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The acceleration of a particle in ms^(-1...

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

Text Solution

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`a=(dv)/(dt)=(3t^(2)+2t+2)dt`
`dv=(3t+2t+2)dt`
`intdv=int(3t^(2)+2t+2)dt`
`v=t^(3)+t^(2)+2t+C`
`c=2m//s,v=18m//s` at t=2s
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