A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds Its velocity at time - Biology - Plant Kingdom

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
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 are required to reach the top?

Ved Prakash Lakhera · Accepted Answer

V elocity of climbing up, vt = 2t3-t; 0<t<3   andvelocity of slipping  vt = -t-36-t; 3<t<6
a) time when velocity  vt =6t -2t2, in order vt to be maximum dvtdt =0, i
 e
6-4t=0  or     t= 32  s
b)corresponding to time t=3/2s and time t= 3/4 s the distances travelled are 9/2m and 9/8m
 therefore total time taken = 3/2 +3/4 =(6+3) /4 = 9/4 s
c) the acceleration is maximum at time t=3/2 second
d) monkey will cover 4
5m height of the pole in first cycle because for first 3s it covers 9m distance and for next 3s it slips 4
5m vertical height
 Again in second and third cycle it covers 4
5m each
 So total vertical distance covered in 3cycles is13
5m
When 4th cycle, starts in the first 3s second or before that the required height of 20m is scaled by the monkey
distance covered in first 3s= S =3t2-23t3= 27-18=9m
and distance  slipped  in next 3 second= ∫36(t2-9t+18)dt  =4
5m