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NCERT Exemplar: Motion in a Straight Line - 1 Notes | Study NCERT Exemplar & Revision Notes for JEE - JEE

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Multiple Choice Questions

Q.1. Among the four graphs figure, 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?

• Average velocity: It is defined as the ratio of displacement to time taken by the body.
• Average Velocity=Displacement / Time Taken
• According to this problem, we need to identify the graph which has the same displacement for two timings.
• When there are two timings for same displacement, the corresponding velocities should be in opposite directions.
• As shown in graph (b), the first slope is decreasing that means particle is going in one direction and its velocity decreases, becomes zero at highest point of the curve and then increasing in backward direction. Hence the particle return to its initial position.
• So, for one value of displacement there are two different points of time and we know that slope of x, x-t graph gives us the average velocity.
• Hence, for one time, slope is positive then average velocity is A also positive and for other time slope is negative then average velocity is also negative. As there are opposite velocities in the interval 0 to T, hence average velocity can only vanish in option (B)
• This can be seen in the figure given alongside.
As shown in the graph,  OA=BT (same displacement) for two different points of time

Q.2. A lift is coming from the 8th floor and is just about to reach the 4th floor. Taking ground floor as the 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

• The time rate of change of velocity of an object is called the acceleration of the object.
• It is a vector quantity. Its direction is the same as that of change in velocity (Not of the velocity).
Table: Possible ways of velocity change
• Here we will take an upward direction positively. As the lift is coming in a downward direction, the displacement will be negative, thus, x < 0. As displacement is in a negative direction, velocity will also be negative, i.e. v < 0. We have to see whether the motion is accelerating or retarding.
• We know that due to downward motion displacement will be negative. When the lift reaches the 4th floor and is about to stop velocity is decreasing with time, hence motion is retarding in nature. Hence, a > 0.
• The motion of the lift will be shown like this

Q.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 – vo T < x < vo T.
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points.

• Instantaneous speed is the speed of a particle at a particular instant of time. When we say “speed”, it usually means instantaneous speed.
• The instantaneous speed is the average speed for an infinitesimally small time interval (i.e., ∆ → 0). Thus, Instantaneous speed .
• By definition, instantaneous velocity, v=ds / dt
where, s = displacement, T= time
⟹ ds = v.dt
Integrating both sides
⟹ s(0) = vo(T−0)
⟹ s = voT
• If the velocity is positive or towards the right of origin,
Then s = +voT, maximum displacement = +voT
But if the velocity is negative or towards left of origin
Then s = −voT, maximum displacement = −voT
Hence, the displacement s in time T satisfies −voT<s<voT
• Note: We should not confuse with the direction of velocities, i.e., in one direction it is taken as positive, and in another direction, it is taken as negative.

Q.4. A vehicle travels half the distance L with speed V1 and the other half with speed V2, then its average speed is:

Time t1 taken in half distance = L/v1
Time t2 taken in half distance = L/v2
Total time (t) taken in the distance

Total distance = L + L = 2L

Q.5. The displacement of a particle is given by x = (t – 2)2 where x is in meters and t in seconds. The distance covered by the particle in the first 4 seconds is
(a) 4 m
(b) 8 m
(c) 12 m
(d) 16 m

Instantaneous velocity is defined as the rate of change of position vector of particles with time at a certain instant of time.

i.e., if x  is given as a function of time, the second time derivative of displacement gives acceleration.

• In such types of problems, we have to analyze whether the motion is accelerating or retarding. When acceleration is parallel to velocity, particle velocity increases with time, i.e. motion is accelerated. And when acceleration is anti-parallel to velocity, the velocity of the particle decreases with time, i.e. motion is retarded. During retarding journey, particles will stop in between.
• According to the problem, displacement of the particle is given as a function of time,
x = (t-2)2
• By differentiating this equation w.r.t. time we get the velocity of the particle as a function of time,
v = dx/ dt = d/dt (t-2)2 = 2(t – 2) m/s
• If we again differentiate this equation w.r.t. time we will get an acceleration of the particle as a function of time,
= [1 - 0] = 2m/s2
When t = 0; v = -4 m/s
t = 2 s; v = 0 m/s
t = 4 s; v = 4 m/s
• That means the particle starts moving towards the negative axis, then at t = 0, with a speed 4 m/s, at t = 2 it stops and start coming backwards and at t = 4 its speed is +4 m/s.
• v-t graph is shown in graph (a) and the speed-time graph of the same situation is shown in graph (b).
Distance travelled = Area of the speed-time graph = area OAC + area ABD

