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

Step-by-step explanation: The equations used are as follows:

`\begin{gathered} x(t)=x_o+v_ot+(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=(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)+(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: