Transportation Practice Problem

[bigray76]

Transportation Problem of the Week (from the review class I took at Rutgers)

A vehicle hits a bridge abutment at a speed estimated by investigation as 15 mph. Skid marks of 100 feet on the pavement (f = 0.35) followed by skid marks on the gravel shoulder approaching the abutment (f = 0.5) are observed. The grade is level. What is the initial speed of the vehicle?

 

[owillis28]

Can you post the answer right away so we can see if our work was correct?

Thanks

owillis28

 

[Suns Den]

Transportation Problem of the Week (from the review class I took at Rutgers)
A vehicle hits a bridge abutment at a speed estimated by investigation as 15 mph. Skid marks of 100 feet on the pavement (f = 0.35) followed by skid marks on the gravel shoulder approaching the abutment (f = 0.5) are observed. The grade is level. What is the initial speed of the vehicle?

50 mph

 

[owillis28]

I didn't get that answer. Not sure where I went wrong? Help!

owillis28

 

[Suns Den]

I didn't get that answer. Not sure where I went wrong? Help!
owillis28

I used 75.24b CERM and Vmiles = 35 mph + 15 mph (impact speed)= 50mph

 

[owillis28]

How did you account for the coefficient of friction of the pavement & the gravel shoulder?

I am guessing that you added the two together in the bottom part of the equation. Reading the question again, I think I was to assume that the 100 foot distance contained both the pavement and gravel shoulder before the abutment. Damn tricky questions.

owillis

p.s. BigRay-Is Suns Dens' answer correct?

 

[Suns Den]

I did average the coefficient of friction...

 

[bigray76]

I will post the solution this afternoon... on my way out the door to go to a presentation.

-Ray

 

[bigray76]

I will post the solution this afternoon... on my way out the door to go to a presentation.
-Ray

Make that tomorrow morning... never made it back to the office after the presentation... at a co-worker's house and we are going drinking the rest of the afternoon. What a brutal client!

-Ray

 

[Undertaker]

Will work on this tonight. Thanks bigray76

 

[bigray76]

I forgot to mention that the gravel shoulder skid marks are 200 feet long. Sorry about that.

Solution:

db = (v2^2 - v1^2) / 30(f +/- g)

db (gravel) = 200 = v^2 - 15^2 / 30(0.5)

3000 = v^2 - 15^2

3225 = v^2

v = 56.8 mph (speed at end of pavement skid, beginning of gravel skid)

db (pavement) = 100 = v^2 - 56.8^2 / 30(0.35)

1050 = v^2 - 3226.24

v^2 = 4276.24

v = 65.4 mph (speed of vehicle just prior to skidding on pavement)