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 :

Options

A. `3A+7B`
B. `(3)/(2)A+(7)/(3)B`
C. `(A)/(2)+(B)/(3)`
D. `(3)/(2)A+4B`

Answer

B

Explanation

Velocity of the particle is given as
`upsilon=AT + Bt^(2)`
where, A and B are constants.
`rArr (dx)/(dt)=At+Bt^(2) " " [because upsilon=(dx)/(dt)]`
`rArr dx=(At+Bt^(2))dt`
Integrating both sides, we get
`int_(x_(1))^(x_(2))dx=int_(1)^(2)(At+Bt^(2))dt`
`rArr Delta x=x_(2)-x_(1)=A int_(1)^(2)t dt+B int_(1)^(2)t^(2)dt`
`=A[(t^(2))/(2)]_(1)^(2)+B[(t^(3))/(3)]_(1)^(2)`
`=(A)/(2)(2^(2)-1^(2))+(B)/(3)(2^(3)-1^(3))`
`therefore` Distance travelled between 1s and 2s is
`Delta x=(A)/(2)x(3)+(B)/(3)(7)=(3A)/(2)+(7B)/(3)`