Question

A car accelerates uniformly from rest and

reaches a speed of 17.9 m/s in 9.6 s. The
diameter of a tire is 38 cm.
Find the number of revolutions the tire
makes during this motion, assuming no slip-
ping.
Answer in units of rev.

Answer 1

Answer: 64.5 rev. ✅

Explanation: The diameter of the tire is 38cm and the car accelerates uniformly from rest to reach a speed of 17.9m/s in 9.6s. The distance covered by the tire can be calculated by using the formula: distance = speed x time.

The distance covered by the tire is 17.9 x 9.6 = 171.84 meters. But when we convert it centimeters it is 17184 cm

The distance covered by the tire is equal to the circumference of the tire, which is equal to 2πd (d is the diameter of the tire). So we can calculate the number of revolutions by dividing the distance covered by the tire by the circumference of the tire.

Number of revolutions = distance covered / Circumference of the tire = 17184/ (2 x π x (38/100)) = 64.5 ✅

Answer 2

Below

Step-by-step explanation:

Find distance vehicle travels during this acceleration

d = vo t + 1/2 a t^2

a = change in velocity / change in time = 17.9/9.6 = 1.86458 m/s^2

vo = 0 ( it starts from rest )

then d = 1/2 ( 1.86458)* 9.6^2 = 85.92 meters

Now divide by the tire circumference to get the number of revs

85.92 m / ( .38 m * pi )/rev = ~ 71.97 revs