NCERT Exemplar (Part - 1) - Motion in a Straight Line Notes | EduRev

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?




Ans. (b)
Solution: 

• We need to identify the graph in which there is one displacement for different timings. It means that these displacements would be in opposite directions, and when we add these opposite displacements, net displacement would be zero or average velocity would be zero. This thing is only possible in the graph (b).
  
• Suppose we draw a line parallel to the time axis from the point (A) on the graph at t = 0 sec. This line can intersect the graph again at B. At this point, The change in displacement (O-T) time is zero, i.e., displacement at A and B are equal so as the change in displacement is zero, so the average velocity of the body vanishes to zero.
  
  

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
Ans. (a)
Solution: 
Key concept: 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).
  In the 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. 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. Thus, x < 0; a > 0. As displacement is in a negative direction, velocity will also be negative, i.e. v < 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 – v_o T < x < v_o T.
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points.
Ans. (b)
Solution: 
Key concept: Instantaneous speed: It 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

• As instantaneous speed is less than maximum speed. Then either the velocity is increasing, or it is decreasing. For maximum and minimum displacement, we have to keep in mind the magnitude and direction of maximum velocity.
• As maximum velocity in a positive direction is v₀, the magnitude of maximum velocity in the opposite direction is also v₀.
• Maximum displacement in one direction = v₀T Maximum displacement in opposite directions = -v₀T
  Hence, -v₀T < x <  v₀T.

Important Point: 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




Ans. (c)
Solution: Time t₁ taken in half distance = t₁ = L/v₁ 
► Time t₂ taken in half distance t₂ = 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
Ans. (b)
Solution: 
Key concept: Instantaneous velocity: 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 particle starts moving towards negative axis, then at t = 0, with a speed 4 m/s, at t = 2 it stops and start coming backward. At t = 4 its speed is +4 m/s.
  
• v-t graph is shown in graph (a) and 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/(t₂ – t₁)
(c) t₁t₂/(t₁ + t₂)
(d) t₁ – t₂
Ans. (c)
Solution: Let L be the length of the escalator.
► Velocity of girl w.r.t. ground v_g = L/t₁
► Velocity of escalator w.r.t. ground v_e = L/t₂   
► Effective Velocity of girl on moving escalator with respect to ground = v_g + v_e


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


Q.7. The variation of quantity A with quantity B, plotted in Fig., describes a particle's motion 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.
Ans. (a, c, d)
Solution: 

► If B represents velocity, then the graph becomes the v-t graph is a straight line, so it is uniformly accelerated motion, so motion is not uniform. Verifies option (a), (d).
► If B represents time and A represents displacement, then the graph becomes (s-t) graph. Here s-t graph is a straight line that represents uniform motion, so verifies the option (c).

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.
Ans. (a, c, e)
Solution: Slope of x-t graph gives v = dx/dt

• A graph (x-t) is parallel to the time axis, 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 B and C is a graph (x-t), that is 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 the 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 time axis, so 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 - 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
Ans. (a, d)
Solution: 

• Position of the particle is given as a function of time, i.e. x = t – sint
  By differentiating this equation w.r.t. time we get the velocity of the particle as a function of time.
  
  
• If we again differentiate this equation w.r.t. time we will get an acceleration of the particle as a function of time.
  
  ► As acceleration a > 0 for all t > 0
  Hence, x(t) > 0 for all t > 0
  ► Velocity v = 1 - cos
  When, cos t = 1, velocity v = 0
  ► v_max = 1 - (cos t)_min = 1 - (- 1) = 2
  ► v_min = 1 - (cos t)_max = 1 - 1 = 0
  Hence, v lies between 0 and 2.
  
  ► When t = 0; x = 0, v = 0, a = 0
  ► When t = π/2 ; x = positive, v = 0, a = -1 (negative)
  ► When t = π, x = positive, v = positive, a = 0
  ► When t = 2π; x = 0, v = 0, a = 0

Important points:
(i) When the sinusoidal function is involved in an expression, we should be careful about sine and cosine functions.
(ii) We should be very careful when calculating the maximum and minimum value of velocity because it is in inverse relation with cost in the given expression.

Q.10. A spring with one end attached to amass and the other to rigid support is stretched and released.
(a) Magnitude of acceleration, when 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.
Ans. (a, c)
Solution: 

• 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

Hence, when just released. x = x_max

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.
Ans. (b, c, d)
Solution: 

• 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 observer. 
• Compared to the velocity of trains (10 m/s) speed of the ball is less (1 m/s). (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 s.
• Since the collision of the ball is perfectly elastic, there is no dissipation of energy; hence total momentum and kinetic energy are conserved.
  Since the train is moving with a constant velocity, it will act as an inertial frame of reference as Earth and acceleration will be the same in both frames.

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.

Very Short Answer Type Questions 

Q.1. Match the following.

Ans. (a) (iii); (b) (ii); (c) (iv);(d) -(i)
Solution: 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).
Ans. 

• 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.
Ans. 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.
  
  If we again differentiate this equation w.r.t. time we will get acceleration of the particle as a function of time.
  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.
Ans. 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?
Ans. 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).