A car accelerates uniformly from rest and reaches a speed of 9.9 m/s in 11.4 s. The diameter of a tire is 86.9 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.

Number of revolutions = 20.71 rev.

Explanation:

Given the following data;

Initial speed, u = 0m/s

Final speed, v = 9.9m/s

Time, t = 11.4secs

Diameter = 86.9cm to meters = 86.9/100 = 0.869m

To find the acceleration;

Acceleration, a = (v - u)/t

Acceleration, a = (9.9 - 0)/11.4

Acceleration, a = 9.9/11.4

Acceleration, a = 0.87m/s²

Now we would find the distance covered by the tire using the second equation of motion.

S = ut + ½at²

S = 0(11.4) + ½*0.87*11.4²

S = 0 + 0.435*129.96

S = 56.53m

The circumference of the tire is calculated using the formula;

Circumference = 3.142 * diameter

Circumference = 3.142 * 0.869

Circumference = 2.73m

Number of revolutions = distance/circumference

Number of revolutions = 56.53/2.73

Number of revolutions = 20.71 rev.

Therefore, the number of revolutions the tire makes during this motion is 20.71 revolutions.

c the car's velocity increases constantly over time.