NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion in a Straight Line

NCERT Exemplar Class 11 Physics Solutions Chapter 3: MCQI

Question:1

Among the four graphs (Fig. 3.1), there is only one graph for which average velocity over the time interval (0, T ) can vanish for a suitably chosen T. Which one is it?

Answer:

The correct answer is the option (b)
Explanation: In graph (b) displacements are in the opposite direction & when we add them, we get net displacement & average velocity as zero. It satisfies the condition of displacement for different timings.

Let us draw a parallel line from A to the time axis at $t=0s$. It intersects the graph at B & the change in displacement time is zero. So, the displacement from A to $B = 0$ & hence the average velocity of the body also vanishes to 0.

Question:2

A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
$(a) x < 0, v < 0, a > 0$
$(b) x > 0, v < 0, a < 0$
$(c) x > 0, v < 0, a > 0$
$(d) x > 0, v > 0, a < 0$

Answer:

The correct answer is the option (a) x < 0, v < 0, a > 0
Explanation: The value of x becomes negative as the lift comes downward, i.e., from 8th to 4th floor, thus, x<0.

Velocity is downward, i.e., negative, thus, v<0.
The lift retard before reaching the 4th floor and hence the acceleration will be upwards, i.e., a>0.

Question:3

In one dimensional motion, instantaneous speed v satisfies $0 \leq v < v_{0} .$
(a) The displacement in time T must always take non-negative values.
(b) The displacement x in time T satisfies – $v_{0}T < x < v_{0}T .$
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points.

Answer:

The correct answer is the option (b) The displacement x in time T satisfies -$v_{0}T < x < v_{0}T .$
Explanation: The magnitude & direction of max. & min. velocity can be used to determine the max. & min. displacement.
v₀ is the maximum velocity in the positive direction as well as in the opposite direction (or we can say minimum velocity).
Thus, v₀T is the maximum displacement in the positive direction & -v0T is the maximum displacement in the opposite direction.
Thus, - $v_{0}T < x < v_{0}T .$

Question:4

A vehicle travels half the distance L with speed $V_{1}$ and the other half with speed $V_{2}$, then its average speed is
$a) \frac{(V_{1}+V_{2})}{2}$
$b) \frac{(2V_{1}+V_{2})}{(V_{1}+V_{2})}$
$c) \frac{(2V_{1}V_{2})}{(V_{1}+V_{2})}$
$d) \frac{L(V_{1}+V_{2})}{V_{1}V_{2}}$

Answer:

The correct answer is the option $c) \frac{(2V_{1}V_{2})}{(V_{1}+V_{2})}$
Explanation: Let $t_{1}$ be the time taken in half distance, $t_{1}=\frac{L}{v_{1}}$
Let $t_{2}$ be the time taken in half distance, $t_{2}=\frac{L}{v_{2}}$
Therefore, the total time taken in distance will be equal to $(L+L)=\frac{L}{v_{1}}+\frac{L}{v_{2}}$
$=\frac{L(v_{1}+v_{2})}{v_{1}v_{2}}$
& the total distance will be equal to $L + L = 2L$
Thus, the average speed will be,
$v_{av}=\frac{Total \; distance}{Total\; time}$
$\\=\frac{2L}{\frac{L(v_{1}+v_{2})}{v_{1}v_{2}}}\\\\=\frac{2V_1V_2}{V_1+V_2}$

Question:5

The displacement of a particle is given by $x = (t-2)^{2}$ where x is in metres and t is seconds. The distance covered by the particle in first 4 seconds is
a) 4 m
b) 8 m
c) 12 m
d) 16 m

Answer:

The answer is the option (b) 8m
Explanation: It is given that, $x = (t-2)^{2}$
Now, we know that,
$V=\frac{dx}{dt}$
$= 2 (t - 2) m/s$
&
$a=\frac{d^{2}x}{dt^{2}}$
$= 2 (1 - 0)$
$=2\; ms^{-2}$
Now,
$v_{0} = 2(0-2) = -4 m/s \; \; \; \; \; \; \;............ (at\; t = 0)$
$v_{2} = 2(2-2) = 0\; m/s \; \; \; \; \; \; \;............ (at\; t = 2)$
$v_{4} = 2(4-2) = 4\; m/s \; \; \; \; \; \; \;............ (at\; t = 4)$
Now, distance is equal to the area between the time-axis graph and the (v-t) graph,

$=\frac{1}{2}(2.4)+\frac{1}{2}(2.4)$
$=8\; m$
Hence, option (b).

Question:6

At a metro station, a girl walks up a stationary escalator in time t1. If she remains stationary on the escalator, then the escalator take her up in time $t_{2}$. The time taken by her to walk up on the moving escalator will be
$a) \frac{(t_{1}+t_{2})}{2}$
$b) \frac{t_{1}t_{2}}{(t_{2}-t_{1})}$
$c) \frac{t_{1}t_{2}}{(t_{2}+t_{1})}$
$d) \; t_{1}-t_{2}$

Answer:

The answer is the option $c) \frac{t_{1}t_{2}}{(t_{2}+t_{1})}$
Explanation: Let us consider L as the length of the escalator,
$V_{g}$ as the velocity of the girl w.r.t. the ground
& $V_{e}$ as the velocity of the escalator w.r.t. the ground
Now, w.r.t. the ground, the effective velocity of the girl will be,
$V_{g}+V_{e}=\frac{L}{t_{1}} + \frac{L}{t_{2}}$
$=L(\frac{1}{t_{1}} + \frac{1}{t_{2}})$
$V_{ge}=L\left [ \frac{t_{1}+t_{2}}{t_{1}t_{2}} \right ]$
$\\\frac{L}{t}=L[\frac{t_1+t_2}{t_1t_2}]\\t=\frac{t_1 t_2}{t_1+t_2}$

NCERT Exemplar Class 11 Physics Solutions Chapter 3 MCQII

Question:7

The variation of quantity A with quantity B, plotted in Fig. 3.2 describes the motion of a particle in a straight line.

(a) Quantity B may represent time.
(b) Quantity A is velocity if motion is uniform.
(c) Quantity A is displacement if motion is uniform.
(d) Quantity A is velocity if motion is uniformly accelerated.

Answer:

The correct answer is the option:
(a) Quantity B may represent time.
(c) Quantity A is displacement if the motion is uniform.
(d) Quantity A is velocity if the motion is uniformly accelerated.
Explanation:
Verification of opt (a) & (d)
If the quantity B would have represented velocity instead of time, then the graph would’ve become a straight line, viz., uniformly accelerated motion, hence the motion is not uniform.
Verification of opt (c)
If A represents displacement and B represents time, then the graph will be a straight line which would represent uniform motion.

Question:8

A graph of x versus t is shown in Fig. 3.3. Choose correct alternatives from below.

(a) The particle was released from rest at t = 0.
(b) At B, the acceleration a > 0.
(c) At C, the velocity and the acceleration vanish.
(d) Average velocity for the motion between A and D is positive.
(e) The speed at D exceeds that at E.

Answer:

The correct answer is the option:
(a) The particle was released from rest at t = 0.
(c) At C, the velocity and acceleration vanish.
(e) The speed at D exceeds that at E.
Explanation: Now, we know that,
Slope of the x-t graph gives us $V=\frac{dx}{dt}$
Verification of opt(a)-
$\frac{dx}{dt}$ is zero or particle is at rest at A since graph (x-t) is parallel to the time axis.
Slope $\frac{dx}{dt}$ increases after A and hence velocity also increases.
Verifying option (c) and rejecting opt (b)-
Now, $\frac{dx}{dt}$ or v = 0 since the tangent at B & C is graph (x-t), viz., parallel to the time axis. Hence, acceleration = 0.
Verifying opt (e)-
Speed at D is greater than speed at E since the slope at D is greater at D than at E.
Rejecting opt (d)-
Average velocity at A is zero as graph (x-t) is parallel to time axis, also displacement is negative at D, which makes it clear that the velocity at D is also negative.

Question:9

For the one-dimensional motion, describe by $x = t - \sin t$
$a)\; x(t)>0 for \; all\; t>0$
$b) v(t)>0 for\; all\; t>0$
$c) a(t)>0\; for \; all\; t>0$
$d) v(t) lies\; between\; 0 \; and \; 2$

Answer:

