In: Physics

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.

Diameter of the tires = D = 0.82 m

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

Initial speed of the car = V_{1} = 88 km/h = 24.44
m/s

Speed of the car after 72 revolutions of the tires =
V_{2} = 60 km/h = 16.67 m/s

Number of revolutions made by the tire in this time =
n_{1} = 72

Initial angular velocity of the tire = _{1}

_{1} =
V_{1}/R

_{1} =
24.44/0.41

_{1} =
59.61 rad/s

Angular velocity of the tire after 72 revolutions of the tire =
_{2}

_{2} =
V_{2}/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} =
2n_{1}

_{1} =
2(72)

_{1} =
452.39 rad

_{2}^{2}
= _{1}^{2}
+ 2_{1}

(40.658)^{2} = (59.61)^{2} + 2(452.39)

= -2.1
rad/s^{2}

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 =

_{3}^{2}
= _{1}^{2}
+ 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/s^{2}

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

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