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Question

A particle is moving along the positive x-axis and at t=0, the particle is at x=0. The acceleration of the particle is a function of time. The acceleration at any time t is given by a=2(1[t]) where [t] is the greatest integer function. Assuming that the particle is at rest initially, the displacement of the particle in 4 s is

A
1 m
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B
2 m
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C
6 m
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D
4 m
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Solution

The correct option is B 2 m
Given that at t=0, x=0 and accelaration a=2(1[t]) where [t] is the greatest integer function.
dvdt=22[t]
v0dv=40(22[t])dt
v=40(22[t])dt
For t=0 to t=1 s,[t]=0, v=102dt=2 m/s=v1
For t=1 s to t=2 s,[t]=1, v2v1=210dt=0v1=v2=2 m/s
For t=2 s to t=3 s,[t]=2, v3v2=322dt=2 m/sv3=2+2=0 m/s
For t=3 to t=4 s,[t]=3, v4v3=434dt=4 m/s=v4



Displacement is given by area under the vt graph
= Area above x-axis - Area below x-axis
=12(3+1)×212×4×1=2 m
Displacement in 4 s=2 m

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