Question

if the velocity of the particle varies as v =2t^3-3t^2 in km/hr if t=0 is taken at 12:00 noon at which time the speed of particle is maximum?

Answer 1

It is given that $v(t)=2t^3-3t^2$.

It is required to find the time at maximum velocity.

At maximum velocity, the derivative of the velocity with respect to time; $\frac{dv(t)}{dt}=0$.

Therefore,

$\frac{dv(t)}{dt}=6t^2-6t=0\\ 6t(t-1)=0\\t=0\:\:or\:\:t=1$.

It is given that t=0 is taken at 12:00 noon.

Therefore, time at maximum speed = 12:00 + 1 hour = 13:00 pm.

Answer 2

The particle will attain maximum value at time 11:59 of the next day.

velocity of the given particle is,

v = 2t³-3t²

clearly v has a maxima at t = ∞

i.e. v will be maximum when t is maximum which is 11 : 59 am of the next day as at 12:00 noon again t = 0

Hence the particle will attain maximum value at time 11:59 of the next day.

I hope it helps.