A car accelerates uniformly from rest and

reaches a speed of 34.3 m/s in 10.4 s. The

diameter of a tire is 33.2 cm.

Find the number of revolutions the tire

makes during this motion, assuming no slipping.

Answer in units of rev.

Question

Physics

Fuzzy

13 May, 20:40

A car accelerates uniformly from rest and

reaches a speed of 34.3 m/s in 10.4 s. The

diameter of a tire is 33.2 cm.

Find the number of revolutions the tire

makes during this motion, assuming no slipping.

Answer in units of rev.

+4

Answers (1)

Know the Answer?

Quintin · Answer 1

171 rev

Explanation:

Given:

ω₀ = 0 rad/s

ω = 34.3 m/s / (0.332 m / 2) = 207 rad/s

t = 10.4 s

Δθ = ½ (ω + ω₀) t

Δθ = ½ (207 rad/s + 0 rad/s) (10.4 s)

Δθ = 1070 rad

Δθ = 171 rev

A car accelerates uniformly from rest andreaches a speed of 34.3 m/s in 10.4 s. Thediameter of a tire is 33.2 cm.Find the number of revolutions the tiremakes during this motion, assuming no slipping.Answer in units of rev.

A car accelerates uniformly from rest and

reaches a speed of 34.3 m/s in 10.4 s. The

diameter of a tire is 33.2 cm.

Find the number of revolutions the tire

makes during this motion, assuming no slipping.

Answer in units of rev.