Answer to Question #100556 in Physics for Nyx

Question

The driver of a car traveling at 17.7 m/s applies the brakes and undergoes a constant
deceleration of 1.38 m/s2.
How many revolutions does each tire make
before the car comes to a stop, assuming that
the car does not skid and that the tires have
radii of 0.42 m?
Answer in units of rev.

Answer 1

The equations of motion with constant degradation are $s(t) = v_0 t - \frac{ a t^2}{2}$, $v(t) = v_0 - a t$. The time it takes the car to stop can be found from $v(t) = 0 \Rightarrow v_0 - a t = 0 \Rightarrow t = \frac{v_0}{a}$. Substituting the last expression for time into the equation for distance above, obtain: $s = \frac{\frac{v_0^2}{a} - \frac{v_0^2}{2 a}}{2 \pi r} = \frac{v_0^2}{2 \pi r}$. This is the distance the car will cover until full stop.

The distance, covered until the car stops is equal to number of revolutions of the wheel times the circumference of the wheel: $s = 2 \pi r \cdot N$, where $r$ is the radius of the wheel.

Hence, the number of revolutions is $N = \frac{s}{2 \pi r} = \frac{v_0^2}{4 \pi r a} \approx 43$.