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The acceleration of a particle moving al...

The acceleration of a particle moving along x-direction is given by equation `a=(3-2t) m//s^(2)`. At the instant `t=0` and `t=6 s`, it occupies the same position.
(a) Find the initial velocity `v_(o)`
(b) What will be the velocity at `t=2 s`?

A

`2m//s`

B

`15m//s`

C

`25m//s`

D

`5m//s`

Text Solution

Verified by Experts

The correct Answer is:
D
By substituting the given equation in equation `a=dv//dt`, we have
`dv=(3-2t)dt rArr underset(v_(o))overset(v)(int)dv=underset(0)overset(t)(int)(3-2t)dtrArr v=v_(o)+3t-t^(2)` …(i)
By substituting eq. (i) in equation `v=dx//dt`, we have
`dx=(v_(o)+3t-t^(2))dt rArr underset(x_(o))overset(x)(int) dx=underset(0)overset(t)(int)(v_(o)+3t-t^(2))dt rArr x=x_(o)+v_(o)t+3/2 t^(2)-1/3 t^(3)` ...(ii)
(a) applying the given condition that the particle occupies the same x coordinate at the instant `t=0` and `t=6s` in eq. (ii), we have
`x_(o)=x_(6)rArr x_(o)=x_(o)+6v_(o)+54-72rArr v_(o)=3m//s`
(b) Using `v_(o)` in eq. (i), we have `v=3+3t-t^(2)rArr v_(2)=5m//s`
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