The correct answer is the option:
$a)\; x(t)>0 for \; all\; t>0$
$d) v(t) lies\; between\; 0 \; and \; 2$
Explanation: Now, $x = t -\sin t$
We know that $v=\frac{dx}{dt}$
$=1-\cos \; t$
Now, $a=\frac{dv}{dt}$
$=\frac{d(1-\cos \; t)}{dt}$
$=+\sin\; t$
v_max will be,
$V_{max}=1-(-1)$
$= 1+1 = 2$
. v_min will be,
$v_{min} = 1-1$
$=0$
Thus, it is clear that v lies between 0 to 2 and option (d) is verified.
Now, $x = t -\sin t$
Thus, sin t lies between 1 and -1 for all t > 0.
Thus, x will always be positive and option (a) is verified.
Now, $v = 1 -\cos \; t$
$v= 0, when \; t = 0$
$v= 1, when\; t = \frac{\pi }{2}$
$v=2, when t = \pi$
& $v= 0, when\; t = 2\pi .$
Hereby opt (b) is discarded.
Now, $a = \sin \; \; t$
$a = 0, when\; t = 0$
$a = 1, when \; t = \frac{\pi }{2}$
$a = 0, when\; t = \pi$
& $a = 1, when\; t = 2\pi .$
Thus, acceleration can be negative as well, and hence opt (c) is also discarded here.

Question:10

A spring with one end attached to a mass and the other to a rigid support is stretched and released.
a) magnitude of acceleration, when just released is maximum
b) magnitude of acceleration, when at equilibrium position is maximum
c) speed is maximum when mass is at equilibrium position
d) magnitude of displacement is always maximum whenever speed is minimum

Answer:

The correct answer is the option:
(a) Magnitude of acceleration, when just released is maximum.
(c) Speed is maximum when mass is at the equilibrium position.
Explanation: Let us consider a spring lying on a frictionless table. Let k be the spring constant, viz., attached to a mass ‘m’ at one end and the other end is fixed at right support.
Now let us stretch the spring to a displacement x by force F, $F= -kx$
Now, P.E. at $A = \frac{1}{2} kx^{2}$
Since restoring force is proportional to x, Simple Harmonic Motion is executed here.
Therefore, $a=\frac{-F}{m}$
Or $a= \frac{-kx}{m}$
$a = 0, when\; x = 0$
& $a= \frac{-kx}{m},at \; x= x$
Thus, when spring is released, the magnitude will be maximum. Hence, opt(a) is verified here.
At x = 0, the speed of mass is maximum.
Hence opt (c) is also verified.
At x = 0, magnitude of a = 0.
Hence, opt (b) is discarded.
The speed of mass may or may not be zero when it is at its maximum displacement.
Hence, opt (d) is also discarded here.

Question:11

A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground,
a) the direction of motion of the ball changes every 10 seconds
b) speed of ball changes every 10 seconds
c) average speed of ball over any 20 seconds intervals is fixed
d) the acceleration of ball is the same as from the train

Answer:

The correct answer is the option:
(b) The speed of the ball changes every 10 seconds.
(c) average speed of the ball over any 20 seconds interval is fixed.
(d) the acceleration of the ball is the same as from the train.
Explanation: If we observe the motion from the ground, we will see that the ball strikes with the wall after every 10 seconds. The direction of the ball is the same since it is moving at a very small speed in the moving train, therefore, it will not change w.r.t. observer from the earth.
The speed of ball can change after a collision, hence, option (a) will be discarded and opt (b) is verified.
Average speed of the ball at any time remain same or is 1 m/s, i.e., it is uniform.
Hence opt (c) is also verified.
When the ball strikes to the wall, initial speed of the ball will be in the direction of the moving train w.r.t. the ground as well as its speed will also change (vTG)
Thus, $V_{TG}= 10+1= 11m/s$
The speed of the ball after collision with a side of the train is in the opposite direction of the train $(v_{BG}) = 10-1 = 9 m/s.$
Thus, the magnitude of acceleration on both the walls of the compartment will be the same, but in opposite direction. Hence, opt (b), (c) & (d) are verified here.

NCERT Exemplar Class 11 Physics Solutions Chapter 3: Very Short Answer

Question:12

Refer to the graphs below and match the following:

Answer:

(i) From the graph (d), it is indicated that the slope is always positive between $0^{o}$ to $90^{o}$ (tan ?).
Hence, (i) $\rightarrow$ (d)
(ii) At point A, v = 0 & a = 0 as the slope is zero; thus, the graph always lies in +x direction.
Hence, (ii)$\rightarrow$ (b).
(iii) There is zero displacement only in graph (a), where y = 0.
Hence, (iii) $\rightarrow$ (a).
(iv) In the graph (c), since v < 0, the slope is negative here.
Hence, (iv) $\rightarrow$ (c).

