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

$v=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t$

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

$0=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t\implies t=13.85\,\mathrm s$

or after about 13.9 s.

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

$x=\left(27.7\dfrac{\rm m}{\mathrm s}\right)(13.85\,\mathrm s)+\dfrac12\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)(13.85\,\mathrm s)^2=191.823\,\mathrm m$

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.