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Question

A particle starts from rest with a time-varying acceleration, a=(2t4). Here t is in s and a in m/s2. After what time (in s) particle comes to rest?

A
1
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B
4
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C
3
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D
2
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Solution

The correct option is B 4
Acceleration – time graph of the particle is a straight line as shown in the figure. Area enclosed by the at graph and the x-axis gives the change in velocity .

Particle will come to rest, when area above time axis and below becomes equal.
Area below time axis, A1=12×4×2=4
Area above time axis, A2=12×t×a=A1=4
at=8(1)
Slope of the line, m=42=2
Hence at=2(2)
Solving (1) and (2), we get
t=2 s
Hence after t+2 i.e. 4 sec, particle comes to rest.

Alternate:
Given, particle starts from rest i.e. at t=0,v=0
As we know that
a=dvdt v(t)=t0adt
v(t)=t0(2t4)dt
v(t)=t24t
The particle will come to rest only if v=0
t24t=0 t(t4)=0
t=0,4
Hence, after 4 sec, particle will come to rest.

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