Help w problem

A car engine exerts a thrust of 1800 lb to accelerate the car (weight = 3400 lb) up a 5% grade speed of 80 mph. Rolling resistance on all wheels is estimated to be 40 lb/ton. Once at peak speed, the engine generates just enough thrust to maintain constant velocity. If the total distance travelled is 0.5 mile, what is the average running speed (mph) of the car?

A. 64

B. 68

C. 71

D. 74

I'm stuck on this and the solution provide just does a bunch of short cuts which are not helpful.

Doing a quick calc, I get B, is that correct?

Yes that is correct. Can you show me your quick calc. I'm trying to come up with a procedure as to how to calculate these type of problems.

M6, sure, no problem. But keep in mind, I know I didn't solve it the "correct" way. But this is how I solved it:

For starters, before you can even begin, you must assume the car starts at 0 mph, even though the problem doesn't state so.

Now, I hate these rolling resistance type problems! To me, they feel more like dynamic/physics problems meant for the FE not the PE. So, in my "quick/cheater" calc, I ignored the rolling resistance. What I did is I assumed an industry average for acceleration. In my experience, a well-written PE problem should always include realistic values. So I assumed that this problem used a realistic value for the acceleration rate. I used 11.2 ft/s2, per AASHTO GDHS. I ignored the 5% grade. Then, the rest is all just simple Uniform Acceleration formulas (see CERM Table 72.1).

There are two stages. The first is accelerating from 0 to 80. The second is traveling at a constant 80 for 0.5mi. Calc the time and distance for each stage:

Time and Distance to accel from 0 to 80mph: (80mph = 117.33 ft/s)

t = (117.33 - 0) / 11.2 = 10.48 s

d = (117.332 - 02) / (2 x 11.2) = 614.57 ft

Time and Distance to travel at a constant 80mph for 0.5mi: (0.5mi = 2640ft)

t = (2 x 2640) / (117.33 + 117.33) = 22.50 s

d = 2640 ft

Total time = 10.48 + 22.50 = 32.98 s

Total distance = 614.57 + 2640 = 3254.57 ft

Running Speed = 3254.57 / 32.98 = 98.68 ft/s = 67.28 mi/hr = Choice B

I am not sure if that helps or not. ? Could you post the book's solution?

Here's the solution:

Grade resistance = 0.05x3400=170 lb

Rolling resistance = 3400 lb x 1 ton/2000 lb x 40 lb/ton= 68 lb

Net thrust = 1800-170-68=1562 lb

Mass=3400/32.2= 105.6 lb-s2/ft

Net acceleration = 1562/105.6= 14.8 ft/s2

Time to accelerate to 80 mph (117.3 fps) from rest = 7.93 s

Distance traveled during acceleration: s = 1/2 at2 = 1/2x14.8x7.93^2= 465 ft

Time to travel the remaining (2640-465)=2175/117.3= 18.54 s

Total time to travel 2640 ft = 7.93 + 18.54 = 26.47 s

Average running speed = 2640/26.47 = 99.7 fps = 68 mph.

M6, I think I may have had the case of two wrongs made a right.

1.) I just realized that the total distance traveled is 0.5mi, not just the distance traveled at a constant 80mph, as I previously misinterpreted.

2.) Also, the more I think about it, the more uncomfortable I feel about assigning acceleration as 11.2ft/s2. I think the 11.2 value that the GDHS uses is strictly for deceleration, not acceleration (acceleration should be less than deceleration – even 50% less).

Please disregard my calcs. Sorry.

If time allows, I’ll try to solve the problem “the real way”.

I was posting when you were posting.

11.2 ft/sec^s is the deceleration time used for braking distance. I was wondering why you used that for acceleration. Makes sense now.

John Q said:

11.2 ft/sec^s is the deceleration time used for braking distance. I was wondering why you used that for acceleration. Makes sense now.

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Yeah, sorry about that. I did not realize accel and decel rates can be quite different. Here is what I found in researching this:

Acceleration rate Deceleration rate

AASHTO GDHS 4 to 4.5 ft/s2 (From 0 to 25-40mph) 11.2 ft/s2 (comfortable)

≥14.8 ft/s2 (emergency)

ITE Traffic Engineering Handbook 3.6 ft/s2 9.8 ft/s2

FHWA 7.84 ft/s2 (sedans) (0-60mph)

12.49 ft/s2 (sport cars) (0-60mph)

Spurr 6.4 ft/s2 (0.16-0.37g; avg: 0.2g)

Haas, Inman, Dixson, Warren 4.5 ft/s2 (for all reached speeds) 5.79 ft/s2 (initial speed 35-40 mph)

That's a nifty little table you have there. I'll be taking that with me!

