A monkey climbs up a slippery pole for 3 seconds and subsequently… - Vidyadip

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 < 6s in m/s. It repeats this cycle till it reaches the height of 20m.
How many cycles (counting fractions) are required to reach the top?

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Answer

Given velocity

$\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$

Distance covered in 0 to 3s = 9m

Distance covered in 3 to 6s

$=\int_3^6(18-9\text{t}+\text{t}^2)\text{dt}=\Big(18\text{t}-\frac{9\text{t}^2}{2}+\frac{\text{t}^3}{3}\Big)^6$

$=18\times6-\frac{9}{2}\times6^2+\frac{6^3}{3}-\Big(18\times3-\frac{9\times3^2}{2}+\frac{3^3}{3}\Big)$

$=108-9\times18+\frac{6^3}{3}-18\times3+\frac{9}{2}\times9-\frac{27}{3}$

$=-4.5\text{m}$

$\therefore$ Total distance travelled in one cycle

$=\text{s}_1+\text{s}_2=9-4.5=4.5\text{m}$

Number of cycles to be covered in total distance $=\frac{20}{4.5}\approx4.44\approx5$

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