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Q.
A particle moves so that its acceleration is always twice its velocity. If its initial velocity is $0.1 \,ms ^{-1},$ its velocity after it has gone $0.1\, m$ is
Laws of Motion
Solution:
$\frac{d v}{d t}=2 v \Rightarrow \frac{d v}{d t}=\frac{2 d x}{d t}$
$d v=2 d x$
$\int_{0.1}^{v} d v=2 \int_{0}^{0.1} d x$
$(y-0.1)=2(0.1) \Rightarrow v=0.3\, ms ^{-1}$