Question

A car initially traveling at 27.7 m/s undergoes a constant negative acceleration of magnitude 2.00 m/s2 after its brakes are applied. (a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.340 m

Answer 1

The car's velocity at time t is given by

`v=27.7(\rm m)/(\rm s)+\left(-2.00(\rm m)/(\mathrm s^2)\right)t`

It comes to a stop when v = 0, which happens when

`0=27.7(\rm m)/(\rm s)+\left(-2.00(\rm m)/(\mathrm s^2)\right)t\implies t=13.85\,\mathrm s`

or after about 13.9 s.

In this time, the car travels a distance x given by

`x=\left(27.7(\rm m)/(\mathrm s)\right)(13.85\,\mathrm s)+\frac12\left(-2.00(\rm m)/(\mathrm s^2)\right)(13.85\,\mathrm s)^2=191.823\,\mathrm m`

or about 192 m.

In one complete revolution, each tire covers a distance equal to its circumference,

`2\pi(0.340\,\mathrm m)\approx2.13628\,\mathrm m`

or about 2.14 m.

This means each tire will complete approximately 192/2.14 ≈ 90 revolutions.