Question

A car accelerates uniformly from rest and reaches a speed of 11.9 m/s in 11.5 s. The diameter of a tire is 89 cm. Find the number of revolutions the tire

makes during this motion, assuming no slip-ping

Answer 1

153 revolutions

Step-by-step explanation:

D = (Vi)t + 1/2(a)(t²)

Since Vi = 0, the first term drops out

a = Δv/t = (11.9 m/s)/(11.5 s) = 1.03 m/s²

D = 1/2(1.03 m/s²)(11.5 s)² = 68.1 m

Circumference of a circle = C = 2πr

r = D/2 = (0.89 m)/2 = 0.445 m

So for every 1 revolution of a tire, the distance traveled is 0.445 m

68.1 m / 0.445 m/rev = 153 revolutions