Calculate the linear acceleration of a car, the 0.270-m radius tires of which have an angular acceleration of 15.0 rad/s2. Assume no slippage and give your answer in m/s2. m/s2 (b) How many revolutions do the tires make in 2.50 s if they start from rest? rev (c) What is their final angular velocity in rad/s? rad/s (d) What is the final velocity of the car in m/s? m/s

A) 4.05 m/s²

B) 7.46 revolutions

C) 37.5 rad/s

D) 10.125 m/s

Explanation:

A) Radius; r = 0.27m

Angular acceleration; α = 15 rad/s²

Now, the formula for linear acceleration is given by;

a = rα

Where r is radius and α is angular acceleration while a is linear acceleration.

Thus,

a = 0.27 x 15

a = 4.05 m/s²

B) First, Let's find the final velocity in rev/s using the equation of motion; V = u + αt

Where α is angular acceleration.

V = 0 + (15 x 2.5) = 37.5 rad/s

Now, we are calculating in revolutions, so let's convert the velocity to rev/s

1 rad/s = 1/2π rev/s

Thus, 37.5 rad/s = 37.5/2π revs/s = 5.968 rev/s

Now,average velocity = (v + u)/2 = (5.968 + 0)/2 = 5.968/2 = 2.984 revs/s

Thus, number of revolutions will be= average velocity x t =

2.984 rev/s x 2.5 s = 7.46 revolutions

C) we will get the final angular velocity from;

ω_f = ω_o + αt

Since it starts from rest, initial angular velocity ω_o = 0

Thus,

ω_f = 15 x 2.5 = 37.5 rad/s

D) The final velocity is given by;

v_f = ω_f•r = 37.5 x 0.27 = 10.125 m/s

the answer is b, have negligible mass