A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. Assuming the diameter of a tire is 58.0 cm, (a) find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?

Explanation:

Given:

Initial velocity, u = 0 m/s (at rest)

Final velocity, v = 22 m/s

Time, t = 9 s

Diameter, d = 58 cm

Radius, r = 0.29 m

Using equation of motion,

v = u + at

a = (22 - 0)/9

= 2.44 m/s^2

v^2 = u^2 + 2a × S

S = (22^2 - 0^2)/2 × 2.44

= 99.02 m

S = r × theta

Theta = 99.02/0.29

= 341.44 °

1 rev = 360°

341.44°,

= 341.44/360

= 0.948 rev

= 0.95 rev

B.

Final angular speed, wf = v/r

= 22/0.29

= 75.86 rad/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

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