Answer to Question #156740 in Mechanics | Relativity for Jake

Question

An object moving in a circular path with a constant speed of 2.0 m/s changes direction by 30⁰ in 3.0 s. What is the magnitude of the average acceleration over the 3.0 second interval?

Answer 1

Let's first find the angular velocity of the object:

$$\omega=\dfrac{\Delta \theta}{\Delta t}=\dfrac{30^{\circ}\cdot\dfrac{\pi}{180^{\circ}}}{3\ s}=0.17\ \dfrac{rad}{s}.$$

Then, we can find the radius of the circular path from the formula:

$$v=\omega r,$$ $$r=\dfrac{v}{\omega}=\dfrac{2.0\ \dfrac{m}{s}}{0.17\ \dfrac{rad}{s}}=11.7\ m$$

Finally, we can find the average acceleration from the formula:

$$a=\dfrac{v^2}{r}=\dfrac{(2.0\ \dfrac{m}{s})^2}{11.7\ m}=0.34\ \dfrac{m}{s^2}.$$

Answer:

$a=0.34\ \dfrac{m}{s^2}.$