Question

Ch.8, 4

The tires of a car make 72 revolutions as the car reduces its speed uniformly from 88.0 km/h to 60.0 km/h. The tires have a diameter of 0.82 m.

Part A

What was the angular acceleration of the tires?

Part B

If the car continues to decelerate at this rate, how much more time is required for it to stop?

Part C

If the car continues to decelerate at this rate, how far does it go? Find the total distance.

Answer 1

Diameter of the tires = D = 0.82 m

Radius of the tires = R = D/2 = 0.41 m

Initial speed of the car = V₁ = 88 km/h = 24.44 m/s

Speed of the car after 72 revolutions of the tires = V₂ = 60 km/h = 16.67 m/s

Number of revolutions made by the tire in this time = n₁ = 72

Initial angular velocity of the tire = ₁

₁ = V₁/R

₁ = 24.44/0.41

₁ = 59.61 rad/s

Angular velocity of the tire after 72 revolutions of the tire = ₂

₂ = V₂/R

₂ = 16.67/0.41

₂ = 40.658 rad/s

Angular acceleration of the tires =

Angle turned by the tires in 72 revolutions = ₁

₁ = 2n₁

₁ = 2(72)

₁ = 452.39 rad

₂² = ₁² + 2₁

(40.658)² = (59.61)² + 2(452.39)

= -2.1 rad/s²

Negative as it is deceleration.

Final angular velocity of the tire = ₃ = 0 rad/s

Additional time required for the car to come to a stop = t

₃ = ₂ + t

0 = 40.658 + (-2.1)t

t = 19.36 sec

Total angle turned by the tires from initial speed to rest =

₃² = ₁² + 2

(0)² = (59.61)² + 2(-2.1)

= 846.03 rad

Total distance traveled by the car before coming to a stop = L

L = R

L = (0.41)(846.03)

L = 346.87 m

A) Angular acceleration of the car = -2.1 rad/s²

B) If the car continues to decelerate additional time required for it to stop = 19.36 sec

C) Total distance traveled by the car before stopping = 346.87 m