For the velocity-time graph shown in the figu | vedclass

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(B) The distance covered is equal to the area enclosed by the velocity-time graph with the time axis.Total distance covered in the interval from $t =

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(B) The distance covered is equal to the area enclosed by the velocity-time graph with the time axis.Total distance covered in the interval from $t = 1 \ s$ to $t = 7 \ s$ (as per the graph,the motion ends at $t = 7 \ s$):The area is a trapezoid with parallel sides of length $(5-3) = 2 \ s$ and $(7-1) = 6 \ s$,and height $10 \ m/s$.Total distance $S = \frac{1}{2} 	imes (2 + 6) 	imes 10 = 40 \ m$.Distance covered in the last two seconds (from $t = 5 \ s$ to $t = 7 \ s$):This is a triangle with base $(7-5) = 2 \ s$ and height $10 \ m/s$.$S_1 = \frac{1}{2} 	imes 2 	imes 10 = 10 \ m$.Therefore,the ratio is $\frac{S_1}{S} = \frac{10}{40} = \frac{1}{4}$.

For the velocity-time graph shown in the figure,the distance covered by the body in the last two seconds of its motion is $S_1$. What is the ratio of $S_1$ to the total distance covered by it?

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