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

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(D) The displacement is the algebraic sum of the areas under the velocity-time graph.Area $1$ ($0$ to $2\,s$): $2 	imes 8 = 16\,m$Area $2$ ($2$ to $4

Vedclass · Accepted Answer

$1: 3$

Vedclass · Answer

(D) The displacement is the algebraic sum of the areas under the velocity-time graph.Area $1$ ($0$ to $2\,s$): $2 	imes 8 = 16\,m$Area $2$ ($2$ to $4\,s$): $2 	imes (-4) = -8\,m$Area $3$ ($4$ to $8\,s$): $4 	imes 4 = 16\,m$Area $4$ ($8$ to $10\,s$): $2 	imes (-4) = -8\,m$Displacement $= 16 - 8 + 16 - 8 = 16\,m$Distance is the sum of the magnitudes of the areas.Distance $= |16| + |-8| + |16| + |-8| = 16 + 8 + 16 + 8 = 48\,m$Ratio of displacement to distance $= \frac{16}{48} = \frac{1}{3}$ or $1: 3$.

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 $10\,s$ is

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