**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.

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