2.Motion in Straight Line
medium

From the $v$ - $t$ graph shown. the ratio of distance to displacement in $25\,s$ of motion

A

$\frac{3}{5}$

B

$\frac{1}{2}$

C

$\frac{5}{3}$

D

$1$

(JEE MAIN-2023)

Solution

Area under the graph from $t=0$ to $t=20\,sec =200\,m$

Area under the graph from $t=20$ to $t=25\,sec =50\,m$

So distance covered $=(200+50) m =250\,m$

Displacement $=(200-50) m =150\,m$

$\frac{250}{150}=\frac{5}{3}$

Standard 11
Physics

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