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. At what time is its average velocity maximum?

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Answer

Given velocity $\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$ From Eq. (i) $\text{v}=6\text{t}-2\text{t}^2$ $\Rightarrow\frac{\text{ds}}{\text{dt}}=6\text{t}-2\text{t}^2$ $\Rightarrow\text{ds}=(6\text{t}-2\text{t}^2)\text{dt}$ where, s is displacement $\therefore$ Distance travelled in time interval 0 to 3s, $\text{s}=\int^3_0(6\text{t}-2\text{t}^2)\text{dt}$ $=\Big[\frac{6\text{t}^2}{2}-\frac{2\text{t}^3}{3}\Big]^3_0=\Big[3\text{t}^2-\frac{2}{3}\text{t}^3\Big]^3_0$ $=3\times9-\frac{2}{3}\times3\times3\times3$ $=27-18=9\text{m}$ $\text{Average velocity}=\frac{\text{Distance travelled}}{\text{Time}}$ $=\frac{9}{3}=3\text{m/s}$ Given, $\text{x}=6\text{t}-2\text{t}^2$ $\Rightarrow3=6\text{t}-2\text{t}^2\Rightarrow2\text{t}^2-6\text{t}-3=0$ $\Rightarrow\text{t}=\frac{6\pm\sqrt{6^2-4\times2\times3}}{2\times2}=\frac{6\pm\sqrt{36-24}}{4}$ $=\frac{6\pm\sqrt{12}}{4}=\frac{3\pm2\sqrt{3}}{2}$ Considering positive sign only $\text{t}=\frac{3+2\sqrt{3}}{2}=\frac{3+2\times1.732}{2}=\frac{9}{4}\text{s}$

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