Physics: 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?

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You are driving down the road at 15 m/s you step on

Kingzion5775

19.01.2022

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

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.

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