 # Motion Graphs Worksheet with Answers

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Question 1. A ball is dropped vertically from a height h above the ground .It hits the ground and bounces up vertically to a height h/2.Neglecting subsequent motion and air resistance ,its velocity v varies with the height h as    Solution(1):
. Before hitting the ground ,the velocity v is given by v2=2gd Which is a quadratic equation and hence parabolic path. Downward direction means negative velocity , After collision ,the velocity becomes positive and velocity decreases Further v12=2g(d/2)=gd Therefore v=v1√2 As the direction is reversed and speed is decreased, Hence (a) is the answer
Question 2. The displacement -time graph of a moving particle is shown below.The instantaneous velocity of the particle is negative at the point a. C
b. D
c. E
d. B

Solution(2):
. The instantaneous velocity is given by the slope of the displacement time graph.
Since slope is negative at point B, instantaneous velocity is negative at B
Hence (d) is correct

Question 3.The velocity -time graph of a moving particle is shown below.Total displacement of the particle during the time interval when there is non zero acceleration and retardation is a. 60m
b. 40m
c. 50m
d. 30m

Solution(3):
. Acceleration is given by the slope of the velocity time graph
Non zero acceleration happened in the time interval 20 to 40 sec as the slope of the graph is non zero in that time interval

Now displacement in that interval is given by the area enclosed the v-t curve during that interval

So area=(1/2)20*3 +20*1=50m
Hence (c) is correct
Question 4. Figure below shows the displacement -time graph of two particles.Mark the correct statement about their relative velocity a. It first increases and then decreases
b. It is a non zero constant
c. it is zero
d. none of the above
Solution(4):
. The instantaneous velocity is given by the slope of the displacement time graph.
As slope of both the particle displacement time graph is constant.That means there individual velocities are constant.
So relative velocity is also constants
so it is a non zero constant as they have different velocity
Hence (b) is correct

Four position -time graph are shown below.
a. b. c. d. Question 5 What all graph shows motion with positive velocities
a. a and c only
b. all the four
c. b and D only
d. b only

Solution(5):
. The instantaneous velocity is given by the slope of the displacement time graph.
Since slope is positive in graph a and c.Positive velocity is there in a and c curve
Hence (a) and (c) are correct
Question 6. What all graph shows motion with negative velocities
a. a and b only
b. all the four
c. a and c only
d. c only

Solution(6):
. The instantaneous velocity is given by the slope of the displacement time graph.
Since slope is negative in graph b and d.Negative velocity is there in b and d curve
Hence (b) and (d) are correct

Question 7.which of the following graph correctly represents velocity-time relationships for a particle released from rest to fall under gravity
a. b. c. d. Solution(7):
.
The velocity will increase with time so b is the correct answer
Hence (b) is correct
Question 8.The v-x graph of a particle moving along a straight line is shown below.Which of the below graph shows a-x graph a. b. c. d. Solution(8):
. The equation for the given graph is
v=-(v0/x0)x + v0 ---(1)
Differentiating both sides we get
dv/dx=-v0/x0 ---(2)
Now
a=v(dv/dx)
or
a=(-v0/x0)[-(v0/x0)x + v0 ]
or
a=mx+c
where m=v02/x0)2
and c=-v02/x0)2

So that means slope is positive and intercept is negative
So (d) is correct