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2.Motion in Straight Line
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A particle moves in a straight line and its position $x$ at time $t$ is given by $x^2=2+t$. Its acceleration is given by
A
$\frac{-2}{x^3}$
B
$-\frac{1}{4 x^3}$
C
$-\frac{1}{4 x^2}$
D
$\frac{1}{x^2}$
Solution
(b)
$x^2=t+2 \Rightarrow \frac{1}{x^2}=\frac{1}{t+2}$
$\Rightarrow x=\sqrt{t+2}$
$\Rightarrow \frac{d x}{d t}=\frac{1}{2}(t+2)^{\frac{1}{2}-1}$
$\Rightarrow \frac{d x}{d t}=\frac{1}{2}(t+2)^{-\frac{1}{2}}$
$\Rightarrow \frac{d^2 x}{d t^2}=\frac{1}{2}\left(-\frac{1}{2}\right)^{(i)}(t+2)^{-\frac{1}{2}-1}$
$\Rightarrow a=-\frac{1}{4}(t+2)^{-\frac{3}{2}}=-\frac{1}{4(t+2)} \times \frac{1}{(t+2)^{\frac{1}{2}}}=-\frac{1}{4} \times \frac{1}{x^2} \times \frac{1}{x}$
$\Rightarrow a=-\frac{1}{4 x^3}$
Standard 11
Physics
