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=2s and
(ii) decelerating phase from t=2s to t=4s.
In accelerating phase, a=3m/s2(From t=0s to t=2s) ⇒v=3t(∵u=0) ∴
Velocity will increase linearly with time
At t=2s,v=6m/s
Since v=3t=dxdt
Integrating above equation, we get x=32t2(1)
Displacement varies parabolically with time.
In decelerating phase, a=−3m/s2(From t=2s to t=4s) ⇒∫dv=∫−3dt ⇒v=−3t+c
On putting t=2s and v=6m/s , we get c=12 ∴v=−3t+12=dxdt
Integrating above equation, x=−32t2+12t+C1
From equation (1), at t=2s,x=6m, C1=−12
Hence x=−32t2+12t−12
Displacement varies parabolically with time.