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 < 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?
(d) How many cycles (counting fractions) are required to reach the top?
To find maximum velocity, we can differentiate the velocity expression and find time at which there will be a maxima or minima.
For 0< t < 3, v(t) = 2t (3-t) = 6t – 2t2
Thus, at t = 1.5 s, there is a maxima, the velocity is maximum at t = 1.5 s.
(b) Average velocity is given by
Now,
This is a quadratic equation, we can solve it using our knowledge of quadratic equations.
(c) Acceleration is the first derivative of velocity
Now this is for the time-period 0 to 3 seconds, so |6-4t| will be maximum either when the negative factor deducts nothing from 6 or when it deducts maximum from 6 which is at t = 0s and t = 3s respectively.
(d) To count how many cycles are required, we need to find the height covered in 1 cycle.
From 0< t< 3s, let the height covered be s1
For 3< t< 6s, let the height covered be s2
Total distance covered in cycle = 4.5 m
Total distance to be covered = 20 m
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