NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion in a Straight Line
Have you not noticed how a car accelerates on a straight line, how a train decelerates at a station or how a falling object falls by the influence of gravity? All these experiences are described in terms of one-dimensional motion, which is the foundation of NCERT Class 11 Physics Chapter 3 - Motion in a Straight Line. This chapter enables the students to be aware of the motion in one direction by applying the basic concepts of kinematics.
This Story also Contains
- NCERT Exemplar Class 11 Physics Solutions Chapter 3: MCQI
- NCERT Exemplar Class 11 Physics Solutions Chapter 3 MCQII
- NCERT Exemplar Class 11 Physics Solutions Chapter 3: Very Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 3: Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 3: Long Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 3: Important Concepts and Formulas
- Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion in a Straight Line
- Approach to Solve Exemplar Questions of Chapter 3 Motion in a Straight Line
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- NCERT Exemplar Solutions Class 11 Subject-wise
- NCERT Solutions for Class 11 Physics Chapter-wise
This chapter (NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion in a Straight Line) has several important areas that it addresses; these include position, displacement, velocity, acceleration, graphical analysis of motion, and equations of motion. The concepts are very important in the analysis of real-life motion and often appear in both the board and competitive tests. The NCERT Exemplar solutions are written in a simple step-by-step manner, thus numerical problems and conceptual problems are easy to interpret and solve. Prepared by experienced subject matter experts, NCERT Exemplar Physics Solutions Class 11 Chapter 3 strictly follow the latest CBSE syllabus and NCERT guidelines. The Exemplar NCERT solutions are all oriented toward gaining a good conceptual clarity, preventing errors, and creating a systematic solution to the problems of kinematics. Regular practice of these questions improves analytical thinking and problem-solving confidence. NCERT Exemplar Class 11 Physics Solutions of Motion in a Straight Line can be a very convenient source of revision and studying material to be used during the preparation of exams due to its high emphasis on real-life examples and logical explanations.
NCERT Exemplar Class 11 Physics Solutions Chapter 3: MCQI
NCERT Exemplar Class 11 Physics Solutions Chapter 3: MCQIs are also aimed at assessing the conceptual learning and formula application of the students in objective-type questions. These multiple-choice questions assist students in pinpointing misperceptions, increasing accuracy, and developing speed, which is very beneficial in the effective preparation of exams and the rapid revision of such questions.
Question:1
Answer:
The correct answer is the option (b)Explanation: In graph (b), displacements are in the opposite direction & when we add them, we get a 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 > 0Explanation: The value of x becomes negative as the lift comes downward, i.e., from the 8th to the 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) 8mExplanation: 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
The Motion in a Straight Line NCERT Exemplar Class 11 Physics Solutions: MCQ II are aimed at testing a higher level of conceptual knowledge by means of multiple-choice and reasoning-based objective questions. These questions assist the students to analyse the statements, prevent frequent mistakes and have a more powerful mechanism of responding to higher-order MCQs, which are put in exams.
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 had 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 the particle is at rest at A, since the 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 option (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 that at E.
Rejecting opt (d)-
Average velocity at A is zero as the graph (x-t) is parallel to the 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, $a=\frac{dv}{dt}$
$=\frac{d(1-\cos \; t)}{dt}$
$=+\sin\; t$
vmax will be,
$V_{max}=1-(-1)$
$= 1+1 = 2$
. vmin will be,
$v_{min} = 1-1$
$=0$
Thus, it is clear that v lies between 0 and 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 a 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 the mass is maximum.
Hence, opt (c) is also verified.
At x = 0, the 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) The average speed of the ball over any 20-second interval is fixed.
(d) The acceleration of the ball is the same as that of the train.
Explanation: If we observe the motion from the ground, we will see that the ball strikes the wall every 10 seconds. The direction of the ball is the same since it is moving at a very low speed in the moving train; therefore, it will not change w.r.t observer from the earth.
The speed of the ball can change after a collision; hence, option (a) will be discarded, and option (b) is verified.
The 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 the wall, the initial speed of the ball will be in the direction of the moving train w.r.t the ground and 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 directions. Hence, opt (b), (c) & (d) are verified here.
NCERT Exemplar Class 11 Physics Solutions Chapter 3: Very Short Answer
The Motion in a Straight Line NCERT Exemplar Class 11 Physics Solutions: Very Short Answer aims at ensuring that students learn to be accurate and to-the-point. These solutions enhance prompt memorisation of definitions, formulas and fundamental concepts, thus useful in rapid revision and self-centred examination preparation.