Q.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 takes her up in time t2. The time taken by her to walk upon the moving escalator will be
(a) (t1 + t2)/2
(b) t1t2/(t– t1)
(c) t1t2/(t+ t2)
(d) t1 – t2

Let L be the length of the escalator.
⇒ Velocity of girl w.r.t. ground vg = L/t1
Velocity of escalator w.r.t. ground ve = L/t2
Effective Velocity of girl on moving escalator with respect to ground = vg + ve

∴ Time t taken by the girl on moving escalator in going up the distance L is

*Multiple options can be correct
Question for NCERT Exemplar: Motion in a Straight Line - 1
Try yourself:Q.7. The variation of quantity A with quantity B, plotted in Fig., describes a particle's motion in a straight line.

Q.8. A graph of x versus t is shown in Fig. 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.

Correct Answer is Option (a), (c) & (e)

Slope of x-t graph gives v = dx/dt

• The graph (x-t) is parallel to the time axis at point A, so dx/dt is zero or the particle is at rest. After A, slope dx/dt increases, so velocity increases. Verifies option (a).
• Tangent at points B and C are parallel to the time axis, so dx/dt = 0 or v = 0.
• It implies that acceleration a = 0 so it discards option (b) and verifies option (c).
• From the graph, the slope at D is greater than at E. So speed at D is greater than at E. Verifies the option (e).
• Velocity at A is Zero as x-t parallel to the time axis, so the average velocity at A is zero. At D displacement or slope is negative. So, the average velocity at D is negative, not positive discards option(d).

Q.9. For the one-dimensional motion, described 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

Correct Answer is Option (a) & (d)

• The position of the particle is given as a function of time, i.e. x = t – sin t.
Now, sin t < t.
Dividing by t on both sides, this wouldn't affect the inequality, since we have assumed t > 0.
Therefore, sin t / t <1
This is always true for any t > 0
This means that x > 0 for all t > 0. Option a is correct.
• By differentiating this equation w.r.t. time we get the velocity of the particle as a function of time.

Velocity is equal to zero when t = 2nπ
When t = 2π, v = 1 - cos2π = 1-1 = 0
Therefore, option b is incorrect.
• If we again differentiate this equation w.r.t. time we will get an acceleration of the particle as a function of time.

When t = 0;  a = 0
When t = π/2; a = 1 (negative)
When t = π; a = 0
When t = 3π/2; a =-1
When t = 2π;  a = 0
Therefore, a is both postive and negative. Option c is incorrect.
• Now  Velocity v = 1 - cos t
When, cos t = 1, velocity v = 0
vmax = 1 - (cos t)min = 1 - (- 1) = 2
vmin = 1 - (cos t)max = 1 - 1 = 0
Hence, v lies between 0 and 2. Option d is correct.

Q.10. A spring with one end attached to abmass and the other to rigid support is stretched and released.
(a) Magnitude of acceleration, when mass just released is maximum.
(b) Magnitude of acceleration, when at the equilibrium position, is maximum.
(c) Speed is maximum when mass is at an equilibrium position.
(d) Magnitude of displacement is always maximum whenever speed is minimum.

Correct Answer is Option (a) & (c)

• As shown in the figure above when spring is stretched by length v, restoring force will be F = -kx (-ve sign shows that the force is always in the direction opposite to displacement x). Then the potential energy of the stretched spring
• The restoring force is central; hence when the particle is released, it will execute Simple Harmonic Motion about an equilibrium position.
⇒ Acceleration will be
• At equilibrium position, x = 0 ⇒ a = 0
⇒ when just released. x = xmax
Hence, acceleration is maximum. Thus option (a) is correct.
• At equilibrium whole PE will be converted to K.E. so KE will be maximum and hence, speed will be maximum. Thus option (c) is correct.

