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Acceleration of a particle in x-y plane ...

Acceleration of a particle in x-y plane varies with time as `a=(2t hati+3t^2 hatj) m//s^2` At time `t =0,` velocity of particle is `2 m//s` along positive x direction and particle starts from origin. Find velocity and coordinates of particle at `t=1s.`

A

`(2t^(2) + 3t^(3)) W`

B

`(2t^(2) + 4t^(4)) W`

C

`(2t^(3) + 3t^(4)) W`

D

`(2t^(3) + 3t^(5)) W`

Text Solution

Verified by Experts

`vecF = 2t hati + 3t^(2) hatj`
`m (d vecv)/(dt) = 2t hati + 3t^(2) hatj (m = 1 kg)`
`rArr underset(0) overset(vecv)intdvecv=underset(0)overset(t)int2thati+3t^(2) hat j " "(m=1kg)`
Power `= vecF. Vecv = (2t^(3) + 3t^(5)) W`
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