Question

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

• A. 5 $m{s}^{-1}$
• B. 2$m{s}^{-1}$
• C. 3.2 $m{s}^{-1}$
• D. 4.3 $m{s}^{-1}$

Answer 1

Answer: (C)

3.2 $m{s}^{-1}$

Here, r=10m, v=5 $m{s}^{-1},a_1=2m{s}^{-2}$,
$a_r=\frac{v^2}{r}=\frac{5\times 5}{10}=2.5m{s}^{-2}$
The net acceleration is
$a=\sqrt{a_r^2+a_t^2}=\sqrt{(2.5{)}^2+2^2}=\sqrt{\left(10.25\right)}=3.2m{s}^{-2}$