Question:13

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

Answer:

When we hit a ball with a bat, the acceleration of the ball decreases, till its velocity becomes zero. Hence, the acceleration will be in the backward direction

After the velocity of the ball has been decreased to zero, it increases in the forward direction. Thus, the acceleration will be negative in the forward direction

Question:14

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

Answer:

(a) Let us consider a motion where
$x(t) = \omega t - \sin \; \omega t$
Thus, $v=\frac{dx}{dt}$
$= \omega - \omega\; \cos \; \omega t$
& $a=\frac{dv}{dt}$
$=\omega ^{2}\; \sin \; \omega t$
$x(t) = 0, v=0 \; and \; a=0; at \; \omega t = 0$
$x(t) = \pi >0, v= \omega -\omega \; \cos \pi = 2\; \omega >0 \; and\; a=0; at \; \omega t = \pi$
$x(t) = 2\pi >0, v=0 \; and\; a=0; at\; \omega t = 2\pi .$
(ii) Let us consider a function of motion where,
$x(t) = -a \sin \; \omega t$
$x(t) = -a \; \sin 0 = 0; at\; t=0$
$x(t) = -a \sin \frac{2\pi }{T}.\frac{T}{4}= -a\sin \frac{\pi }{2}=-a;at\; t=\frac{T}{4}$
$x(t) = -a \sin \frac{2\pi }{T}.\frac{3T}{4}=-a\; \sin \frac{3\pi}{2}$
$= -a \sin (\pi +\frac{\pi}{2}) = -a ( -\sin \frac{\pi}{2}) = a ; at\; t = \frac{3T}{4}$
$x(t) = -a \sin \frac{2\pi}{T}.\frac{T}{2} = -a \sin \pi = 0; at\; \; t = \frac{T}{2}$
$x(t) = -a \sin \frac{2\pi}{T}.T = -a\; \sin 2\pi = +0; at\; \; t = T$
Thus the displacement of the particle is in negative direction and it comes to rest periodically.
Thus,
$-a \sin \; \omega t$ is a periodic function.
$V=v=\frac{dx(t)}{dt}$
$=\frac{d}{dt}.(-a\; \sin\; \omega t)$
$=-a\; \omega \cos\; \omega t$
Now, $v=a\omega \; \cos\; 0^{o}=-\omega a;at\; t=0$
$v = -\omega\; a \; \cos \frac{2\pi}{T}.\frac{T}{4}=-\omega\; a\; \cos\; \frac{\pi}{2}=0;at\; t=\frac{T}{4}$
$v = -\omega\; a \; \cos \frac{2\pi}{T}.\frac{T}{4}=-a\; \omega\; \cos\; \pi=+\omega\; a;at\; t=\frac{T}{2}$
$v = -\omega\; a \; \cos \frac{2\pi}{T}.\frac{3T}{4}=-\omega\;a \cos\; (\pi+\frac{\pi}{2})at\; t=\frac{3T}{4}$
$v = -a\; \omega\; \cos \frac{2\pi}{T}.T=-\omega\;a \cos\; 2\pi=-\omega\;a\; at\; t=T$
After zero displacement, velocity changes periodically.
Thus, $x(t) = -a \sin \omega\; t$ is the function required.
(i) Now, let us consider a function
$x(t) = a \sin \omega\: t.$
$x(0) = 0$
$x(\frac{T}{4}) = a\; \sin\; \frac{2\pi}{T}.\frac{T}{4}=a\; \sin\frac{\pi}{2}=a$
$x(\frac{T}{2}) = a\; \sin\; \frac{2\pi}{T}.\frac{T}{2}=a\; \sin\; \pi=a$
$x(\frac{3T}{2}) = a\; \sin\; \frac{2\pi}{T}.\frac{3T}{2}=a\; \sin\;\left ( \pi+\frac{\pi}{2} \right )=-a$
$x(T) = a\; \sin\; \frac{2\pi}{T}.T=a\; \sin\;2\pi=0$
Thus, the particle moves in a positive direction, periodically with zero displacements.
Hence $x(t) = a \sin \omega t$ is the required function.

Question:15

Give example of a motion where $x>0, v<0, a>0$ at a particular instant.

