Question 88 of 308

Form the $v-t$ graph shown, the ratio of distance to displacement in $25s$ of motion is:

A$1$B$\dfrac{1}{2}$C$\dfrac{5}{3}$D$\dfrac{3}{5}$

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Answer: (C) $\dfrac{5}{3}$

The area under the velocity time graph gives the displacement. The area for $t=0s\text{to}t=25s$is
$(Area)=\left(\dfrac{1}{2}\times 5\times 10\right)+\left(5\times 10\right)+\left(\dfrac{1}{2}\times 30\times 5\right)+\left(\dfrac{1}{2}\times 20\times 5\right)-\left(\dfrac{1}{2}\times 20\times 5\right)$
$\Rightarrow Area=25+50+75+50-50=150m$
Hence, the displacement $=150m$
The distance is defined as the path taken by the object to travel from initial to final position. Also note that displacement is a vector quantity, but distance is a scalar quantity.
Distance $=\left(\dfrac{1}{2}\times 5\times 10\right)+\left(5\times 10\right)+\left(\dfrac{1}{2}\times 30\times 5\right)+\left(\dfrac{1}{2}\times 20\times 5\right)+\left(\dfrac{1}{2}\times 20\times 5\right) =(200+50)=250m$
Therefore, required ratio
$\dfrac{\text{distance}}{\text{displacement}}=\dfrac{250}{150}=\left(\dfrac{5}{3}\right)$

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.