Motion In A Straight Line
Class 11th Physics NCERT Exemplar Solution
Mcq I
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?
A. a
B. b
C. c
D. d
Answer:
Average velocity is the ratio of the total displacement covered to the total time taken to cover the displacement. In the graphs, we need to find one in which there is same displacement at any two timings which is evidently in graph (b).
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 upward direction is positive, the lift is moving downwards, hence, the displacement and the velocity will be negative. Acceleration is the change of velocity. Since the lift is about to stop, there will be retardation/ negative acceleration in downward direction which is same as positive acceleration in upward direction. Thus, acceleration will be positive.
Question 3.
In one dimensional motion, instantaneous speed v satisfies 0 ≤ v < v0.
A. The displacement in time T must always take non-negative values.
B. The displacement x in time T satisfies – v0 T < x < v0 T.
C. The acceleration is always a non-negative number.
D. The motion has no turning points.
Answer:
The instantaneous speed varies from 0 to v0. This means, the instantaneous velocity can vary from -v0 to +v0 (considering one direction as positive and the other direction as negative). Thus, the displacement will vary from -v0T to +v0T.
Question 4.
A vehicle travels half the distance L with speed V1 and the other half with speed V2, then its average speed is
A. 
B. 
C. 
D. 
Answer:
Average speed is the ratio of total distance covered to total time taken to cover that distance.
For first half, let the time taken to cover the distance L be t1
For the second half of the motion, let the time taken to cover the distance L be t2.
Question 5.
The displacement of a particle is given by x = (t – 2)2 where x is in metres and t in 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:
Velocity (v) of the particle is given by derivative of the displacement with respect to time t.
(The modulus operator will account for adding negative displacement to the total distance as well.)
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 t2. The time taken by her to walk up on the moving escalator will be
A. (t1 + t2)/2
B. t1t2/ (t2 – t1)
C. T1t2/ (t2 + t1)
D. t1 – t2
Answer:
Let the height to which the escalator takes the girl be h. Then, the velocity of the girl with respect to the ground = vg= h/t1
The velocity of the escalator with respect to the ground = ve = h/t2
Let the relative velocity of the girl with respect to the escalator while moving up be v.
Then,
v can also be written as h/t where t will be the total time taken by the girl to reach the top while walking on a moving escalator.
Mcq Ii
Question 1.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:
Option (a) is clearly correct as Quantity B represents time.
Option (b) is incorrect because a velocity – time graph with a positive slope does not represent uniform motion.
Option (c) is correct because if Quantity A represents displacement, the motion is uniform because the object is moving with constant velocity (slope of displacement - time graph).
Option (d) is correct because the slope of the graph is same/constant, which in case of a velocity-time graph gives that acceleration (=slope of a velocity time graph) is uniform/constant.
Question 2.
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:
Option (a) is correct because at t=0, the slope of the x-t graph which represents velocity is zero.
Option (b) is incorrect because slope is negative which implies that velocity is decreasing which in turn means that the object is retarding/decelerating/negatively accelerating, thus, a < 0.
Option (c) is correct because the slope of the graph is zero again over here.
Option (d) is incorrect because the total displacement between A and D is negative which means the average velocity is negative.
Option (e) is correct because, the slope of the graph at D is steeper as compared to that at E.
(Note that the option (e) says speed and not velocity, so the sign of the slope is not important only the magnitude is important.)
Question 3.
For the one-dimensional motion, described by x = t–sint
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:
Option (a) is correct. For x(t) > 0 for t > 0
Dividing by t on both sides, this wouldn’t affect the inequality, since we have assumed t > 0.
This is always true for any t > 0.
Option (b) is incorrect. v(t) = dx/dt = 1 – cos(t) which is equal to zero at t = 2nП
Option (c) is incorrect. a(t) = dv/dt = - sin(t) which can have both positive and negative values.
Option (d) is correct. v(t) = 1 – cos(t). cos(t) lies between 1 and -1, so v(t) lies between 0 and 2 (when cos(t) = 1 and when cos(t) = -1 respectively).
Question 4.
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:
When the mass is released, it will undergo simple harmonic motion. The force on the block before it is released is F = -kx.