Answer:

Let x(t) be the function of motion,
$x(t) = A + Be^{-\gamma t} \; \; \; \; \; ........ (i)$
Here, $\gamma$ & A are constant & B is the amplitude.
At time t the displacement is x(t),
Here $A>B \; and\; \gamma > 0$
Thus, $v(t)=\frac{dx(t)}{dt}$
$=0+(-\gamma )Be^{-\gamma t}$
$=-\gamma Be^{-\gamma t}$
Now, $a(t)=\frac{d}{dt[v(t)]}$
$=\frac{d}{dt}(-\gamma \; B\; exp^{-\gamma t})$
$=(\gamma B^{2}exp^{-\gamma t})$
Thus, we get,
x > 0, i.e., x is always positive, since A>B
v<0, i.e., v is always negative, since v<0
& a>0, i.e., a is always positive.
The value of $\gamma Be^{-\gamma t}$ varies from 0 to $+\infty$

Question:16

An object falling through a fluid is observed to have acceleration given by $a = g - bv$ where g = gravitational acceleration and b is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?

Answer:

The velocity becomes constant after a long time from when it is released, thus,
$\frac{dv(t)}{dt} = 0\; \; or\; \; a=0$
$a = g - bv \: \: \: \: \: \: ........(given)$
Thus, $0 = g - bv$
$bv = g$
$v=\frac{g}{b}$
Thus, $(\frac{g}{b})$ is the constant speed after a long time of release.

NCERT Exemplar Class 11 Physics Solutions Chapter 3: Short Answer

Question:17

A ball is dropped and its displacement vs time graph is as shown in the figure where displacement x is from the ground and all quantities are positive upwards.

a) Plot qualitatively velocity vs time graph
b) Plot qualitatively acceleration vs time graph

Answer:

If we observe the graph we know that the displacement (x) is always positive. The velocity of the body keeps on increasing till the displacement becomes zero, after that the velocity decreases to zero in the opposite direction till the maximum value of x is reached, viz., smaller than earlier. When the body reaches towards x=0, the velocity increases and acceleration is in the downward direction. And when the body’s displacement is $x>0$, i.e., the body moves upwards, the direction will be downwards and velocity will decrease, i.e., $a=-g$.
(a) Velocity time graph

(b) Acceleration time graph.

Question:18

A particle executes the motion described by $x(t) = x_{0} (1 - e^{-\gamma t})$ where $t \geq 0, x0 > 0$
a) Where does the particles start and with what velocity?
b) Find maximum and minimum values of $x(t), v(t), a(t)$. Show that $x(t)$ and $a(t)$ increase with time and $v(t)$ decreases with time.

Answer:

Here, $x(t)=x_{0}[1-e^{-\gamma t}]$
So, $v(t)=\frac{dx(t)}{dt}$
$=\frac{d}{dt}[x_{0}(1-e^{-\gamma t})]$
$=+x_{0}\gamma e^{-\gamma t}\; \; \; \; \; \; ..........(i)$
&
+ $a(t)=\frac{dv}{dt}$
$=-x_{0}\gamma ^{2}e^{-\gamma t}\; \; \; \; \; \; ..........(ii)$
$(i) x(0) = x_{0} [1 - e^{0}]$
$= x_{0} (1 - 1) =0$
$v(0) = x_{0}\gamma e^{0}$
$=x_{0}\gamma$
Thus, $x=0$ is the starting point of the particle and its velocity is $v_{0}=x_{0}\gamma$
(b) $x (t)\ is,$
Maximum at $t = \infty$ since $t = \infty$ $[x(t)]_{max} = \infty$
Minimum at $t = 0$ since at $t = 0$,$[x(t)]min = 0$
v(t) is,
maximum at $t = 0$ since $t=0, v(0) = x_{0}\gamma$
minimum at $t = \infty$ since, $t = \infty$, $v(\infty ) = 0$
a(t) is,
maximum at $t = \infty$ since at $t = \infty$, $a(\infty )=0$
minimum at $t = 0$ since at $t = 0$, $a(0)=-x_{0}\gamma ^{2}$

Question:19

A bird is tossing between two cars moving towards each other on a straight road. One car has a speed of 18 m/h while the other has the speed of 27 km/h. The bird starts moving from first car towards the other and is moving with the speed of 36 km/h and when the two cars were separated by 36 km. What is the total distance covered by the bird? What is the total displacement of the bird?

Answer:

We can find out the relative speed of the cars by adding the speed of the two cars
$= 27 + 18$
$= 45 km/hr$
$Time \; taken \: to \: meet = \frac{Distance \; between \; cars}{relative \; speed \; of \; the \; cars}$
$=\frac{36}{45}=\frac{4}{5}$ hours
Hence, the distance that the bird will cover in $\frac{4}{5}$ hours $=36(\frac{4}{5})=28.8\; km.$

Question:20

A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 m/s, the distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building?

