A 2000 kg car moves along a horizontal road at speed vo 15.9 m/s. The road is

wet, so the static friction coefficient between

the tires and the road is only pls 0.105 and

the kinetic friction coefficient is even lower,

= 0.0735.

The acceleration of gravity is 9.8 m/s2.

What is the shortest possible stopping dis-

tance for the car under such conditions? Use

g= 9.8 m/s² and neglect the reaction time of

the driver.

Answer in units of m.

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

(Eq. 1)

(Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

(Eq. 3)

Where:

- Deceleration of the car, measured in meters per square second.

- Kinetic coefficient of friction, dimensionless.

- Gravitational acceleration, measured in meters per square second.

If we know that and , then deceleration of the car is:

The stopping distance of the car (), measured in meters, is determined from the following kinematic expression:

(Eq. 4)

Where:

- Initial speed of the car, measured in meters per second.

- Final speed of the car, measured in meters per second.

If we know that , and , stopping distance of the car is:

The shortest possible stopping distance of the car is 175.319 meters.

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graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2