Question

A car accelerates uniformly from rest and reaches a speed of 21.9 m/s in 8.99 s. Assume the diameter of a tire is 57.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs.

Answer 1

Answer:

Explanation:

initial velocity, u = 0 m/s

time, t = 8.99 s

final velocity, v = 21.9 m/s

diameter of tyre, D = 57.5 cm

(a) Let a is the acceleration, and n be the number f revolutions made by the tyre.

Use first equation of motion

v = u + at

21.9 = 0 + a x 8.99

a = 2.44 m/s²

Use third equation of motion

v² = u² + 2 a s

21.9 x 21.9 = 0 + 2 x 2.44 x s

s = 98.28 m

Circumference of wheel, S = π x D = 3.14 x 0.575 = 1.8055 m

number of revolutions, n = distance / circumference

n = 98.28 / 1.8055

n = 54.4