Answer:

In vertical motion,
$u_{y} = 0$
$a = 10 m/s^{2}$
$s = 9m$ & $t=t$
now, $s = u_{y}t+\frac{1}{2}at^{2}$
thus, $9=0(t)+\frac{1}{2}(10)(t^{2})$
Therefore,
$t=\sqrt{\frac{9}{5}}$
$=\frac{3}{\sqrt{5}}sec$
Now, the horizontal distance covered by the person will be,
$u_{x}\times\; t=9\left ( \frac{3}{\sqrt{5}} \right )$
$=\frac{27}{\sqrt{5}}=\frac{27}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}$
$=27\times\frac{2.236}{5}$
$=12.07\; m$
Therefore, the person will reach the building which s next farther the first edge by $12.7 - 10 = 2.07\; m$.

Question:21

A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.

Answer:

For the first ball-
$U = 0, h = 45m$
$a = g, t=t \; and\; V = v_{1} =?,$
we know that,
$v = u + at$
$v1 = 0 + gt = gt\; downward$
(for second ball)
$V = v_{2}$
$u = 40 m/s, a = -g,$
$t =t$
now, $V = u + at$
$v_{2} = (40 - gt) \; upward$
Now relative velocity of the 1st ball w.r.t. 2^nd
$v_{12} = v_{1} - v_{2} = -gt -(40 - gt) = - gt - 40 + gt = - 40\; m/s$
The speed of one ball increases and the speed of the other decreases with the same rate due to acceleration.
Hence, relative speed = 40 m/s.

Question:22

The velocity-displacement graph of a particle is shown in the figure.

a) Write the relation between v and x.
b) Obtain the relation between acceleration and displacement and plot it.

Answer:

a) Consider the point P(x,v) at any time t on the graph such that angle ABO is $\theta$ such that
$\tan \theta = \frac{AQ}{QP} = \frac{(v_{0}-v)}{x} = \frac{v_{0}}{x_{0}}$

When the velocity decreases from $v_{0}$ to zero during the displacement, the acceleration becomes negative.
$v_{0}-v=\left ( \frac{v_{0}}{x_{0}} \right )x$
$v=v_{0}(1-\frac{x}{x_{0}})$
is the relation between v and x.
$b) a = \frac{dv}{dt} = (\frac{dv}{dt})(\frac{dx}{dx}) = \left ( \frac{dv}{dx} \right )\left ( \frac{dx}{dt} \right )$
$a=\frac{-v_{0}}{x_{0}}v$
$a=\left ( \frac{v{_{0}}^{2}}{x{_{0}}^{2}} \right )x-\left ( \frac{v{_{0}}^{2}}{x_{0}}\right )$

$At \; x = 0$
$a=\frac{-v{_{0}}^{2}}{x^{0}}$
$At \; a = 0$
x=x₀
The points are
$(0, \frac{-v{_{0}}^{2}}{x_{0}})$and $B(x_{0},0)$

NCERT Exemplar Class 11 Physics Solutions Chapter 3: Long Answer

Question:23

It is a common observation that rain clouds can be at about a kilometre altitude above the ground.
a) If a raindrop falls from such a height freely under gravity, what will be its speed? Also, calculate in km/h
b) A typical raindrop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.
c) Estimate the time required to flatten the drop.
d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two raindrops is 5 cm.

Answer:

