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-): 0<?<3 and v (O)=(t-3)(6-t) for 3 <t< 6 s in
m/s. It repeats this cycle till it reaches the height of 20 m.
(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?​

Answer 1

Answer:

Explanation:

=> It is given that,

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

v (O)=-(t-3)(6-t) for 3 <t< 6 s in  m/s.

(a)  Time when velocity is maximum:

time when velocity v(t) = 6t - 2t², in  order v(t) to be maximum dv(t) / dt = 0.

6 - 4t = 0

4t = 6

t = 6/4

t = 3/2 s.

(b)  Time when average velocity is maximum:

correlated with time t = 3/2 s and t = 3/4s, the distance covered are 9/2m and 9/8m.

Thus, total time taken = 3/2 + 3/4

= 6 + 3 / 4

= 9 / 4 s.

(c) Time when its acceleration is maximum in magnitude:

=>time when acceleration is maximum, t = 3/2 sec.