A particle is moving on a circular path of 10 m radius. At any instant of time its speed is $5 \math

Question

A particle is moving on a circular path of 10 m radius. At any instant of time its speed is $5 \mathrm{~m} \mathrm{~s}^{-1}$ and the speed is increasing at a rate of $2 \mathrm{~m} \mathrm{~s}^{-2}$. The magnitude of net acceleration at this instant (in $\mathrm{m} \mathrm{s}^{-2}$ ) is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

(3.2): Here, $r=10 \mathrm{~m}, v=5 \mathrm{~m} \mathrm{~s}^{-1}, a_t=2 \mathrm{~m} \mathrm{~s}^{-2}$ $ \  a_r=\frac{v^2}{r}=\frac{5 	imes 5}{10}=2.5 \mathrm{~ms}^{-2} $ \ The net acceleration is $a=\sqrt{a_r^2+a_t^2}=\sqrt{(2.5)^2+2^2} =\sqrt{10.25}=3.2 \mathrm{~m} \mathrm{~s}^{-2}$