{0ex}}\mathrm{Negative}\mathrm{sign}\mathrm{shows}\mathrm{that}\mathrm{\theta}\mathrm{lies}\mathrm{in}\mathrm{the}2\mathrm{nd}\mathrm{Quadrant}.\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}\mathrm{sin}\mathrm{\theta}=\sqrt{1-{\mathrm{cos}}^{2}\mathrm{\theta}}=\frac{4}{5}\phantom{\rule{0ex}{0ex}}\therefore H=\frac{5-3\mathrm{cos}\mathrm{\theta}}{6\mathrm{sin}\mathrm{\theta}}\phantom{\rule{0ex}{0ex}}=\frac{5-3\left({\displaystyle \frac{3}{5}}\right)}{6\times {\displaystyle \frac{4}{5}}}=\frac{25-9}{24}\phantom{\rule{0ex}{0ex}}=\frac{16}{24}=\frac{2}{3}\mathrm{km}$

#### Page No 54:

#### Answer:

Given:

Distance between points A and B = 500 km

B from A is 30˚ east of north.

Speed of wind due north, *v*_{w} = 20 m/s

Airspeed of the plane, *v*_{a} = 150 m/s

Let $\overrightarrow{R}$ be the resultant direction of the plane to reach point B.

(a) Using the sine formula in ∆ACB, we get:

$\frac{20}{\mathrm{sin}A}=\frac{150}{\mathrm{sin}30\xb0}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}A=\frac{20}{150}\mathrm{sin}30\xb0\phantom{\rule{0ex}{0ex}}=\frac{20}{150}\times \frac{1}{2}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow A={\mathrm{sin}}^{-1}\left(\frac{1}{15}\right)$

Direction of the aeroplane is ${\mathrm{sin}}^{-1}\left(\frac{1}{15}\right)$ east of line AB.

(b) Time taken by the plane to reach point B from point A:

${\mathrm{sin}}^{-1}\left(\frac{1}{15}\right)=3\xb048\text{'}$

⇒ 30° + 3°48' = 33°48

$R=\sqrt{{A}^{2}+{B}^{2}+2AB\mathrm{cos}\theta}$

$R=\sqrt{150+20+2\left(150\right)\left(20\right)\mathrm{cos}\left(33\xb048\text{'}\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{27886}=167\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{Time}=\frac{\mathit{s}}{\mathit{v}}=\frac{500000}{167}\phantom{\rule{0ex}{0ex}}=2994\mathrm{s}=49.0\approx 50\mathrm{minutes}$

#### Page No 54:

#### Answer:

Given:

Distance between A and B = *x*

Velocity of sound in air = *v*

Velocity of wind = *u*

First Case:

When A beats the drum from his original position:

Resultant velocity of sound = *u* + *v*

⇒ (*v* + *u*)*t*_{1} = *x*

Here, *t*_{1} is the time at which the sound of the drum is heard by B.

$\Rightarrow v+u=\frac{x}{{t}_{1}}...\left(\mathrm{i}\right)$

Second Case:

After interchanging the positions:

Resultant velocity of sound = *v* − *u*

∴ (*v* − *u*)*t*_{2} = *x*

$\Rightarrow v-u=\frac{x}{{t}_{2}}...\left(\mathrm{ii}\right)$

From (i) and (ii), we get:

$2v=\frac{x}{{t}_{1}}+\frac{x}{{t}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{x}{2}\left(\frac{1}{{t}_{1}}+\frac{1}{{t}_{2}}\right)$

From (i), we get:

$u=\frac{x}{{t}_{1}}-v$

Putting the value of *v* in the above equation, we get:

$u\Rightarrow \frac{x}{2{t}_{1}}-\frac{x}{2{t}_{2}}=\frac{x}{2}\left(\frac{1}{{t}_{1}}-\frac{1}{{t}_{2}}\right)$

∴ Velocity of sound, $v=\frac{x}{2}\left(\frac{1}{{t}_{1}}+\frac{1}{{t}_{2}}\right)$

Velocity of wind, $u=\frac{x}{2}\left(\frac{1}{{t}_{1}}-\frac{1}{{t}_{2}}\right)$

#### Page No 54:

#### Answer:

Given:

Distance between A and B = *x*

Let *v *be the velocity of sound in the direction along line AC.

Let *u** *be the velocity of air in the direction along line AB.

Angle between *v* and *u* =* θ* > $\frac{\mathrm{\pi}}{2}$

Resultant velocity of sound and air that will reach B = $\overrightarrow{\mathrm{AD}}=\sqrt{\left({v}^{2}-{u}^{2}\right)}$

Here, the time taken by light to reach B is neglected.

∴ Time lag between seeing and hearing = Time taken to hear the sound of the drum

$t=\frac{\mathrm{Displacement}}{\mathrm{Velocity}}\phantom{\rule{0ex}{0ex}}=\frac{x}{\sqrt{{v}^{2}-{u}^{2}}}$

#### Page No 54:

#### Answer:

A regular hexagon has a side *a.* Six particles situated at the corners of the hexagon are moving with a constant speed *v.*

As per the question, each particle maintains a direction towards the particle at the next corner. So, particles will meet at centroid O of triangle PQR. Now, at any instant, the particles will form an equilateral triangle PQR with the same centroid O.

We know that P approaches Q, Q approaches R and so on.

Now, we will consider the motion of particle P. Its velocity makes an angle of 60˚.

This component is the rate of decrease of distance PO.

Relative velocity between P and Q:

${\overrightarrow{v}}_{\mathrm{PQ}}={\overrightarrow{v}}_{\mathrm{P}}-{\overrightarrow{v}}_{\mathrm{Q}}=\overrightarrow{v}-\overrightarrow{v}\mathrm{cos}60\xb0\phantom{\rule{0ex}{0ex}}=\overrightarrow{\mathit{v}}\mathit{-}\frac{\overrightarrow{\mathit{v}}}{\mathit{2}}\mathit{=}\frac{\overrightarrow{\mathit{v}}}{\mathit{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Time},t=\frac{\mathrm{Displacement}}{\mathrm{Velocity}}\phantom{\rule{0ex}{0ex}}=\frac{\mathit{a}}{\mathit{v}\mathit{/}\mathit{2}}\mathit{=}\frac{\mathit{2}\mathit{a}}{\mathit{v}}$

Hence, the time taken by the particles to meet each other is $\frac{2a}{v}$*.*

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