Question 92 of 308

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

A$1:1$B$1:2$C$1:3$D$1:4$

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Answer: (C) $1:3$

Area of upper rectangles, which will give us magnitude of the area under the velocity-time graph,
${A}_{1}=8\times 2+4\times 4=32$
Area of lower rectangles, which will give us magnitude of the area under the velocity-time graph,
${A}_{2}=4\times 2+4\times 2=16$
Therefore,
$Displacement={A}_{1}-{A}_{2}=32-16=16 Distance={A}_{1}+{A}_{2}=32+16=48$
Hence, the required ratio would be
$=\dfrac{16}{48}=\dfrac{1}{3}$

JEE Main Physics practice question on Kinematics (Graphs of motion in one dimension), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.