Thus, the acceleration will be a = -kx/m. Magnitude of the acceleration will thus depend on x and will be maximum when x is maximum. Thus, option (a) is correct.
At equilibrium position, x = 0, thus, acceleration will be zero. Thus, option (b) is incorrect. At equilibrium position, all the potential energy will be converted to kinetic energy, thus, the speed will be maximum. Option (c) is correct.
Option (d) is correct because the speed is minimum at maximum compression or maximum elongation of the spring.
Question 5.
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 second interval is fixed.
D. the acceleration of ball is the same as from the train.
Answer:
The speed of the ball when travelling in the same direction as that of the train is (10+1) m/s = 11 m/s. The speed of the ball just after collision is 9 m/s in the opposite direction. Thus, it changes after the collision. Time taken by the ball travelling from one end to other to collide = 10/1= 10s. Thus, option (b) is correct.
For the first 10 seconds the speed of the ball is 11 m/s. For the next 10 seconds the speed of the ball is 9 m/s. The total distance covered by the ball is 20 m. The total time taken by the ball to do that is 20 seconds. Thus, average speed during these 20 seconds = 1 m will be uniform for every 20 seconds. Thus, option (c) is correct.
The train is not accelerating itself and is moving with a constant speed. It thus acts as an inertial frame of reference. Earth/ground is also an inertial frame of reference. Thus, the acceleration will be same in both the frames. Option (d) is correct.
Vsa
Question 1.Refer to the graphs in Fig 3.1. Match the following.
Graph Characteristic
(a) (i) has v > 0 and a < 0 throughout.
(b) (ii) has x > 0 throughout and has a point with v = 0 and a point with a = 0.
(c) (iii) has a point with zero displacement for t > 0. (d) (iv) has v < 0 and a > 0.
Answer:
(a) - (iii)
(b) – (ii)
(c) – (iv)
(d) – (i)
(a) - (iii) has a point with zero displacement for t > 0.
(b) – (ii) has x > 0 throughout and has a point with v = 0 and a point with a = 0.
x > 0 throughout in this graph and since the slope in changing. Thus, there is a point of a = 0 and a point of v = 0.
(c) – (iv) has v < 0 and a > 0.
Since, the slope is negative throughout, v < 0 and since the graph is concave upwards (second derivative is greater than zero), a > 0.
(d) – (i) has v > 0 and a < 0 throughout.
The slope is positive throughout the graph, thus, v > 0. The graph is concave downwards (second derivative of x is negative), thus, acceleration is negative.
Question 2.
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 time. (Take acceleration in the backward direction as positive).
Answer:
We shall be considering only the horizontal motion of the ball (no effect due to gravity). The ball is moving uniformly before hitting the bat, thus, it will have no/zero acceleration. When the bat hits the ball an impulsive force is generated which provides acceleration to the ball. The ball returns with the same speed thereafter.
Question 3.
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:
Since, the question talks about periodic motion, it can be easily related to sine and cosine functions as they are periodic in nature.
(a) x = t – sin(t), this particle will move in positive x-direction if and only if t > sin(t).
When t = 0, x(t) = 0
When t = π , x(t) = π
When t = 2π , x(t) = 2π
(b) x = cos(t)
At t = 0, x(t) = 1
At t = π /2, x(t) = 0
At t = π , x(t) = -1
Question 4.
Give example of a motion where x > 0, v < 0, a > 0 at a particular instant.
Answer:
Consider the motion described by x(t) = t2 – t +1/4.
Then velocity v(t) can be written as
Acceleration a(t) can be written as
We can clearly see that for all time instants less than 0.5 seconds, the required conditions i.e. x > 0, v < 0 and a > 0 are satisfied.
(Important note: To find such an equation we can do reverse calculation, like considering acceleration as a positive constant and then integrating to find velocity and displacement according to the given conditions.)
Question 5.
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 acceleration of the object in the fluid is a = g – bv.
We know that acceleration is the derivative of velocity with respect to time.
Integrating with proper limits
As t approaches infinity, the exponential factor reaches zero. Thus, the object has a constant speed of g/b.
Alternative Way:
After a long time of release, speed is constant (say v’) which implies there is no acceleration. Acceleration is given by
This implies, the constant speed v’= g/b.
