Question

The tires of a car make 70 revolutions as the car reduces its speed uniformly from 86.0 km/h to 65.0 km/h . The tires have a diameter of 0.90 m . Part A: What was the angular acceleration of the tires? Express your answer using two significant figures. Part B: If the car continues to decelerate at this rate, how much more time is required for it to stop? Express your answer using two significant figures and include the appropriate units. Part C: How far does the car go? Find the total distance. Express your answer to three significant figures and include the appropriate units.

Answer 1

A)
radius of the wheel, r = d/2 = 0.9/2 = 0.45 m

v1 = 86 km/h = 86*5/18 = 23.9 m/s

v2 = 65 km/h = 65*5/18 = 18.06 m/s

linear dispalcement during this time, s = 70*pi*0.9

= 197.92 m

let a is the linear acceleration

Apply, v2^2 - v1^2 = 2*a*s

==> a = (v2^2 - v1^2)/(2*s)

= (18.06^2 - 23.9^2)/(2*197.92)

= -0.619 m/s^2

angular acce;eration, alfa = a_tan/r

= 0.619/0.45

= -1.38 rad/s^2 <<<<<<------------Answer

B) let t is the time taken to stop.

v3 = 0

t = (v3 - v2)/a

= (0 - 18.06)/(-0.619)

= 29 s <<<<<<------------Answer

C) Total distance travelled by the car = (v3^2 - v1^2)/(2*a)

= (0^2 - 23.9^2)/(2*(-0.619))

= 461.4 m <<<<<<------------Answer