For the following velocity-time graph, the av | vedclass

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(B) As we know that, distance is equal to the area under the velocity-time graph. Since average speed is defined as the total distance traveled divide

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$5 \, m/s$

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(B) As we know that, distance is equal to the area under the velocity-time graph. Since average speed is defined as the total distance traveled divided by the total time taken, we must consider the magnitude of the area for both positive and negative velocity intervals.Total distance traveled in $80 \, s$ is the sum of the absolute areas of the triangles formed:$Distance = |Area_{OAB}| + |Area_{BCD}|$Area of triangle $OAB$ (from $t=0$ to $t=40 \, s$):$Area_{OAB} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 40 \, s 	imes 10 \, m/s = 200 \, m$Area of triangle $BCD$ (from $t=40$ to $t=80 \, s$):$Area_{BCD} = \frac{1}{2} 	imes 	ext{base} 	imes |	ext{height}| = \frac{1}{2} 	imes 40 \, s 	imes |-10 \, m/s| = 200 \, m$Total distance $= 200 \, m + 200 \, m = 400 \, m$Total time taken $= 80 \, s$Average speed $= \frac{	ext{Total distance}}{	ext{Total time}} = \frac{400 \, m}{80 \, s} = 5 \, m/s$

For the following velocity-time graph, the average speed for the motion during first $80 \, s$ is

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