From the given v-t graph,find the ratio of di | vedclass

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(C) The distance is the total area under the $v-t$ graph (taking all areas as positive),and displacement is the net area (taking areas below the time

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$\frac{5}{3}$

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(C) The distance is the total area under the $v-t$ graph (taking all areas as positive),and displacement is the net area (taking areas below the time axis as negative).$1$. Area from $t=0$ to $t=5\,s$ (triangle): $\frac{1}{2} 	imes 5 	imes 10 = 25\,m$.$2$. Area from $t=5$ to $t=10\,s$ (rectangle): $5 	imes 10 = 50\,m$.$3$. Area from $t=10$ to $t=15\,s$ (trapezium): $\frac{1}{2} 	imes (10 + 20) 	imes 5 = 75\,m$.$4$. Area from $t=15$ to $t=20\,s$ (triangle): $\frac{1}{2} 	imes 5 	imes 20 = 50\,m$.$5$. Area from $t=20$ to $t=25\,s$ (triangle below axis): $\frac{1}{2} 	imes 5 	imes (-20) = -50\,m$.Total distance $= 25 + 50 + 75 + 50 + |-50| = 250\,m$.Total displacement $= 25 + 50 + 75 + 50 - 50 = 150\,m$.Ratio of distance to displacement $= \frac{250}{150} = \frac{5}{3}$.

From the given $v-t$ graph,find the ratio of distance to displacement in $25\,s$ of motion.

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