A car accelerates uniformly from rest andreaches a speed of 32.1 m/s in 13.8 s. Thediameter of a tire is 84.7 cm.Find the number of revolutions the tiremakes during this motion, assuming no slipping.Answer in units of rev.

Respuesta :

Given that the speed is v = 32.1 m/s and time is t = 13.8 s.

Also, the diameter of the tire is d = 84.7 cm

We have to find the number of revolutions of the tire.

As the diameter is 84.7, so the radius will be

[tex]\begin{gathered} r=\frac{d}{2} \\ =\frac{84.7}{2} \\ =0.423\text{ m} \end{gathered}[/tex]

The circumference of the circle will be

[tex]C=2\pi r[/tex]

Substituting the values, the circumference will be

[tex]\begin{gathered} C=2\times3.14\times0.423 \\ =2.656\text{ m} \end{gathered}[/tex]

The distance can be calculated using speed and time,

[tex]\begin{gathered} s=v\times t \\ =32.1\times13.8 \\ =442.98\text{ m} \end{gathered}[/tex]

To find the number of revolutions, the formula will be

[tex]n\text{ =}\frac{\text{total distance}}{circumference}[/tex]

Substituting the values, n will be

[tex]\begin{gathered} n=\frac{442.98}{2.656} \\ =166.78\text{ rev} \end{gathered}[/tex]