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A particle is moving along x-y plane. It...

A particle is moving along x-y plane. Its x and y co-ordinates vary with time as `x=2t^2 and y=t^3` respectively. Here, x and y are in metre and t is in seconds. Find average acceleration between the time interval from `t=0` to `t=2 s.`

A

`=(6 hati+6 hatj) m//s^2`

B

`=(4 hati+8 hatj) m//s^2`

C

`=(4 hati+6 hatj) m//s^2`

D

`=(1 hati+6 hatj) m//s^2`

Text Solution

Verified by Experts

The correct Answer is:
C
The position vector of the particle at any time t can be given as
`r=x hati+y hatj=2t^2 hati+t^3 hatj`
The instantaneous velocity is `v=(dr)/dt=(4t hati+3t^2 hatj)`
Now, `a_(av)=(Delta v)/(Delta t)=(v_f-v_i)/(Delta t)=(v_(2 sec)-v_(0 sec))/(2-0)`
`([(4)(2) hat i+(3)(2)^2 hat j]-[(4)(0) hat i+(3)(0)^2 hat j])/2`
`=(4 hati+6 hatj) m//s^2`
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