A monkey climbs up a slippery pole for 3 seconds and subsequently… - Vidyadip

Question

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

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Answer

Given velocity
$\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$
For maximum velocity $\frac{\text{dv(t)}}{\text{dt}}=0$
$\Rightarrow\frac{\text{d}}{\text{dt}}(6\text{t}-2\text{t}^2)=0$
$\Rightarrow6-4\text{t}=0$
$\Rightarrow\text{t}=\frac{6}{4}=\frac{3}{2}\text{s}=1.5\text{s}$

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