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A particle moves in x-y plane and at tim...

A particle moves in x-y plane and at time t is at the point `(t^2, t^3 -2 t),` then which of the following is/are correct?

A

At `t = 0,` particle is moving parallel to y-axis

B

At `t = 0,` direction of velocity and acceleration are perpendicular

C

At `t = sqrt(2/3)`, particle is moving parallel to x-axis

D

At `t=0,` particle is at rest

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`x=t^2 rArr v_x=(dx)/(dt)=2t`
`rArr a_x=(dv_x)/(dt)=2`
`rArr y=t^3-2t`
`rArr v_y=(dy)/(dt)=3t^2-2`
`rArr a_y=(dv_y)/(dt)=6t`
At `t=0 v_x=0,v_y=-2, a_x=2` and `a_y=0`
`:. v=-2 hatj` and `a=2 hati`
or `v bot a`
At `t=sqrt(2/3), v_y=0,v_x!=0.`
Hence, the particle is moving parallel to x-axis.
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