$\\\; h=1\; km=1000\; m\\g=10\; m/s^{2}\\d=4\; mm\; and\; u=0\; m/s$
Thus, $\\r=\frac{4}{2}\; mm\\=2\times10^{-3}m$
(a) Let's find out the velocity of raindrop on the ground’
$\\v^{2}=u^{2}-2as\\=u^{2}-2g(-h)\\=u^{2}+2gh\\=0^{2}+2(10)(1000)$
Thus,
$\\v=100\sqrt{2}\; m/s\\v=100\sqrt{2}\left ( \frac{18}{5} \right )km/hr$
$\\=360\sqrt{2}\; km/hr\\=510\; km/hr$
(b) Momentum of the raindrop when it touches the ground
mass of drop(m) = Vol. × density
$=\frac{4}{3}\pi \; r^{3}\rho$ $(\rho = density\; of\; water)$
Now, density of water$=10^{3}kg/m^{3}$
Thus, $M=\frac{4}{3}\pi (2\times10^{-3})^{3}\times1000$
$=\frac{4}{3}\times\frac{22}{7}\times2\times2\times2\times10^{-9}\times10^{3}$
$=\frac{704}{21}\times10^{-6}kg$
$=3.35\times10^{-5}kg$
Now, we know that Momentum (p) = mv
$p=3.35\times10^{-5}\times100\sqrt{2}$
$=4.7\times10^{-3}kg\; ms^{-1}$
(c) Time required for a drop to be flattened-
$Time=\frac{distance}{speed}$
$=\frac{4\times10^{-3}}{100\sqrt{2}}m\; \; \; \; \; \; .....(distance=4mm=4\times10^{-3})$
$=4\frac{(\sqrt{2})}{100(2)}\times10^{-3}$
$=\frac{2(1.414)}{100}\times10^{-3}$
$=2.8\times10^{-5}\; sec$
(d) Now, we know that,
$Force=\frac{dp}{dt}$
$=\frac{mv-0}{t-0}$
$Force=\frac{4.7\times10^{-3}}{2.8\times10^{-5}}$
$=1.68(10^{2})$
$=168\; N$
(e) Here,
Radius of umbrella $=\frac{1}{2}m\; \; \; \; \; \; .......(since\; its\; \; diameter=1m)$
Thus, Area of umbrella$=\pi R^{2}$
$=\frac{22}{7}.\frac{1}{2}.\frac{1}{2}m^{2}$
$=\frac{11}{14}m^{2}$
Now, the square area covered by one drop
$= (5 \times 10^{-2})^{2}$
$= 25 \times 10^{-4} m^{2}$
Therefore, no. of drops falling on the umbrella $= \frac{\pi R^{2}}{25} \times 10^{-4}$
$=\frac{11(10)^{4}}{14(25)}$
$=0.0314\times10^{4}$
Therefore 314 drops fell on the umbrella.
Thus, the net force on the umbrella $= 314 \times 168N = 52752 N.$

Question:24

A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3 s while for a truck this time interval is 5 s. On a highway the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at an emergency. At what distance the car should be from the truck so that it does not bump onto the truck. Human response time is 0.5 s.

Answer:

Given : (for truck)
$U=72\; km/hr=2\times\frac{5}{18}m/s$
$=20\; m/s$
$V = 0, t = 5s$
$a =?$
We know that,
$V= u + at$
i.e., $0 = 20 + a (5)$
thus, $a = \frac{-20}{5} = \frac{-4m}{s}$
Given : (for car)
$U = 20 \; m/s$
$V =0, t = 3s$
$a = a_{c}$
Again, $v = u + at$
$0 = 20 + a{_{c}}^{3}$
$a_c = \frac{-20}{3} \; m/s^2$
A human takes atleast 0.5 seconds to respond, thus time taken by the car driver to respond is $(t-0.5)$ sec …. (car takes t time to stop)
$Vc = u + a{_{c}}{t}$
$0 = 20-\frac{20}{3}.(t - 0.5) \; \; \; \; \; \; .......... (i)$
There is no responding time for the truck driver so he applies breaks with passing signal to car back side, hence,
$V = u + at$
$0 = 20 - 4t \; \; \; \; \; \; .......... (ii)$
From (i) & (ii),
$20 - 4t = 20 - \frac{20}{3} ( t - 0.5)$
$-4t = - \frac{-20}{3} ( t - 0.5)$
$\\12t = 20t - 10\\-20 + 12t = -10\\$
Thus, $-8t = -10$
Thus, $t=\frac{10}{8}$
$=1.25\; sec$
Now the distance covered by the car & the truck in $\frac{5}{4}$ sec will be,
$S = 20 \left ( \frac{5}{4} \right ) + 0.5(-4) \left ( \frac{5}{4} \right )\left ( \frac{5}{4} \right )$
……. $( because \; s = ut + \frac{1}{2} at^{2})$
$= 25 - 3.125$
$=21.875 m$
First 0.5 seconds the car moves with uniform speed but after responding brakes are applied for 0.5 sec and the retarding motion of the car starts.
$S_{c} = (20 \times 0.5) \times 20 (1.25 - 0.5) + \frac{1}{2} \left ( \frac{-20}{3} \right ) (1.25 - 0.5)^{2}$
$= 25 - 1.875$
$=21.875 m$
Thus, $s_{c} - s = 23.125 - 21.875$
$= 1.25 m$
Therefore, the car must be 1.25 m behind the truck to avoid bumping into it.

