Kinetic Friction To Find Acceleration

[tater08]

Homework Statement

You have been called to testify as an expert witness in a trial involving an automobile accident. The speed limit on the highway where the accident occurred was 40 mph. The driver of the car slammed on his brakes, locking his wheels, and left skid marks as the car skidded to a halt. You measure the length of these skid marks to be 219 ft, 9 in., and determine that the coefficient of kinetic friction between the wheels and the pavement at the time of the accident was 0.400.

How fast was this car traveling (to the nearest number of mph) just before the driver hit his brakes?

Homework Equations

(x1-x0)=(vf^2-Vi^2)/2a a=ug

The Attempt at a Solution

219.75 feet/5280 feet = 0.04 miles
-0.04miles=-vi^2/2(0.40 *9.8)
-0.04=-vi^2/7.84 vi=0.56 miles/second 2016 miles per hourt

this answer does not make sense. I am totally lost on this problem. Help please!

 

[Rake-MC]

Look at your units. You're using miles and feet, but your value for g is 9.8 (m/s/s)

 

[tater08]

yeah uhm, I'm still getting the wrong answer.

 

[Perillux]

have you tried converting the 9.8 m/s^2 into miles/hour^2 like Rake-MC said?
if you have tried it, then please give us an update.

 

[LowlyPion]

Homework Statement

You have been called to testify as an expert witness in a trial involving an automobile accident. The speed limit on the highway where the accident occurred was 40 mph. The driver of the car slammed on his brakes, locking his wheels, and left skid marks as the car skidded to a halt. You measure the length of these skid marks to be 219 ft, 9 in., and determine that the coefficient of kinetic friction between the wheels and the pavement at the time of the accident was 0.400.

How fast was this car traveling (to the nearest number of mph) just before the driver hit his brakes?

Homework Equations

(x1-x0)=(vf^2-Vi^2)/2a a=ug

The Attempt at a Solution

219.75 feet/5280 feet = 0.04 miles
-0.04miles=-vi^2/2(0.40 *9.8)
-0.04=-vi^2/7.84 vi=0.56 miles/second 2016 miles per hourt

this answer does not make sense. I am totally lost on this problem. Help please!

Converting everything to meters (.3048m = 1 foot) to then calculate meters/sec using your formula, which looks ok I get 36.1 m/s and converting with 2.237 m/s to m/h yields 80.8 or 81mph to the nearest mph.

 

[tater08]

Thanks Rake, Perrilux and Lowly Pion. I ended up getting the right answer. Thanks.

 

[SuPaFly43218]

I am also trying to figure this out. The answer so happens to be 51.2 mph, but I have NO idea how to figure this out. PLEASE HELP!

 

[Perillux]

If you write out the equation for friction you will see that the mass cancels and you are left with the acceleration of friction and the acceleration of the normal (which should equal the acceleration of gravity in this case). So you can find how much it is decelerating (due to friction), then just figure out at what velocity he needs to be going to travel 219ft, 9in. with the deceleration you found.

Make sure your units are correct, I would recommend making sure all distances are in meters and all times are in seconds. so velocity should be given in m/s and acceleration m/s^s. Do these conversions before you start your calculations.