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Motion in a Straight Line — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line3 Marks
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?
✓

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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