The velocity (v) versus time (t) plot of a pa | vedclass

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(C) Distance is the total path length covered,which is the sum of the areas under the $|v|$ vs $t$ graph.For the interval $0$ to $20 	ext{ s}$,the ar

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$100 	ext{ m}$ and zero

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(C) Distance is the total path length covered,which is the sum of the areas under the $|v|$ vs $t$ graph.For the interval $0$ to $20 	ext{ s}$,the area of the triangle is $A_1 = 1/2 	imes 20 	imes 5 = 50 	ext{ m}$.For the interval $20$ to $40 	ext{ s}$,the magnitude of velocity is $5 	ext{ m/s}$,so the area is $A_2 = 1/2 	imes 20 	imes 5 = 50 	ext{ m}$.Total distance $= A_1 + A_2 = 50 + 50 = 100 	ext{ m}$.Displacement is the net change in position,which is the algebraic sum of the areas under the $v$ vs $t$ graph.Displacement $= A_1 - A_2 = 50 - 50 = 0 	ext{ m}$.Average velocity $= 	ext{Displacement} / 	ext{Total time} = 0 / 40 = 0 	ext{ m/s}$.

The velocity $(v)$ versus time $(t)$ plot of a particle is shown in the figure,for a time interval of $40 \text{ s}$. The total distance travelled by the particle and the average velocity during this period are,respectively . . . . . . .

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