 # A monkey climbs up a slippery pole for 3 seconds `
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

(a) At what time is its velocity maximum?

(b) At what time is its average velocity maximum?

(c) At what times is its acceleration maximum in magnitude?

(d) How many cycles are required to reach the top?

Solution:

(a) For maximum velocity v(t)

dv(t)/dt = 0

Substituting the value for v, we get

t = 1.5 seconds

(b) For average velocity = total distance/time taken

Average velocity = 3 m

And the average velocity is maximum when time t = 2.36 sec

(c) When the acceleration is maximum in a periodic motion, the time is maximum when the body returns to the mean position when v = 0

v(t) = 6t – 2t2

When t = 3 second, acceleration is maximum

(d) Distance covered between 0-3 second

s = 9m

v(t) = -(t-3) (6-t)

ds/dt = (t-3)(t-6)

Integrating the equation from 3 to 6, s = -4.5m

The net distance = 9 – 4.5 = 4.5m

Height of climb in three cycle = (4.5)(3) = 13.5m

The remaining height = 20 -13.5 = 6.5 m

Therefore, no.of cycles is 20 when the height of the pole is 4.