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Displacement of a particle moving along ...

Displacement of a particle moving along X-axis is given by `x=at^(2)-bt^(3)`.

A

Particle starts from rest and again comes to rest after time a/3b.

B

Particle starts from origin and again returns to origin after time t=3a/b.

C

Particle starts with zero acceleration and acceleration again becomes zero after time t=a/3b.

D

Acceletration of the particle becomes zero at t=a/3b.

Text Solution

Verified by Experts

The correct Answer is:
D
`x = at^(2) - bt^(3)`
`rArr " " v = (dx)/(dt) = 2at - 3bt^(2)`
`rArr " " a = (dv)/(dt) = 2a - 6bt`
`x = 0 rArr ` the `t^(2) (a-bt) rArr t - 0 and t= a//b`
Hence the particle returns to origin after time t - a/b.
`v = 0 rArr 2at - 3bt^(2) = 0 rArr t(2a-3bt) = 0 rArr t = 0 ` and `t = 2a//3b`
Hence the particle comes to rest after time 2a/3b. a = `0 rArr 2a - 6bt = 0 rArr t = a//3b`.
Hence acceleration of the particle becomes zero after time t = a/3b.
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