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Question

A particle is moving along positive x axis with velocity v=ABt, 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=ABt

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=A2BA22B=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=A22B2A2B=3A22B and y=0

at t=3AB sec,
(x,y)=(3A22B,0)

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