Question

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

• 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 $a_c$

Given the speed of the particle is $v=4\text{m/s}$.
speed is increasing at $3\text{m/s}$ every second $a_t=\frac{dv}{dt}=\frac{3}{1}=3{\text{m/s}}^2$

Centripetal acceleration
$a_c=\frac{v^2}{R}=\frac{4^2}{4}=4{\text{m/s}}^2$



Net acceleration,
$a_{\text{net}}=\sqrt{a_c^2+a_t^2}=\sqrt{4^2+3^2}=5{\text{m/s}}^2$
Now angle between net acceleration and tangential acceleration is $\cos \theta =\frac{a_t}{a_{\text{net}}}=\frac{3}{5}$
$\therefore \theta ={53}^{\circ }$