Question

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

• A. $3.2m/s^2$
• B. $2m/s^2$
• C. $2.5m/s^2$
• D. $4.3m/s^2$

Answer 1

Answer: (A)

$3.2m/s^2$

Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration thus find both of them as follows$:$

$A_c-$ be centripetal acceleration$.$

$A_t-$ be tangential acceleration

$A_{net}=\sqrt{\left(A_c^2+A_t^2\right)}$

Now then$,$

$A_c$ is $\frac{V^2}{R}=\frac{25}{10}=2.5m/s^2$

$A_t$ is given as $2.m/s^2,$ thus

$A_{net}=\sqrt{\left({2.5}^2+2^2\right)}$

$=3.20m/s^2.$

Hence,

option $\left(A\right)$ is correct answer.