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Car accelerating up a hill

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What's mostly missing is the common sense step of verifying what the high school homework problem actually is, determining what elements should be included, and if there's any variables that are missing or need clarification.
 

haruspex

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Thanks for posting the original exactly.
This suggests to me that the acceleration quoted is from some genuine data for the vehicle, hence is for the flat, and you do need to discount for the gradient.
However, the same view of the top speed number leads to the likelihood that the top speed is reached before the hilltop is attained. That would leave us with insufficient data to solve the problem.
 

haruspex

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If we removed all the curves, making a straight line from top to bottom, how quickly could the car make it to the top?
Making it a straight line means it is a lot less than 12 miles. Shouting doesn't change that.
you use the formulas to figure all of that out.
Without knowing the distance, how? You seem to be arguing it surely wouid reach top speed before hill top, but then we have insufficient data to answer the question.
 

Tom.G

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12.42 miles are the total real course mileage including 156 turns...the mileage to be ran its assumed to be shorter
I offer an alternative assumption, that the curves are large enough radius that full speed can be maintained around them. (Since there was no map included, that still fits the description.)
 

haruspex

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I offer an alternative assumption, that the curves are large enough radius that full speed can be maintained around them. (Since there was no map included, that still fits the description.)
I do not see the relevance of that observation. The curvy course of 12.42 miles is mere background and of no significance in the question being asked.
 

collinsmark

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then i guess you can't answer that problem then, can you?
He's is not trying to do your homework for you. He's trying to help you. Politely keep that in mind, please.
 
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Could we assume that in this case 203mph is reached at the moment the car crosses the finish line and consider it Vf..? Could the time from "a" to "b" be determined then....?
 
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Screenshot_20191108-184229_Gallery.jpg

Theres Pikes Peak. No curves..forget curves...straight shot from point A to point B...with an incline of 6.4% thrown in that line
 
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BTW..fastest time ever ran is 7:57..That's throught the whole 12.42miles...Just FYI if curious...nuthin to do with the problem...👍
 

collinsmark

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The problem statement, as written, has a few ambiguities in it. The main one regards what is meant by "if we removed all the curves ... ."

Interpretation A:
Keep the 3-dimensional starting location and finish location exactly where they are, and then draw a new, straight line between them, and that is the new track.

However, this will increase the slope above 6.4 %, if that even matters.

Without knowing more information the relative position between the starting location and finish location, there isn't enough information to solve this problem, in this interpretation.

Interpretation B:
Grab the starting position and pull it out horizontally until the track becomes a taught, straight line. In this interpretation, the length of the track is still 12.42 miles and the slope is still 6.4% (if the slope even matters).

Other ambiguities:
The maximum acceleration is specified as [itex] 2.93 \ \mathrm{\frac{ft}{s^2}} [/itex]. But is this the maximum acceleration with respect to the road when climbing a 6.4% slope, or [itex] 2.93 \ \mathrm{\frac{ft}{s^2}} [/itex] total?

You see, initially when the vehicle is accelerating, it will be fighting the acceleration of gravity somewhat due to the fact that it's on a slope. Should this be taken into account, or is it already taken into account in the [itex] 2.93 \mathrm{\frac{ft}{s^2}} [/itex] specification?

Also, I looked up the Pikes Peak International Hill Climb. Its track length is 12.42 Miles with an elevation change (from start to finish) of 4725 ft. That produces an average slope gradient of 7.22%, although this problem says to use only 6.4%, and honestly, I'm not certain whether the slope is relevant to solving this problem at all, given the other ambiguities.
 
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collinsmark

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If I had to guess though (and this is just a guess -- I could be wrong), ignore the slope altogether, and just treat this problem as a car driving on a straight, level, 12.42 mile track.

The vehicle accelerates at [itex] 2.93 \ \mathrm{{\frac{ft}{s^2}}} [/itex] until it reaches its top speed of 203 mph. After that, it continues at 203 mph until reaching the finish line.

But that's just my guess. The question is ambiguous enough such that there's a good chance my interpretation is wrong.
 
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Guess the author of this problem expects for the "made up or not slope" of 6.4% to be considered...just go with it..he didnt include friction nor mass throughout the slope so assume the given acceleration is "average" as stated from Zero to 230mph at finish all throughout the incline....you guys are getting way to complicated...I'm surprised you guys haven't questioned if the driver has checked the proper PSI on the tires..😁
 

jbriggs444

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so assume the given acceleration is "average" as stated
An "average" acceleration is of little use. The distance covered in a given time with a front-loaded acceleration will be greater than that covered in a given time with a more constant acceleration, even though the "average" is the same in both cases.
 
