Question

A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 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 0.325 m

Answer 1

Answer:

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

Explanation:

We can use the following equation:

$\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta$ (1)

The angular acceleration is:

$a_{tan}=\alpha R$

$\alpha=\frac{1.9}{0.325}$

$\alpha=5.85\: rad/s^{2}$

and the initial angular velocity is:

$\omega_{i}=\frac{v}{R}$

$\omega_{i}=\frac{27.2}{0.325}$

$\omega_{i}=83.69\: rad/s$

Now, using equation (1) we can find the revolutions of the tire.

$0=83.69^{2}-2*25.85 \Delta \theta$

$\Delta \theta=135.47\: rad$

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

I hope it helps you!