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The speed of particles moving in a circle of radius $2 \mathrm{~m}$ varies with time as $\mathrm{v=t^2}$ , where $\mathrm{v}$ in $\mathrm{m} / \mathrm{s}$ and time in second. The net acceleration at $\mathrm{2 \sec}$ is -

Option: 1
$\sqrt{60} \mathrm{~m} / \mathrm{s}^2$

Option: 2
$\sqrt{160} \mathrm{~m} / \mathrm{s}^2$

Option: 3
$\sqrt{65} \mathrm{~m} / \mathrm{s}^2$

Option: 4
$\sqrt{80} \mathrm{~m} / \mathrm{s}^2$

Answers (1)

best_answer

$\mathrm{ v=t^2 }$

$\mathrm{ a^r=r \omega^2=\frac{v^2}{r}=\frac{(4)^2}{2}=8 \mathrm{~m} / \mathrm{sec}^2 }$

$\mathrm{ v=(2)^2=4 \mathrm{~m} / \mathrm{sec} }$

$\mathrm{ a^t=r \alpha=\frac{d v^2}{d t}=2 t=2 \times 2=4 \mathrm{~m} / \mathrm{s}^2 }$

$\mathrm{ a=\sqrt{\left(a^\gamma\right)^2+(a ^{t})^2} }$

$\mathrm{ a=\sqrt{(8)^2+(4)^2} }$

$\mathrm{ a=\sqrt{64+16} }$

$\mathrm{ a=\sqrt{80} \mathrm{~m} / \mathrm{sec}^2 }$

Hence option 4 is correct.

Posted by

chirag

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