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

A

3 A +7B

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
B
Velocity of the particle,
v = At +`Bt^(2)`
or, `(ds)/(dt) = At+Bt^(2)`
or, `int_(0)^(s)ds=int_(1)^(2)(At+Bt^(2))dt`
or, `s=[(At^(2))/(2)+(Bt^(3))/(3)]_(1)^(2)=2A+(8B)/(3)-(A)/(2)-(B)/(2)=(3A)/(2)+(7B)/(3)`
`:.` The distance travelled by it between 1 s and 2s
`=(3A)/(2)+(7B)/(3)`
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Knowledge Check

  • The charge flowing through a resistance R varies with time t as Q=at-bt^(2) , where a and b are positive constants The total heat produced in R is - (A) (a^3R)/(3b) (B) (a^3R)/(2b) (C) (a^3R)/(b) (D) (a^3R)/(6b)

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