Question:25

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

Answer:

(a) for the velocity to be maximum,
$\frac{dv(t)}{dt}=0$
$d\frac{[2t(3-t)]}{dt}=0$
$d\frac{6t-2t^{2}}{dt}=0$
$6 - 4t = 0$
$4t = 6$
Thus, $t=\frac{3}{2}$
$= 1.5 s$
(b)Now, average velocity
$V=\frac{x}{t}$
$=6t-2t^{2}$
$\frac{ds(t)}{dt}=6t-2t^{2}$
$ds=(6t-2t^{2})dt$
Let us integrate both the sides from 0 to 3
$\int_{0}^{s}ds=\int_{0}^{3}(6t-2t^{2})dt$
$s=\left [ 6\frac{t^{2}}{2}-2\times\frac{t^{3}}{3} \right ]_{0}^{3}=\left [ 3t^{2}-\frac{2}{3}t^{3} \right ]_{0}^{3}$
$=\left [ 3\times9-\frac{2}{3}\times2 \right ]=27-18$
Thus, $s = 9m$
Now, average velocity $(V_{av})=\frac{9}{3}=\frac{3m}{s}$
$V(t) = 6t - 2t^{2} .......... since \; 0 < t < 3$
$3 = 6t - 2t^{2} .......... since V_{av}=3$
$2t^{2} - 6t + 3 = 0 \; \; \; \; \; \; .............. since ( a=2, b=-6 \; and\; c=3)$
$t=2.36s$
Thus, the average velocity is maximum at 2.36 seconds.
(c) When the body returns at its mean position or changes direction in periodic motion, time for acceleration is maximum.

$Here, v=0$

$V(t) = 6t - 2t^{2}$
$0 = 6t - 2t^{2}$
$2t (3-t) = 0$
$t\neq 0$
Thus, the acceleration is maximum at t = 3 sec.
(d)Now, for 3 to 6 sec
$V(t) = -(t-3) (6-t)$
$\frac{ds}{dt}=(t-3)(6-t)$
$ds=(t^{2}-9t+18)dt$
Integrate from 3 to 6 s
$s_{2}=\int_{3}^{6}(t^{2}-9t+18)dt=\left [ \frac{t^{3}}{3}-\frac{9}{2}t^{2}+18t \right ]_{3}^{6}$
$=\frac{(6)^{3}}{3}-\frac{9}{2}(6)^{2}+18\times6-\left [ \frac{(3)^{3}}{3}-\frac{9}{2}(3)^{2}+18\times3 \right ]$
$=\frac{6\times6\times6}{3}-\frac{9\times6\times6}{2}+108-\frac{3\times3\times3}{2}-54$
$=180-162-63+40.5=18-22.5$
$S_{2}=-4.5\; m$ .............because distance is in downward direction
Thus, net distance $=4.5\; m$
Thus, in three cycle $=4.5(3)$
$=13.5\; m$
Remaining height will be
$20 -13.5 = 6.5 m$
The monkey can climb up to 9m without slipping but in the 4^th cycle it will slip and the height remaining to climb will be 6.5 m.
Net no. of cycle = 4.

Question:26

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

Answer:

Let $u_{1}$ be the speed of the 1^st ball, $u_{1} = 2u\; m/s$
& $u_{2}$ be the speed of the second ball, $u_{2} = u \; m/s$
Let $h_{2}$ be the height of the two balls before coming to rest & $h_{1}$ be the height covered by 1 ball before coming to rest.
Now, we know that,
$V^{2}=u^{2}+2gh$
$H=\frac{v^{2}}{2g}$
Thus, $h_{1}=\frac{u{_{1}}^{2}}{2g}$
$=\frac{4u^{2}}{2g}$
and $h_{2}=\frac{u^{2}}{2g}$
Now, $h_{1}-h_{2}=15\; \; \; \; \; \; \; \; ........(given)$
Thus, $u=10\; m/s$
$h_{1}=20\; m\; and\; h_{2}=5\; m$
Calculating time for 1^st ball,

$V_{1} = u_{1} + gt$

$0 = 20 - 10t_{1}$

Thus, $t_{1} = 2 s$
Now, calculating time for second ball,
$V_{2} = u_{2} + gt_{2}$
$0 = 10 - 10t_{2}$
This, $t_{2}=1\; s$
Thus, time intervals between these two balls will be,
$=t_{1} - t_{2}$
$=(2-1)$
$= 1\; second.$