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 acceleration maximum in magnitude?

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Answer

Given velocity$\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$
In a periodic motion when velocity is zero acceleration will be maximum putting v = 0 in Eq. (i).$0=6\text{t}-2\text{t}^2\Rightarrow0=\text{t}(6-2\text{t})$
$=\text{t}\times2(3-\text{t})=0\Rightarrow\text{t}=0\ \ \text{or}\ \ 3\text{s}$

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