Sa
Question 1.A ball is dropped and its displacement vs time graph is as shown Fig. 3.4 (displacement x is from ground and all quantities are +ve upwards). (a) Plot qualitatively velocity vs time graph. (b) Plot qualitatively acceleration vs time graph.
Answer:
The ball is dropped which means it has zero initial velocity and acceleration is –g i. e acceleration due to gravity (upwards direction is +ve).
The velocity of the ball increases when it goes downwards and decreases when it goes upwards.
Seeing the displacement graph, we see that the slope is first positive, then zero and then negative. This gives us the nature of velocity - time graph.
The acceleration is ‘-g’ (-ve sign because it’s always in downward direction).
Question 2.
A particle executes the motion described by X(t) =X0(l - e-yt) t ≥ 0,
(a) Where does the particle 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:
x(t) = x0( 1 – e-yt)
(a) The particle starts at t = 0 that is at x = 0.
For velocity, we need to differentiate the given equation with respect to time.
At t = 0, v = 
. This is the starting velocity of the object.
(b) x(t) will be maximum when ‘t’ approaches infinity and will be equal to x0. It is increasing from zero to x0 with time.
v(t) will be maximum at t=0 and will be x0y. It will be minimum as ‘t’ approaches infinity. Clearly it is decreasing with time.
a(t) will be maximum as ‘t’ approaches infinity and will be equal to 0. Its minimum value will be (–x0y2) at t = 0. We can note that acceleration is increasing (from (–x0y2) to 0) with increasing time.
Question 3.
A bird is tossing (flying to and fro) 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 27km/h. The bird starts moving from first car towards the other and is moving with the speed of 36km/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:
Speed of car 1 = 18 km/h
Speed of car 2 = 27 km/h
Relative speed of the cars = (18+27) = 45 km/h. (since they both are moving in the same direction)
Total distance between the cars = 36 km/h.
Time taken to cover the distance = Total distance / Relative Speed = 36/45 = 0.8 hours
Distance covered by the bird in 0.8 hours = Speed x Time = 36 x 0.8 = 28.8 km
Displacement will be the distance between the starting point and the point of meeting of the cars.
Since the cars meet in 0.8 hours, the distance from the starting point will be 18 x 0.8 = 14.4 km. (The bird starts from the car running at 18 km/h). Thus, the displacement of the bird is 14.4 km.
Question 4.
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 (horizontal) 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? (take g = 10 m/s2)
Answer:
For horizontal motion
Initial velocity in the vertical direction (uV) = 0 m/s
Initial velocity in the horizontal direction (uH) = 9 m/s
Distance between the two buildings = x(t) = 10m
Height difference between two buildings = y(t) = 9m
Acceleration (a) = g = 10 ms-2{Downward direction is assumed to be positive}
The motion in the vertical direction.
Horizontal distance travelled in time (t)
For horizontal motion
Initial velocity (uH) = 9 ms-1
Horizontal distance covered (H) in time t,
This is more than 10 m (the horizontal difference between the buildings) and thus the man will be able to make the jump.
Question 5.
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 ball that has been dropped down (taking upwards direction as positive)
u (initial velocity) = 0
a (acceleration) = -g
v (final velocity ) = u+at
For the ball that has been thrown up with a velocity of 40 m/s
u’ (initial velocity) = 40
a’ (acceleration) = -g
v’ (final velocity ) = u+a’t
Question 6.
The velocity-displacement graph of a particle is shown in Fig. 3.5.
(a) Write the relation between v and x.
(b) Obtain the relation between acceleration and displacement and plot it.
Answer:
The graph is a straight line with intercepts of the axis given. Using the intercept form of straight lines, we can write the equation for this graph.
Plotting the above equation on graph, we have
La
Question 1.It is a common observation that rain clouds can be at about a kilometre altitude above the ground.
(a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in km/h. (g = 10m/s2)
(b) A typical rain drop is about 4mm 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 rain drops is 5 cm. (Assume that umbrella is circular and has a diameter of 1m and cloth is not pierced through !!)
