Question

Find the angle between the net acceleration vector and tangential acceleration vector for a particle moving in a circle of radius $4m$ and its speed is increasing at $3m/s$ every second, when the speed of the particle is $4m/s$.

• A. ${37}^{\circ }$
• B. ${53}^{\circ }$
• C. ${60}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B ${53}^{\circ }$

Let the tangential acceleration be $a_t$
and centripetal acceleration be $a_c$
$a_c=\frac{v^2}{R}=\frac{4^2}{4}=4m/s^2$
$a_t=3m/s^2$
$a_{net}=\sqrt{a_c^2+a_t^2}$
$=\sqrt{4^2+3^2}=5m/s^2$


Now given that $\theta$ is the angle between the net acceleration and tangential acceleration.
Hence,
$cos\theta =\frac{a_t}{a_{net}}=\frac{3}{5}$
$\therefore \theta ={53}^{\circ }$