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Question

A particle starts from rest and has acceleration a=2t4 in m/s2. The displacement of the particle as a function of time is
[Take x=0 at t=0]

A
x(t)=2t332t2
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B
x(t)=t33+t2
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C
x(t)=t332t2
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D
x(t)=t32t2
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Solution

The correct option is C x(t)=t332t2
Given, a(t)=2t4
As we know that a=dvdt v0dv=t0a(t)dt
and also v=dxdt x0dx=t0v(t)dt
Thus, v=t0a(t)dt=t0(2t4)dt
v(t)=t24t
also, x=t0v(t)dt=t0(t24t)dt
x(t)=t332t2
Hence, displacement of the particle as a function of time is x(t)=t332t2

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