Question

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

• A. $5m/s^2$
• B. $2m/s^2$
• C. $3.2m/s^2$
• D. $4.3m/s^2$

Answer 1

The correct option is C $3.2m/s^2$

$a_{centripetal}=\frac{V^2}{r}=\frac{5\times 5}{10}=2.5m/s^2{a}_{tangential}=2m/s^2{a}_{net}=\sqrt{a_c^2+a_t^2}=\sqrt{(2.5{)}^2+(2{)}^2}{a}_{net}=3.2m/s^2$