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(A) Let the total time be $T$ and the maximum velocity be $v$. The velocity-time graph is a trapezoid.The area under the $v-t$ graph gives the total d

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(A) Let the total time be $T$ and the maximum velocity be $v$. The velocity-time graph is a trapezoid.The area under the $v-t$ graph gives the total distance $S$.$S = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height} = \frac{1}{2} 	imes (T + (T - 2t)) 	imes v = \frac{1}{2} 	imes (2T - 2t) 	imes v = (T - t)v$,where $t$ is the time taken for acceleration.The average speed $v_{av} = \frac{S}{T} = \frac{(T - t)v}{T}$.Given $v_{av} = \frac{7v}{8}$,we have $\frac{(T - t)v}{T} = \frac{7v}{8}$.$1 - \frac{t}{T} = \frac{7}{8} \implies \frac{t}{T} = \frac{1}{8}$.The time spent with constant velocity is $T_{const} = T - 2t$.The fraction of total time is $\frac{T - 2t}{T} = 1 - 2(\frac{t}{T}) = 1 - 2(\frac{1}{8}) = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$.

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