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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

1) `frac(2)(3)A+frac(7)(3)B`

B

2) `frac(A)(2)frac(B)(3)`

C

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

D

4) `3A+7B`

Text Solution

Verified by Experts

The correct Answer is:
A
Velocity of the particle is `V=At+Bt^(2)`
`(ds)/(dt)= At+Bt^(2),intds=int` `(AT+Bt^(2))dt`
`Therefore s = (At^*(2)/(2)+ B(t^(3)/(3)+C`
`s(t=1s) = (A)/(2)+(B)/(3)+ C,s(t=2s)=2A+ (8)/(3)B+C`
Required distance =s(t=2s)=2A+(8)/(3)B+C`
`=(2A+(8)/(3)B+C)-((A)/(2)+(B)/(3)+C)=(3)/(2)A+(7)/(3)B`
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