Answer:
Height = 1km = 1000m
Initial Speed = 0 (freely falling under gravity)
Acceleration (a) = g (freely falling under gravity)
Let final speed be v m/s
Using equation of motion, we have
(b) Diameter of the rain drop (d) = 4 mm = 4 x 10-3 m
Volume of the rain drop (considering it to be spherical) = 
Mass of the rain drop = density of water x volume of drop = 
Mass = 3.349 x 10-5 kg
Momentum of the drop = Mass x velocity = 3.349 x 10-5 x 140 = 4.68 x 10-3 kg-m/s
(c) Time required to flatten the drop will be same as the time it takes to cover a distance equal to its diameter close to the ground.
(d) Force exerted = Change of momentum/time
(e) Radius of the umbrella = � m = 0.5 m
Area of the umbrella = 
Lateral separation between rain drops = 5 cm = 5 x 10-2 m
Number of rain drops falling on the umbrella = 
Total force exerted = (314 x 168) = 52752 N
Question 2.
A motor car moving at a speed of 72km/h cannot come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a highway the car is behind the truck both moving at 72km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5s.
(Comment : This is to illustrate why vehicles carry the message on the rear side. “Keep safe Distance”)
Answer:
Time required by the truck to reach 0 km/h (t1) = 5s
Initial speed of the truck (u) = 72 km/h = 20 m/s
Final speed (v) = 0 m/s
Deceleration of the truck ( �atruck) can be calculated using equations of motion.
(Negative sign denotes deceleration/retardation)
For the car, time required to stop = 3s
We can find the deceleration of the car similarly to how we found of the truck above.
The human response time is 0.5s. This means after the truck gives signal, the car will begin to retard after 0.5 seconds. For these 0.5 seconds, the car will move at a uniform speed.
Let the car be at a distance ‘x’ when the truck gives the signal and ‘t’ be the time taken to cover this distance.
Velocity of car (v,) after time t will be
Velocity of truck after time t (v,,)
For avoiding the car to bump into the truck
On solving for t, we get
Distance travelled by the truck in t = 1.25 seconds (s1)
Distance travelled by the car in time t is the sum of the distance travelled by the car with uniform velocity for 0.5 seconds and the distance travelled by the car while retarding for (t – 0.5) seconds (s2)
Safe distance = s2 – s1 = (23.125 – 21.875) m = 1.25 m
Question 3.
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 < 6 s in 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 time is its acceleration maximum in magnitude?
(d) How many cycles (counting fractions) are required to reach the top?
Answer:
To find maximum velocity, we can differentiate the velocity expression and find time at which there will be a maxima or minima.
For 0< t < 3, v(t) = 2t (3-t) = 6t – 2t2
Thus, at t = 1.5 s, there is a maxima, the velocity is maximum at t = 1.5 s.
(b) Average velocity is given by
Now,
This is a quadratic equation, we can solve it using our knowledge of quadratic equations.
(c) Acceleration is the first derivative of velocity
Now this is for the time-period 0 to 3 seconds, so |6-4t| will be maximum either when the negative factor deducts nothing from 6 or when it deducts maximum from 6 which is at t = 0s and t = 3s respectively.
(d) To count how many cycles are required, we need to find the height covered in 1 cycle.
From 0< t< 3s, let the height covered be s1
For 3< t< 6s, let the height covered be s2
Total distance covered in cycle = 4.5 m
Total distance to be covered = 20 m
Question 4.
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 (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at t = 2 s. 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:
The distance between the two balls can be constant if and only if they have the same velocity. For the balls to have the same velocity, the velocity of the first ball released should clearly be same as the second ball when it is released at time (t) otherwise this scenario won’t be possible.
Let the initial velocity of the first ball be (u).
Let the second ball be released at time t = t seconds.
Then the initial velocity of the second ball and the final velocity of the first ball at time (t) is u/2
For the first ball
(v -> final velocity, u -> initial velocity, t->time).
(g is acceleration due to gravity)
The first ball would have gone up by 15 m till now because then only they will maintain a constant distance (s) of 15 m thereafter.
Taking (g) as 10m/s2, we get u = 20 m/s. Putting the value of (u) to get (t) which is equal to u/2g, we get t = 1 s.
Thus, the first ball was thrown at 20 m/s.
The second ball was thrown at 10m/s.
The time interval was 1 second.