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
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 an 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
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
NCERT Exemplar Class 11 Physics Solutions Chapter 3: Short Answer are the solutions that present sharp and correct answers to conceptually based questions in a systematic way. Such responses assist the students in exercising clarity in presentation, emphasis of important points and also in preparation to score well on examinations.
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 x=0, the velocity increases, and the 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 the 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
Answer:
We can find out the relative speed of the cars by adding the speeds 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
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 is next farther the first edge by $12.7 - 10 = 2.07\; m$.
Question:21
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 the relative velocity of the 1st ball w.r.t. the 2nd
$v_{12} = v_{1} - v_{2} = -gt -(40 - gt) = - gt - 40 + gt = - 40\; m/s$
The speed of one ball increases and the speed of the other decreases at 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
NCERT Exemplar Solutions Chapter 11 Physics Solutions Chapter 3: Long Answer provides in-depth solutions to in-depth questions in a step-by-step manner. The solutions make students learn how to think logically, extract their answers correctly and how to present them clearly, which form the basis of a good score in descriptive tests.
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
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 at least 0.5 seconds to respond; the 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 the brakes with a passing signal to car's 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 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 to its mean position or changes direction in periodic motion, the 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 $=4.5(3)$
$=13.5\; m$
The remaining height will be
$20 -13.5 = 6.5 m$
The monkey can climb up to 9m without slipping, but in the 4th cycle, it will slip, and the height remaining to climb will be 6.5 m.
Net no. of cycle = 4.
Question:26
Answer:
Let $u_{1}$ be the speed of the 1st 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 1st ball,
$V_{1} = u_{1} + gt$
$0 = 20 - 10t_{1}$
Thus, $t_{1} = 2 s$
Now, calculating time for the second ball,
$V_{2} = u_{2} + gt_{2}$
$0 = 10 - 10t_{2}$
This, $t_{2}=1\; s$
Thus, the time intervals between these two balls will be,
$=t_{1} - t_{2}$
$=(2-1)$
$= 1\; second.$
NCERT Exemplar Class 11 Physics Solutions Chapter 3: Important Concepts and Formulas
Important Concepts and Formulas section of the NCERT Exemplar Class 11 Physics Chapter 3 allows the students to recapitulate major concepts and mathematical relationships on one-dimensional motion in a short time. It enhances the clarity of concepts and enhances the pace of solving problems, which is crucial in both numerical and theoretical questions in exams.
Important Concepts
1. Position, Distance, and Displacement
Position defines the location of an object with respect to a reference point. Distance is the total path covered, while displacement is the shortest distance between the initial and final positions with direction.
2. Speed and Velocity
Speed is the rate of change of distance and is a scalar quantity, whereas velocity is the rate of change of displacement and has both magnitude and direction.
3. Acceleration
Acceleration is the rate of change of velocity with time. It can be positive, negative (retardation), or zero, depending on the nature of the motion.
4. Uniform and Non-Uniform Motion
In uniform motion, velocity remains constant, while in non-uniform motion, velocity changes with time due to acceleration.
5. Graphical Analysis of Motion
Position–time, velocity–time, and acceleration–time graphs help analyse motion visually. The slope and area under these graphs have physical significance.
6. Equations of Motion
These equations describe motion with constant acceleration and are widely used to solve numerical problems.
7. Relative Motion (One Dimension)
Relative motion describes the motion of one object with respect to another moving object along a straight line.
Important Formulas
- Average speed:
$
\text { Average speed }=\frac{\text { Total distance }}{\text { Total time }}
$
- Average velocity:
$
\text { Average velocity }=\frac{\text { Total displacement }}{\text { Total time }}
$
- Acceleration:
$
a=\frac{v-u}{t}
$
- First equation of motion:
$
v=u+a t
$
- Second equation of motion:
$
s=u t+\frac{1}{2} a t^2
$
- Third equation of motion:
$
v^2=u^2+2 a s
$
- Relative velocity:
$
v_{A B}=v_A-v_B
$
Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion in a Straight Line
One of the initial chapters is Motion in a Straight Line where students begin solving numerical problems in physics. Class 11 Physics NCERT Exemplar Solutions Chapter 3 Motion in a Straight Line are specially designed to make these basics strong by explaining concepts in a clear and practical way. They make the students learn about motion rationally rather than memorizing the formula.
- The chapter provides the foundation of kinematics. The Exemplar solutions clarify each minor step so that the students could build a solid foundation that they can later apply in all subsequent physics chapters.
