Velocity of a particle varies as `v=2t^3-3t^2` in `(km)/(hr)` If `t=0` is taken at 12:00 noon
Q. Find the time between 12:00 noon and 1:00 pm which speed is maximum

Options

A. 12:00 noon
B. 0.54166666666667
C. 0.45833333333333
D. 0.58333333333333

Answer

A

Explanation

`v=2t^3-3t^2`
`(dv)/(dt)=a=6t(t-1)`
`(dv)/(dt)=0`
`t=0`, 1sec
`(d^2v)/(dt^2)=12t-6implies((d^2v)/(dt^2))_(t=0)=-6`
`((d^2v)/(dt^2))_(t=1)=6`
At `t=0`, The sigh of double derivative is negative hence velocity will be maximum at 12:00 noon. At `t=1`, the sign of double derivative is positive hence the speed will be minimum at 1:00
P.M.