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 < 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?

Answer 1

Answer on Question #74247-Physics-Classical Mechanics

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 $20 \, \text{m}$.

(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?

Solution

a)

$$\frac{dv}{dt} = 6 - 4t = 0$$ $$t = 1.5 \, \text{s}.$$

b)

$$v = 6t - 2t^2$$ $$s = \int_{0}^{3} (6t - 2t^2) \, dt = 9 \, \text{m}.$$ $$v_{av} = \frac{s}{t} = \frac{9}{3} = 3 \, \frac{\text{m}}{\text{s}}.$$

Thus,

$$6t - 2t^2 = 3$$ $$t = \frac{3 + \sqrt{3}}{2} = 2.31 \, \text{s}.$$

c) Acceleration maximum will be when velocity is zero:

$$(6t - 2t^2) = 0$$ $$t = 3 \, \text{s}.$$

d)

$$s_1 = \int_{0}^{3} (6t - 2t^2) \, dt = 9 \, \text{m}$$ $$s _ {2} = \int_ {3} ^ {6} (- (t - 3) (6 - t)) d t = - 4. 5 m$$

Total distance travelled in 1 cycle is

$$d = s _ {1} + s _ {2} = 9 - 4. 5 = 4. 5 m.$$

We need

$$N = \frac {2 0}{4 . 5} = \frac {4 0}{9} = 4 \frac {4}{9} \approx 4. 4 4 \text{ cycles}.$$

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