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Motion in straight Line





Motion in straight Line Solved Examples Part 1


Four position -time graph are shown below.
a.Position time graph 1

b.Position time graph 2

c.Position time graph 3

d.Position time graph 4


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.Graph 1 velocity-time relationships for a particle released from rest to fall under gravity

b.Graph 2 velocity-time relationships for a particle released from rest to fall under gravity

c.Graph 3 velocity-time relationships for a particle released from rest to fall under gravity

d.Graph 4 velocity-time relationships for a particle released from rest to fall under gravity


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


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