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Question

A particle is placed at the origin. It starts from rest at t=0 and undergoes an acceleration a as shown in the figure. Which of the following options correctly represent velocity v vs. time t and displacement x vs. time t plots?

A
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B
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C
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D
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Solution

The correct option is A
Given motion has two phases
(i) accelerating phase from t=0 to t=2 s and
(ii) decelerating phase from t=2 s to t=4 s.
In accelerating phase,
a=3 m/s2 (From t=0 s to t=2 s)
v=3t(u=0)

Velocity will increase linearly with time
At t=2 s,v=6 m/s
Since v=3t=dxdt
Integrating above equation, we get
x=32t2(1)
Displacement varies parabolically with time.
In decelerating phase,
a=3 m/s2 (From t=2 s to t=4 s)
dv=3 dt
v=3t+c
On putting t=2 s and v=6 m/s , we get
c=12
v=3t+12=dxdt
Integrating above equation,
x=32t2+12t+C1
From equation (1), at t=2 s, x=6 m,
C1=12
Hence x=32t2+12t12
Displacement varies parabolically with time.

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