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Question

# Acceleration versus velocity graph of a particle moving in a straight line starting from rest is as shown in figure. The corresponding velocity-time graph would be:

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 D We, know that a=dvdt...(1) From the given a−v graph, we have, a=−mv+C...(2) m & C are positive constants. So, from equation (1) and (2), we have −mv+C=dvdt Integrating on both sides we get, ∫t0dt=∫v0dvC−mv ⇒−1m[ln|C−mv|]v0=t ⇒lnC−mvC=−mt ⇒C−mvC=exp(−mt) ⇒C−mv=Cexp(−mt) ⇒mv=C(1−exp(−mt) ∴v=C(1−exp(−mt))m Hence, the velocity varies exponentially with time. When, t=∞, v=vmax=Cm and the curve gets flatten. This is possible only for option (d). Hence, option (d) is the correct alternative.

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