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(B) The motion is represented by a velocity-time graph. The area under the velocity-time graph gives the total distance traveled.Let $v$ be the maximu

Vedclass · Accepted Answer

$0.6$

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(B) The motion is represented by a velocity-time graph. The area under the velocity-time graph gives the total distance traveled.Let $v$ be the maximum velocity reached.For the first part (acceleration): Distance $S = \frac{1}{2} 	imes t_1 	imes v \Rightarrow t_1 = \frac{2S}{v}$.For the second part (uniform motion): Distance $2S = v 	imes t_2 \Rightarrow t_2 = \frac{2S}{v}$.For the third part (retardation): Distance $3S = \frac{1}{2} 	imes t_3 	imes v \Rightarrow t_3 = \frac{6S}{v}$.Total distance $D = S + 2S + 3S = 6S$.Total time $T = t_1 + t_2 + t_3 = \frac{2S}{v} + \frac{2S}{v} + \frac{6S}{v} = \frac{10S}{v}$.Average velocity $v_{avg} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{6S}{10S/v} = 0.6v$.Therefore,the ratio $\frac{v_{avg}}{v} = 0.6$.

$A$ body starts from rest and travels a distance $S$ with uniform acceleration,then moves uniformly a distance $2S$ and finally comes to rest after moving further $3S$ under uniform retardation. The ratio of the average velocity to maximum velocity is

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