A monkey climbs up a slippery pole for 3 and subsequently slips for 3 . Its velocity at time $t$ is given by $v(t)=2 t(3-t)$; $0 < t < 3$ and $v(t)=-(t-3)(6-t)$ for $3 < t < 6 \mathrm{~s}$ in $\mathrm{m} / \mathrm{s}$. It repeats this cycle till it reaches the height of $20 \mathrm{~m}$.
(a) At what time is its velocity maximum?
(b) At what time is its average velocity maximum?
(c) At what time is its acceleration maximum in magnitude?
(d) How many cycles (counting fractions) are required to reach the top?

Options:

None

Solution:
To calculate maximum velocity we will use $\frac{d v}{d t}=0$, then the time corresponding to maximum velocity will be obtained.
As given that,
Velocity, $v(t)=2 t(3-t)=6 t-2 t^2$
(a) For maximum velocity $\frac{d v(t)}{d t}=0$
So, $\frac{d}{d t}\left(6 t-2 t^2\right)=0 \Rightarrow 6-4 t=0$
So, $t=\frac{6}{4}=\frac{3}{2} \mathrm{~s}=1.5$ seconds
(b) As given that $v=6 t-2 t^2$ $\frac{d s}{d t}=6 t-2 t^2 \Rightarrow d s=\left(6 t-2 t^2\right) d t$ where, $s$ is displacement. $\therefore$ Distance travelled in time interval 0 to $3 \mathrm{sec}$.
$$
\begin{aligned}
\int_0^{s_1} s &=\int_0^3\left(6 t-2 t^2\right) d t \\
s &=\left[\frac{6 t^2}{2}-\frac{2 t^3}{3}\right]_0^3=\left[3 t^2-\frac{2}{3} t^3\right]_0^3 \\
&=\left[3 \times 9-\frac{2}{3} \times 3 \times 3 \times 3\right] \\
s_1 &=27-18=9 \mathrm{~m} .
\end{aligned}
$$
For average velocity
$$
=\frac{\text { Distance travelled }}{\text { Total time }}=\frac{9}{3}=3 \mathrm{~m} / \mathrm{s}
$$
Given, $v(t)=6 t-2 t^2$
$$
\begin{aligned}
\Rightarrow 3=6 t-2 t^2 \\
\Rightarrow 2 t^2-6 t+3=0 \\
\Rightarrow t &=\frac{6 \pm \sqrt{6^2-4 \times 2 \times 3}}{2 \times 2}=\frac{6 \pm \sqrt{36-24}}{4} \\
&=\left[\frac{3+\sqrt{3}}{2}\right]
\end{aligned}
$$
Taking positive sign only
$$
\begin{aligned}
t &=\frac{3+\sqrt{3}}{2}=\frac{3+1.732}{2} \\
&=\left(\frac{4.732}{2}\right)=2.31 \mathrm{sec}
\end{aligned}
$$
So average velocity is maximum at $2.31 \mathrm{sec}$.
(c) In a periodic motion when velocity is zero acceleration will be maximum when body returns at its mean position or change the direction of motion putting $v$ $=0$ in eq. (i)
$$
\begin{aligned}
&0=6 t-2 t^2 \\
&\Rightarrow 0=t(6-2 t) \Rightarrow t \times 2(3-t)=0 \Rightarrow t=0
\end{aligned}
$$
$\therefore$ at $t=3$ second. So acceleration is maximum.
(d) Distance covered in 0 to $3 \mathrm{~s}$,
$$
\begin{aligned}
&\int_0^s d s=\int_0^3\left(6 t-2 t^2\right) d t \\
&s_1=9 \text { from part (b). } \\
&\text { For } 3 \text { to second, } u(t)=-(t-3)(6-t) \\
&\frac{d s}{d t}=(t-3)(6-t) \\
&d s=(t-3)(6-t) d t \\
&d s=\left(18-9+t^2\right) d t
\end{aligned}
$$
Distance covered in 3 to $6 \mathrm{~s}$,
$$
\begin{aligned}
&\left(\int_0^{s_2} d s\right)=\int_3^6\left(18-9 t+t^2\right) d t \\
&s_2=\left(18 t-\frac{9 t^2}{2}+\frac{t^3}{3}\right)_3^6 \\
&\quad=18 \times 6-\frac{9}{2} \times 6^2+\frac{6^3}{3}-\left(18 \times 3-\frac{9 \times 3^2}{2}+\frac{3^3}{3}\right) \\
&\quad=108-9 \times 18+\frac{6^3}{3}-18 \times 3+\frac{9}{3} \times 9-\frac{27}{3} \\
&=108-18 \times 9+\frac{216}{3}-54+4.5 \times 9-9 \\
&=-4.5 \mathrm{~m} .
\end{aligned}
$$
$\therefore$ Total distance travelled in one cycle
$$
=s_1+s_2=9-4.5=4.5 \mathrm{~m}
$$
Height climb up in three cycles $=4.5 \times 3=13.5 \mathrm{~m}$.
Now remaining height $=20-13.5=6.5 \mathrm{~m}$.
Number of cycles covered in total distance to be
$$
\text { covered }=\frac{20}{4.5}=4.44 \approx 5 \text {. }
$$

Topic: Physics, Motion In One Dimension