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

$v\left(t\right)=2t(3-t)$; $0<t<3s$ 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 what time is its average velocity maximun?

Answer 1

Step 1: Find the distance travelled for the time interval 0 to 3 sec.

Given:

$v\left(t\right)=2t(3-t):0<t<3$

$v\left(t\right)=2t(3-t)$

$v\left(t\right)=6t-2t^2$

As we know,

$v=\frac{ds}{dt}$ where, s is displacement

Therefore, from the given equition

$\frac{ds}{dt}=6t-2t^2$

$ds=(6t-2t^2)dt$

Now, to find distance travelled in time interval $0to3s$,

$s_1=$${\int }_0^3(6t-2t^2)dt$

$s_1=$${[\frac{6t^2}{2}-\frac{2t^3}{3}]}_0^3={[3t^2-\frac{2}{3}{t}^3]}_0^3$

$s_1=3\times 9-\frac{2}{3}3\times 3\times 3$

$s_1=27-18=9m$

Step2:Find the distance travelled for the time interval 3 to 6 sec.

Distance coverd in 3 to 6 s

$s_2=$${\int }_3^6(18-9t+t^2)dt$

$s_2={(18t-\frac{9t^2}{2}+\frac{t^3}{3})}_3^6$

$s_2=18\times 6-\frac{9}{2}\times 6^2+\frac{6^3}{3}-(18\times 3-\frac{9\times 3^2}{2}+\frac{3^3}{3})$


$s_2=108-9\times 18+\frac{6^3}{3}-18\times 3+\frac{9}{2}-\frac{27}{3}$

$s_2=-4.5m$

Step 3:Find the number of cycles(n).

Total distance travelled in one cycle
$s=s_1+s_2$

=9-4.5

=4.5m

Height covered in 3 cycles

=$4.5\times 3$

=13.5

The remaining 6.5 will be covered in 4th cycle

Final Answer:n=4