question_answer 53)

                  A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by \[\upsilon (t)=2t\,(3t);\text{ }0<t<3\] and \[\upsilon (t)=(t3)(6-t)\] for \[3<t<6\] in m/s. It repeats this cycle till it reaches the height of 20 in.                 (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?                 How many cycles (counting fractions) are required to reach the top?                

Answer:

                  \[\upsilon (t)=2t(3t);\text{ }0<t<3\]                 and \[\upsilon (t)=(t3)(6-t);\text{ }3<t<6\]                 For maximum velocity; \[{{\left. \frac{d\upsilon (t)}{dt} \right|}_{\max }}=0\]                 or \[6-4t=0\]                 or \[t=\frac{3}{2}\,s\].                 Average velocity will be maximum at \[t=\frac{9}{4}\,s\].