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A car is moving with a speed of $30 \mathrm{~ms}^{-1}$ on a circular path of radius $500 \mathrm{~m}$. If its speed is increasing at the rate of $2 \mathrm{~ms}^{-2}$, then find its acceleration.
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Verified Answer
The correct answer is:
$2.7 \mathrm{~ms}^{-2}$
Given, initial speed, $u=30 \mathrm{~ms}^{-1}$
Radius, $r=500 \mathrm{~m}$
Linear acceleration, $a_T=2 \mathrm{~ms}^{-2}$
Let acceleration be $a$
$$
\because \quad a=\sqrt{a_T^2+a_r^2}
$$
As, $a_r=\frac{u^2}{r}=\frac{(30)^2}{500}$
$$
=\frac{900}{500}=\frac{9}{5} \mathrm{~ms}^{-2}
$$
$$
\begin{aligned}
\therefore \quad a & =\sqrt{2^2+\left(\frac{9}{5}\right)^2}=\sqrt{4+\frac{81}{25}} \\
& =2.69 \mathrm{~ms}^{-2} \simeq 2.7 \mathrm{~ms}^{-2}
\end{aligned}
$$
Radius, $r=500 \mathrm{~m}$
Linear acceleration, $a_T=2 \mathrm{~ms}^{-2}$
Let acceleration be $a$
$$
\because \quad a=\sqrt{a_T^2+a_r^2}
$$
As, $a_r=\frac{u^2}{r}=\frac{(30)^2}{500}$
$$
=\frac{900}{500}=\frac{9}{5} \mathrm{~ms}^{-2}
$$
$$
\begin{aligned}
\therefore \quad a & =\sqrt{2^2+\left(\frac{9}{5}\right)^2}=\sqrt{4+\frac{81}{25}} \\
& =2.69 \mathrm{~ms}^{-2} \simeq 2.7 \mathrm{~ms}^{-2}
\end{aligned}
$$
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