A particle is moving in a straight line such that its distance `s` at any time `t` is given by `s=(t^4)/4-2t^3+4t^2-7.` Find when its velocity is maximum

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

We have,
`s=(t^(4))/(4)-2t^(3)+4t^(2)-7`
`implies (ds)/(dt)=t^(3)-6t^(2)+8t and (d^(2)S)/(dt^(2))=3t^(2)-12t+8`
`implies v=t^(3)-6t^(2)+8t and a=3t^(2)-12t +8`, where v = velocity and a = acceleration
Now, `a=3t^(2)-12t+8 implies (da)/(dt)=6t-12 and (d^(2)a)/(dt^(2))=6`
For maximum or minimum value of a, we must have
`(da)/(dt) = 0 implies t=2`
Clearly, `(d^(2)a)/(dt^(2))=6 gt 0` for all t
Hence, acceleration is minimum when t = 2.