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Question

# A particle is moving along positive xâˆ’ axis with velocity v=Aâˆ’Bt, where A and B are positive constants and 't' is in seconds. If the particle is at the origin initially, then find the coordinates of the position of the particle at t=3AB.

A
(0,3A22B)
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B
(3A22B,0)
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C
(0,3A22B)
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D
(3A22B,0)
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Solution

## The correct option is D (−3A22B,0) Given that , v=A−Bt When vx=0 m/s , t=AB sec Comparing this with v=u+at, we get ux=A m/s , ax=−B m/s2 Using kinematic relation, s1=ut+12at2 x1=uxt+12axt2 , y1=0 [because particle is moving along x-axis] At t=AB sec, we get x1=A(AB)−B2(AB)2 ⇒x1=A2B−A22B=A22B ......(1) After which the particle will retrace its path. Now, to find the position of particle after t′=3AB, we again use s2=ut+12at2 ⇒x2=vxt+12axt2 , y2=0 x2=0+12×(−B)×(2AB)2=−2A2B .....(2) Total displacement, s=s1+s2 ⇒x=x1+x2 , y=y1+y2 Using (1) and (2), we get x=A22B−2A2B=−3A22B and y=0 ∴ at t=3AB sec, (x,y)=(−3A22B,0)

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