Question

horizontal highway, which we assume extends from our left to our right. The wheels of the car have radii of 64.7 cm. If the car speeds up with an acceleration of 2.83 m/s 2 for 6 s, find the number of revolutions of the wheels during this period. Answer in units of rev.

Answer 1

Answer:

$N = 12.5 rev$

Explanation:

linear acceleration of the car is given as

$a = 2.83 m/s^2$

radius of the wheel is

$R = 64.7 m$

now we know that for pure rolling with slipping the angular acceleration of the wheel is given as

$\alpha = \frac{a}{R}$

so we have

$\alpha = \frac{2.83}{0.647}$

$\alpha = 4.37 rad/s^2$

now the number of revolutions of the wheel is given as

$N = \frac{1}{4\pi}\alpha t^2$

$N = \frac{1}{4\pi}(4.37)(6^2)$

$N = 12.5 rev$