V -t graph of a rectilinear motion is shown in adjoining figure. distance from starting p

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2.Motion in Straight Line

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The $v -t$ graph of a rectilinear motion is shown in adjoining figure. The distance from starting point after $8$ seconds is ..........$metre$

A

$18 $

B

$12$

C

$8$

D

$6$

Solution

Distance $=$area under the graph…

$\mathrm{So}$

Distance $=[(4 \times 1) / 2]+(2 \times 4)+[(4 \times 1) / 2]$$+[(-2 \times 1) / 2]+[(-2) \times 2]+[(-2 \times 1) / 2]$

$=2+8+2-1+-4+-1$

$=12-6$

$=6 \mathrm{m}$

Standard 11

Physics

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