A car accelerates uniformly from rest and
reaches a speed of 32.2 m/s in 8.6 s. The
diameter of a tire is 60.1 cm.
Find the number of revolutions the tire
makes during this motion, assuming no slipping.
Answer in units of rev.

Respuesta :

Answer:

Around 73 revolutions

Explanation:

Tangential acceleration = Angular acceleration x radius

The tangential acceleration is the final speed divided by the time as it was initially at rest and therefore it gives a value of approximately 3.74 m/s^2. The angular acceleration becomes 12.46 rad/s^2 as of the above formula (remember to change the unit of the radius to meters).

Finally, to find the angular displacement we use the rotational equivalent of one of the equations used for uniformly accelerated rectilinear motion. i.e.

[tex]x \: = \frac{1}{2}at {}^{2} [/tex]

We use the angular displacement in place of x and angular acceleration in place of a to get an angular displacement of 460.8 rad or 73.3 revolutions.

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