Question

In a recent test of its braking system, a Volkswagen Passat traveling at 31.2 m/s came to a full stop after an average negative acceleration of magnitude 2.10 m/s².

(a) How many revolutions did each tire make before the car comes to a stop, assuming the car did not skid and the tires had radii of 0.350 m?

(b) What was the angular speed of the wheels (in rad/s) when the car had traveled half the total stopping distance?

Answer 1

a) According to the kinematic equation 4 (see https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas), the distance travelled by the car to full stop is (taking the final speed, $v$, to be 0):

$$d =\dfrac{v_0^2}{2a}$$

where $v_0 = 31.2m/s$ is the initial speed, $a = 2.10m/s^2$ is its acceleration.

In one turn tire travels the distance equal to the length of its circumference:

$$l = 2\pi r$$

where $r = 0.350m$ is its radius. Thus, the total number of turns is:

$$N = \dfrac{d}{l} = \dfrac{ v_0^2}{4\pi ra}\\ N = \dfrac{(31.2m/s)^2}{4\pi \cdot 0.35m\cdot2.1m/s^2} \approx 105$$

b) The angular speed is given as follows:

$$\omega = \dfrac{v}{r}$$

where $v$ is the linear speed. Using the same kinematic equation, find the speed at $d/2$:

$$v = \sqrt{v_0^2 - ad} = \sqrt{v_0^2 - a\cdot \dfrac{v_0^2}{2a}} = \dfrac{v_0}{\sqrt{2}}$$

where '-' sign was used, because the acceleration is negative. Thus, obtain:

$$\omega = \dfrac{v_0}{r\sqrt{2}} = \dfrac{31.2m/s}{0.35m\cdot \sqrt{2}} \approx 63.0rad/s$$

Answer. a) 105, b) 63.0 rad/s.