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A particle is moving with a velocity of ...

A particle is moving with a velocity of `v=(3+ 6t +9t^2) m/s`. Find out
(a) the acceleration of the particle at `t=3 s.`
(b) the displacement of the particle in the interval `t=5s` to `t=8 s.`

A

`60`

B

`20`

C

`10`

D

`15`

Text Solution

Verified by Experts

The correct Answer is:
A
(a) Accleration of particle, `a=(dv)/(dt)=(6+18t) cm//s^2`
At `t=3s,`
`a=(6+8xx3) cm//s^2`
`=60 cm//s^2`
Given, `v=(3+6t+9t^2) cm//s`
or `(ds)/(dt)=(3+6t+9t^2)`
or `ds=(3+6t+9t^2)dt`
`:. int_0^s ds=int_5^8(3+6t+9t^2)dt`
`:. s=[3t+3t^2+3t^3]_5^8`
or `s=1287 cm`
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