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Question

# A particle starts from position x=âˆ’x0 at time t=0 in positive x direction according tot he give velocity-time (v-t) graph. The position (x) of particle at any time t can be represented as :

A
x=x0+v0t+12v0t0t2
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B
x=x0+12v0t0t2
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C
x=x0+12v0t0t2
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D
x=x012v0t0t2
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Solution

## The correct option is B x=−x0+12v0t0t2Given: x=−x0;t=t0Use equation of motion: v=u+atv0=(0)+at0v0=at0a=v0t0......Eq:01Now, use equation of motion s=ut+12at2 for distance/position (x) at any time (t);x=(0)t+12at2x=12at2Put the value of 'a' from Eq:01;x=12(v0t0)t2Required position of particle at any time (t): x=−x0+12(v0t0)t2Option (B) is correct.

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