    Question

# A particle starts from rest with a time-varying acceleration, a=(2t−4). 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 4Acceleration – time graph of the particle is a straight line as shown in the figure. Area enclosed by the a−t 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(2t−4)dt ⇒ v(t)=t2−4t The particle will come to rest only if v=0 ⇒ t2−4t=0⇒ t(t−4)=0 ⇒ t=0,4 Hence, after 4 sec, particle will come to rest.  Suggest Corrections  0      Similar questions  Explore more