Q.11. A ball is bouncing elastically with a speed of 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 the ball changes every 10 seconds.
(c) Average speed of the ball over any 20-second interval is fixed.
(d) The acceleration of the ball is the same as from the train.

Correct Answer is Option (b), (c) & (d)

• In this problem, we have to observe the motion from different frames. Here the problem can be solved by the frame of the observer, but here we must be clear that we are considering the Motion from the ground, so we just keep in mind the motion from the frame of the observer. Compared to the velocity of trains (10 m/s) speed of the ball is less (1 m/s).
• The speed of the ball before the collision with the side of the train is 10 + 1 = 11 m/s Speed after collision with the side of train = 10 - 1 = 9 m/s. As speed is changing after travelling 10 m and speed is 1 m/s; hence time duration of the changing speed is 10 . Thus, option (b) is correct.
• For the first 10 seconds the speed of the ball is 11m/s. For the next 10 seconds the speed of the ball is 9m/s. The total distance covered by the ball is 20m. The total time taken by the ball to do that is 20 seconds. Thus, average speed during these 20 seconds =1m 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.
Remember: We should not confuse with a non-inertial and inertial frame of reference. A frame of reference that is not accelerating will be inertial.

Q.1. Match the following.

Correct Answer is (a) - (iii); (b) - (ii); (c) - (iv);(d) - (i)
Let us pick graphs one by one.

• In graph (a):
There is a point (B) on the curve for which displacement is zero. So curve, (a) matches with (iii).
• In graph (b):
In this graph, x is positive (> 0) throughout and at point B, the highest point of the curve the slope of curve is zero. It means at this point v = dx/dt = 0 . Also at point C the dt curvature changes, it means at this point, the acceleration of the particle should be zero or a = 0, So curve (b) matches with (ii).
• In graph (c):
In this graph the slope is always negative, hence velocity will be negative or v < 0. Also the x-t graph opens up, it represents positive acceleration. So curve (c) matches with (iv).
• In graph (d):
In this graph the slope is always positive, hence velocity will be positive or v > 0. Also x-t graph opens down, it represents negative acceleration. So curve (d) matches with (i).

Q.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).

• Impulsive Force is generated by the bat: If we ignore the effect of gravity just by analyzing the motion of the ball in the horizontal direction only, then ball moving uniformly will return back with the same speed when a bat hits it.

• Acceleration of the ball is zero just before it strikes the bat. When the ball strikes the bat, it gets accelerated due to the applied impulsive force by the bat.
The variation of acceleration with time is shown in the graph:

Q.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 backwards.

The equation which contains sine and cosine functions is periodic in nature.

(a) The particle will be moving along positive x-direction only if t > sin t
We have displacement as a function of time, x(t) = t – sin t
By differentiating this equation w.r.t. time we get velocity of the particle as a function of time i.e.
If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time i.e. acceleration a(t) = dv/dt = sin t
when t = 0: x (t) = 0
when t = π;x (t) = π > 0
when t = 0; x (t) = 2π > 0

(b) Equation can be represented by: x(t) = sin t,
At t = 0; x = 0, v = 1 (positive) and a = 0
At t = π/2 ; x = 1 (positive), v = 0 and a = -1 (negative)
At t = π; x = 0, v = -1 (negative) and a = 0
At t = 3π/2 ; x = -1 (negative), v = 0 and a = + 1 (positive)
At t = 2π, x = 0, v = 1 (positive) and a = 0
Hence the particle moving along positive x-direction comes to rest periodically and moves backwards.
As displacement and velocity are involving sin t and cos t, hence these equations represent periodic nature.

Q.4. Give example of a motion where x > 0, v < 0, a > 0 at a particular instant.

Let the motion is represented by: x(t) = A + Be-γt
Let A > B and y > 0     ...(i)

Suppose we are considering any instant of time t, then from Eq. (i), we can say that: x(t) > 0; v(t) < 0 and a > 0

Q.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?

After a long time of released the velocity becomes constant, i.e.,

Given acceleration is a = g - bv...(i)
0 = g - bv [from (i)]  bv = g
v = g/b
Hence, the constant speed after a long time of release is (g/b).

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