Question

A monkey climbs up a slippery pole for 3 second and subsequenty slips for 3 second.its velocity at time t is given by
$v\left(t\right)=2t(3-t)$; $0<t<3$ and
$v\left(t\right)=-(t-3)(6-t)$; $3<t<6s$
in $m/s$. It repeats this cycle till it reaches the height of $20m$. At whaat times is its velocity maximun?

Answer 1

Hint: condition for maximum velocity
Formula Used:$\frac{dv\left(t\right)}{dt}=0$

Given:
$v\left(t\right)=2t(3-t)$; $0<t<3$
Therefore,
$v\left(t\right)=2t(3-t)=6t-2t^2$

To find the time, we have to apply the condition of maximum velocity.
For maximum velocity,
$\frac{dv\left(t\right)}{dt}=0$
$\frac{d}{dt}(6t-2t^2)=0$
$6-4t=0$
$t=\frac{6}{4}=\frac{3}{2}s$
$=1.5s$
Final answer:$t=1.5s$