The velocity-time graph of a body moving in a | vedclass

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(D) The displacement is the area under the velocity-time graph considering the sign of velocity. Displacement = (Area from $0$ to $2 	ext{ s}$) + (Ar

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$3$

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(D) The displacement is the area under the velocity-time graph considering the sign of velocity. Displacement = (Area from $0$ to $2 	ext{ s}$) + (Area from $2$ to $4 	ext{ s}$) + (Area from $4$ to $8 	ext{ s}$) + (Area from $8$ to $10 	ext{ s}$) Displacement = $(2 	imes 8) + (2 	imes -4) + (4 	imes 4) + (2 	imes -4) = 16 - 8 + 16 - 8 = 16 	ext{ m}$. The distance is the total area under the velocity-time graph considering the magnitude of velocity. Distance = $|2 	imes 8| + |2 	imes -4| + |4 	imes 4| + |2 	imes -4| = 16 + 8 + 16 + 8 = 48 	ext{ m}$. The ratio of displacement to distance = $16 / 48 = 1/3$.

The velocity-time graph of a body moving in a straight line is shown in the figure. The ratio of displacement to distance travelled by the body in time $0$ to $10 \text{ s}$ is

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