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If the velocity of a particle is v = At ...

If the velocity of a particle is `v = At + Bt^2`, where `A` and `B` are constant, then the distance travelled by it between `1 s` and `2 s` is :

A

`(3)/(2)A+4B`

B

`3A+7B`

C

`(3)/(2)A+(7)/(3)B`

D

`(A)/(2)+(B)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
`v=At+Bt^(2)`
`(dx)/(dt)=At+Bt^(2)`
`int_(0)^(x)dx=int_(1)^(2)(At+Bt^(2))dt`
`x=(A)/(2)(2^(2)-1^(2))+(B)/(3)(2^(3)-1^(3))=(3A)/(2)+(7B)/(3)`
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