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2.Motion in Straight Line
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The acceleration of a particle which moves along the positive $x-$axis varies with its position as shown. If the velocity of the particle is $0.8 m/s$ at $x = 0$ , the velocity of the particle at $x = 1.4$ is(in $m/s$)
A$1.6$
B$1.2$
C$1.4$
Dnone of these
Solution
$a=v \frac{d v}{d t} \Rightarrow \int_{u}^{v} v d v=\int_{0}^{1.4} a d x$
$\Rightarrow \frac{v^{2}-u^{2}}{2}=$ Areaofa $-x \operatorname{graph}$
$\Rightarrow v^{2}=(0.8)^{2}=2(0.4) \Rightarrow v=1.2$
$\Rightarrow \frac{v^{2}-u^{2}}{2}=$ Areaofa $-x \operatorname{graph}$
$\Rightarrow v^{2}=(0.8)^{2}=2(0.4) \Rightarrow v=1.2$
Standard 11
Physics
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