Question

A car accelerates uniformly from rest and reaches a speed of 24.8 m/s in 10.1 s. The diameter of a tire is 38.9 cm.

Find the number of revolutions the tire makes during this motion, assuming no slip-ping.
Answer in units of rev.

Answer 1

409.66

Step-by-step explanation:

Given:

\omega_0=0\: \rm rad/sω0=0rad/s

v=24.8\: \rm m/sv=24.8m/s

t=10.1\: \rm st=10.10s

d=38.9\:\rm cmd=38.9cm

The final angular speed

\omega=\frac{v}{R}=\frac{24.8}{0.389/2}=127.5\:\rm rad/sω=Rv=0.389/49.6.0=127.5rad/s

(a)

\phi=\frac{\omega+\omega_0}{2}=\rm \frac{127.5+0.00}{2}\times 10.1= 2575\: radϕ=2ω+ω0=127.5+0.00×10.1=2575.5rad

N=\frac{\phi}{2\pi}=\frac{2575}{2\pi}=409.66N=2πϕ=2575=409.66