$A$ monkey climbs up a slippery pole for $3 \ s$ and subsequently slips for $3 \ s$. Its velocity at time $t$ is given by $v(t) = 2t(3 - t)$ for $0 < t < 3$ and $v(t) = -(t - 3)(6 - t)$ for $3 < t < 6$ in $m/s$. It repeats this cycle until it reaches the height of $20 \ m$.
$(a)$ 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?

(N/A) For $0 < t < 3$,$v(t) = 6t - 2t^2$. For $3 < t < 6$,$v(t) = -t^2 + 9t - 18$.
$(a)$ For $0 < t < 3$,$\frac{dv}{dt} = 6 - 4t = 0 \implies t = 1.5 \ s$. At $t = 1.5 \ s$,$v = 4.5 \ m/s$. For $3 < t < 6$,$v$ is negative,so max velocity is $4.5 \ m/s$ at $t = 1.5 \ s$.
$(b)$ Average velocity $v_{avg} = \frac{1}{t} \int_0^t v(t) dt$. For $0 < t < 3$,$v_{avg} = \frac{1}{t} (3t^2 - \frac{2}{3}t^3) = 3t - \frac{2}{3}t^2$. Setting $\frac{dv_{avg}}{dt} = 3 - \frac{4}{3}t = 0 \implies t = 2.25 \ s$.
$(c)$ Acceleration $a(t) = \frac{dv}{dt}$. For $0 < t < 3$,$a = 6 - 4t$. For $3 < t < 6$,$a = -2t + 9$. At $t=0, |a|=6$. At $t=3, |a|=6$. At $t=6, |a|=3$. Max magnitude is $6 \ m/s^2$ at $t=0 \ s$ and $t=3 \ s$.
$(d)$ Displacement in one cycle ($0$ to $6 \ s$): $x_1 = \int_0^3 (6t - 2t^2) dt + \int_3^6 (-t^2 + 9t - 18) dt = 9 + (-4.5) = 4.5 \ m$. To reach $20 \ m$,cycles needed: $20 / 4.5 \approx 4.44$ cycles.