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2.Motion in Straight Line
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The velocity of a particle moving on the $x-$ axis is given by $v = x^2 + x$ where $v$ is in $m/s$ and $x$ is in $m$ . Find its acceleration in $m/s^2$ when passing through the point $x = 2m$
A$0$
B$5$
C$11$
D$30$
Solution
$\mathrm{V}=\mathrm{x}^{2}+\mathrm{x} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{dx}}{\mathrm{dt}}$
$\Rightarrow \mathrm{a}=(2 \mathrm{x}+1) \mathrm{V}=(2 \mathrm{x}+1)\left(\mathrm{x}^{2}+\mathrm{x}\right)$
$\therefore$ When $\mathrm{x}=2 \mathrm{m}$
$a=(2 \times 2+1)\left(2^{2}+2\right)=5 \times 6=30 \mathrm{m} / \mathrm{s}^{2}$
$\Rightarrow \mathrm{a}=(2 \mathrm{x}+1) \mathrm{V}=(2 \mathrm{x}+1)\left(\mathrm{x}^{2}+\mathrm{x}\right)$
$\therefore$ When $\mathrm{x}=2 \mathrm{m}$
$a=(2 \times 2+1)\left(2^{2}+2\right)=5 \times 6=30 \mathrm{m} / \mathrm{s}^{2}$
Standard 11
Physics