- Many students struggle with motion graphs. These solutions explain position–time and velocity–time graphs in a very simple manner, making it easier to understand slopes and areas.
- By showing how to apply equations of motion step by step, the solutions remove confusion and fear related to numerical questions.
- Instead of jumping to formulas, the solutions teach students to analyse the situation first. This improves reasoning and helps in choosing the correct method to solve a problem.
- The Exemplar answers clearly highlight sign conventions, units, and conditions of motion, which helps students avoid common errors in exams.
- Well-structured solutions make last-minute revision easy. Important formulas and concepts can be recalled quickly without reading long explanations.
- Once students are confident with straight-line motion, they find advanced topics like motion in a plane and laws of motion much easier to understand.
Approach to Solve Exemplar Questions of Chapter 3 Motion in a Straight Line
Chapter 3 focuses on motion along a single direction, but Exemplar questions often test clarity of concepts rather than lengthy calculations. A calm and methodical approach helps students solve these questions accurately and score full marks. Following simple steps can make even tricky problems easy to handle.
- Determine whether the object is moving at a constant rate, constant acceleration or variable acceleration. The knowledge of this will guide you in determining whether you will use equations of motion or graphical methods.
- Identify a point of reference and determine the direction that is positive. This is a critical step to prevent errors of signs, particularly on questions on retardation or return motion.
- Record all the quantities provided, such as initial velocity, final velocity, time and distance. The conversion of units at this point averts mistakes in the future.
- Do not try all equations blindly. Select the equation of motion that directly connects the known and unknown quantities.
- If the question includes a graph or asks about area or slope, carefully read the scale and labels. Many Exemplar questions check understanding of graphs rather than calculations.
- Write each step clearly, even if the question looks simple. This helps in self-checking and also ensures step-wise marks in exams.
- Before writing the final answer, quickly check whether the direction, sign, and unit make physical sense.
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise Links provide students with a structured and organised way to study physics concepts chapter by chapter. These links make it easy to access well-explained solutions for numericals, conceptual questions, and derivations as per the latest NCERT and CBSE guidelines. They help students revise efficiently, strengthen conceptual understanding, and prepare confidently for board and competitive examinations.
NCERT Exemplar Solutions Class 11 Subject-wise
NCERT Exemplar Solutions Class 11 Subject-Wise Links offer a convenient and organised way for students to access solutions for Physics, Chemistry, and Mathematics in one place. The links enable the students to study each topic in a systematic manner with correct step-by-step directions that strictly adhere to the latest NCERT syllabus. They are suitable for rapid revision, clarity of the concepts and proper preparation in exams in all science subjects in Class 11.
NCERT Solutions for Class 11 Physics Chapter-wise
NCERT Solutions for Class 11 Physics Chapter-Wise Links help students study physics in a well-organised and systematic manner. By accessing solutions chapter by chapter, learners can easily understand concepts, numericals, and derivations as prescribed by the latest NCERT syllabus. These solutions support effective revision, strengthen fundamentals, and assist students in preparing confidently for school exams and competitive examinations.
Also, Read NCERT Solution subject-wise -
- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology
Check NCERT Notes subject-wise -
- NCERT Notes Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology
Also, Check NCERT Books and NCERT Syllabus here
Frequently Asked Questions (FAQs)
Q: What is the difference between average speed and average velocity?
A:
Average speed is the total distance traveled divided by the total time taken, and it is always a positive scalar quantity.
Average velocity is the total displacement (straight-line distance from the starting point to the endpoint) divided by the total time taken, and it is a vector quantity with both magnitude and direction.
Q: What is the significance of the displacement-time graph?
A:
The displacement-time graph shows how displacement changes with time. The slope represents velocity. A straight line indicates constant velocity, while a curve shows changing velocity (acceleration).
Q: Why is acceleration negative in retardation cases?
A:
Negative acceleration (retardation) occurs when an object's velocity decreases over time, meaning it is slowing down. This is represented by a negative value for acceleration.
Q: How to interpret velocity-time graphs for different types of motion?
A:
Velocity-time graphs:
Constant velocity: A straight, horizontal line.
Acceleration: A straight, sloped line (positive slope for speeding up, negative slope for slowing down).
Uniform deceleration: A straight line with a negative slope.
Non-uniform motion: A curved line indicates changing acceleration.
Q: What are the common mistakes students make in Chapter 3 Motion in a Straight Line?
A:
- Common mistakes include:
- Confusing speed and velocity.
- Mixing up displacement and distance.
- Incorrect use of kinematic equations.
- Misinterpreting graphs.
- Ignoring direction in calculations.
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