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The acceleration a in ms^-2 of a particl...

The acceleration a in `ms^-2` of a particle is given by `a=3t^2+2t+2`, where t is the time. If the particle starts out with a velocity `v=2ms^-1` at `t=0`, then find the velocity at the end of `2s`.

A

`12 ms ^(-1)`

B

`18 ms ^(-1)`

C

`27 ms ^(-1)`

D

`36 ms ^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given , `a = (dv)/(dt) =3 t^(2) + 2t + 2`
or `dv = (3t^2+2t +2) dt`
Integrating both sides , we get
`int_(u)^(v)dv = int_(0)^(t)(3t^() +2t+2)dt " or " v - u = ((3t^(3))/3+(2t^(3))/3+2t)_(0)^(t)`
or `v = u +t^(3) +t^(2) +2t`
At ` t = 2 , s ,v = 2 + (2) ^(3) + (2)^(2) +2 xx 2 = 18 ms ^(-1)`
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