MARIO650 said:

Here's the solution:

Grade resistance = 0.05x3400=170 lb

Rolling resistance = 3400 lb x 1 ton/2000 lb x 40 lb/ton= 68 lb

Net thrust = 1800-170-68=1562 lb

Mass=3400/32.2= 105.6 lb-s2/ft

Net acceleration = 1562/105.6= 14.8 ft/s2

Time to accelerate to 80 mph (117.3 fps) from rest = 7.93 s

Distance traveled during acceleration: s = 1/2 at2 = 1/2x14.8x7.93^2= 465 ft

Time to travel the remaining (2640-465)=2175/117.3= 18.54 s

Total time to travel 2640 ft = 7.93 + 18.54 = 26.47 s

Average running speed = 2640/26.47 = 99.7 fps = 68 mph.

Click to expand...

I wave the white flag on this one, sorry M6. I am cool to run with it once the acceleration is known but I am lost with all that resistance and thrust stuff leading up to the calculation of a.

That's ok. I think I will do your approach it's an approximation I know. Also I will be taking structural for the pm so hopefully nothing this complicated will show up in the am.

MARIO650 said:

That's ok. I think I will do your approach it's an approximation I know. Also I will be taking structural for the pm so hopefully nothing this complicated will show up in the am.

Click to expand...

I think that's a bad idea, both here and in general... while you can often take a shortcut by neglecting small factors, it's a bad habit if you don't understand each and every one made, and the impact to process.

What's tripping you up here? There are only two preliminary steps to this problem: 1) Drawing a free body diagram and 2) F=ma.

For the free body diagram, the forces that retard the thrust of the car are gravity (what's been called grade resistance) and friction (rolling resistance). In the real-world, you'd need to add drag (air resistance) to the friction, but as you're not given the fluid density, cross-sectional area, and drag coefficient, you cannot reasonably add here. Drag would also complicate the solution as it's proportional to the square of the velocity (so acceleration wouldn't be constant) and you'd have to integrate. In any case, sum up the forces parallel to the direction of travel and you're done.

Anyway, back to the problem: once you get the net thrust (F), you can use it to calculate the constant acceleration (a). Again, you are forced to assume constant acceleration because you don't have enough to factor in air resistance. Then it's a simple application of the constant acceleration formulas (v=u+at and s=ut+0.5at^2) to get you to the answer.

Make sense?

I never liked FBDs.

IlPadrino said:

MARIO650 said:

That's ok. I think I will do your approach it's an approximation I know. Also I will be taking structural for the pm so hopefully nothing this complicated will show up in the am.

Click to expand...

 I think that's a bad idea, both here and in general...  while you can often take a shortcut by neglecting small factors, it's a bad habit if you don't understand each and every one made, and the impact to process.

What's tripping you up here?  There are only two preliminary steps to this problem:  1) Drawing a free body diagram and 2) F=ma.

For the free body diagram, the forces that retard the thrust of the car are gravity (what's been called grade resistance) and friction (rolling resistance).  In the real-world, you'd need to add drag (air resistance) to the friction, but as you're not given the fluid density, cross-sectional area, and drag coefficient, you cannot reasonably add here.  Drag would also complicate the solution as it's proportional to the square of the velocity (so acceleration wouldn't be constant) and you'd have to integrate.  In any case, sum up the forces parallel to the direction of travel and you're done.

Anyway, back to the problem:  once you get the net thrust (F), you can use it to calculate the constant acceleration (a).  Again, you are forced to assume constant acceleration because you don't have enough to factor in air resistance.  Then it's a simple application of the constant acceleration formulas (v=u+at and s=ut+0.5at^2) to get you to the answer.

Make sense?

Click to expand...

What I don't understand is why don't they add the distance up hill and down hill IE 2640 + 465 = 3105 ft then divide by the total time 34s to get 102 ft/s which is 69.5 mph.

I understand now I was adding when I should have subtracted the initial distance it took to get to 80mph. I usually get stuck on how to factor in the friction (incline grade, friction factor, and air resistance) into the equation. But I will draw the FBD to help me see this. From that I will get the force acting on the car and then find my acceleration from the equation F=ma then use that a to go into table 71.1 in CERM 11th (uniform acceleration formulas). Thank you for your help.

MARIO650 said:

I understand now I was adding when I should have subtracted the initial distance it took to get to 80mph. I usually get stuck on how to factor in the friction (incline grade, friction factor, and air resistance) into the equation. But I will draw the FBD to help me see this. From that I will get the force acting on the car and then find my acceleration from the equation F=ma then use that a to go into table 71.1 in CERM 11th (uniform acceleration formulas). Thank you for your help.

Click to expand...

You got it! "TOTAL distance traveled is 0.5 miles"

And for what it's worth, this is an FE type problem - you won't see it on the PE exam.