Question

A particle is revolving in a circular path having radius $‘r^{\prime }=1m$ at a speed of $2m/s$. It is increasing its speed at the rate of $3m/s^2$. The magnitude of total acceleration of the particle at that instant is

• A. $2m/s^2$
• B. $3m/s^2$
• C. $7m/s^2$
• D. $5m/s^2$

Answer 1

The correct option is D $5m/s^2$


Total acceleration $\left(\overrightarrow{a}\right)=\overrightarrow{a_c}+\overrightarrow{a_t}$
$|\overrightarrow{a}|=\sqrt{a_c^2+a_t^2}$
Here, $a_c=\frac{v^2}{r}=\frac{(2{)}^2}{1}=4m/s^2$
and given, $a_t=3m/s^2$
$\therefore |\overrightarrow{a}|=\sqrt{(4{)}^2+(3{)}^2}=\sqrt{25}=5m/s^2$