Question

The driver of a car traveling at 31.3 m/s applies the brakes and undergoes a constant deceleration of 1.6 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 0.31 m?Answer in units of rev.

Answer 1

Answer:

$R=156.99\operatorname{Re}vs$

Explanation: The equations used are as follows:

$\begin{gathered} x(t)=x_o+v_ot+\frac{1}{2}at^2\Rightarrow(1) \\ v(t)=v_o+at\Rightarrow(2) \end{gathered}$

By using equation (2), the time needed for the car to come to rest is calculated as follows:

$\begin{gathered} v(t)=(31.3ms^{-1})_{}+(-1.6ms^{-2})t=0 \\ t=\frac{31.3ms^{-1}}{1.6ms^{-2}}=19.56s \\ t=19.563s \end{gathered}$

By using equation (1), The total distance traveled in that time would be as:

$\begin{gathered} x(19.563s)=_{}(31.3ms^{-1})\cdot(19.563s)+\frac{1}{2}(-1.6ms^{-2})\cdot(19.563s)^2\Rightarrow(1) \\ x(19.563s)=612.31-306.17=306.14m \\ \therefore\Rightarrow \\ x(19.563s)=306.14m \end{gathered}$

The revolutions taken by the tire before the car comes to rest would be:

$\begin{gathered} C=2\pi\cdot(0.31m)=1.95m \\ R=\frac{x(19.563s)}{C}=\frac{306.14m}{1.95m}=156.99\operatorname{Re}v \\ R=156.99\operatorname{Re}vs \end{gathered}$