Question

You are driving down the road at 15 m/s you step on the gas and speed up with uniform acceleration of 2.5 m/s^2 for 0.80 seconds. If your tires have a radius of 34 cm, what is their angular displacement during this period of acceleration?

Answer 1

Answer:

$x_f = 15 m/s *0.8 s + \frac{1}{2} 2.5 m/s^2 (0.8 s)^2 = 12.8 m$

And we have the following relation between the angular displacement and the linear displacement:

$x = r \theta$

Where x represent the linear displacement and $\theta$ the angular displacement, if we solve for $\theta$ we got:

$\theta= \frac{x}{R}= \frac{12.8 m}{0.34 m}=37.65 rad$

Explanation:

For this case we have the following data given:

$v_i = 15 m/s$ represent the initial speed

$a = 2.5 m/s^2$ represent the acceleration

$t = 0.8 s$ represent the time

$R = 34 cm = 0.34 m$ represent the radius

First we can calculate the linear displacement with the following formula from kinematics:

$x_f = x_i + v_i t + \frac{1}{2} at^2$

And replacing we have:

$x_f = 15 m/s *0.8 s + \frac{1}{2} 2.5 m/s^2 (0.8 s)^2 = 12.8 m$

And we have the following relation between the angular displacement and the linear displacement:

$x = r \theta$

Where x represent the linear displacement and $\theta$ the angular displacement, if we solve for $\theta$ we got:

$\theta= \frac{x}{R}= \frac{12.8 m}{0.34 m}=37.65 rad$

And that would be the angular displacement during the period of acceleration.