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Ok maybe im going to be wrong to put it this way but seems that distance travelled is unimportant here to the original author..what if he just wants to know Time from a to b at with that given acceleration and the made up or not slope to add to the fun...and could care less how far the car actually travelled to reach the 203mph at the end..if that average acceleration doesnt change can it eventually reach the 203mph?
 
I am thinking that at a 6.4% slope the straight road would be longer than the curved road.
Possibly 73916.5 feet or 13.999 miles long.
I used a right triangle with a height of 4721 feet and calculated a base of 73765.625 feet based on the height and slope. Then Pythagoras did the rest.
 

collinsmark

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The problem statement, as written, has a few ambi
I am thinking that at a 6.4% slope the straight road would be longer than the curved road.
Possibly 73916.5 feet or 13.999 miles long.
I used a right triangle with a height of 4721 feet and calculated a base of 73765.625 feet based on the height and slope. Then Pythagoras did the rest.
For what it's worth, the actual, real-world track length is 12.42 miles. That's one detail that matches the problem description given here.

But that's not the displacement between the two points, it's the total distance, which includes curves, hills, and all. The real-world displacement will is significantly shorter than 12.42 miles, not longer.

If your approach is to find the straight-line displacement of the real-world start and finish locations, you will need to look up their GPS coordinates on the Internet (presumably), and use geometry to calculate their 3-dimensinal displacement (and effective slope, if that's even relevant.) Keep in mind that these necessary details were not given in the problem statement, as it was described. This leads me to believe that this is probably not the approach the author of the problem intended. (Then again, this problem statement is so full of ambiguities, it's anybody's guess.)
 
The problem statement, as written, has a few ambi

For what it's worth, the actual, real-world track length is 12.42 miles. That's one detail that matches the problem description given here.

But that's not the displacement between the two points, it's the total distance, which includes curves, hills, and all. The real-world displacement will is significantly shorter than 12.42 miles, not longer.

If your approach is to find the straight-line displacement of the real-world start and finish locations, you will need to look up their GPS coordinates on the Internet (presumably), and use geometry to calculate their 3-dimensinal displacement (and effective slope, if that's even relevant.) Keep in mind that these necessary details were not given in the problem statement, as it was described. This leads me to believe that this is probably not the approach the author of the problem intended. (Then again, this problem statement is so full of ambiguities, it's anybody's guess.)
I interpreted the slope given in the problem (6.4%) as a constant. A constant which would have to be ignored in order to use straight line displacement. This is not homework, it is a qualifying question for a contest.
here is the text;
The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from top to bottom, how quickly could a Challenger SRT® Hellcat Redeye reach the top?

ProTips

Slope of 6.4%

Redeye top speed 203 mph

A stock Redeye has a peak acceleration of over 1g. Assume this Redeye will accelerate at an average of .091 g (2.93 ft/s2)

The 12.42 mile long Pike’s Peak course has over 156 turns

What you might need:

A calculator

The internet (unit conversions)

A smart friend

Some kinematic equations:

d=v^2/2a

t=v_f/a

t=d_remaining/v_f
 
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About 12 minutes, including turns.
 

haruspex

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this is the answer to the previous problem
So quite easy once it is explained that the straightening of the road was supposed to preserve the length, not the slope - despite that in the question statement the slope is given prominence, while the distance seems to be just background on the actual course.
 
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yep..wouldve saved us all a bunch of headaches...
 

jbriggs444

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So quite easy once it is explained that the straightening of the road was supposed to preserve the length, not the slope - despite that in the question statement the slope is given prominence, while the distance seems to be just background on the actual course.
Although the 6.4% slope is re-iterated on the solution slide, it is never actually used in the calculation. Despite the prescription of the 0.91 g acceleration being average, the proffered calculation treats it as constant instead.
 
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The problem - as given - would barely qualify as a primary school homework assignment.

As stated earlier, the actual time for the run was about 12 minutes : 60mph average, give or take.

It might be interesting to use the car and tire specs, and include the slope, hairpins, and road-surface data. But, mostly that would just show up the car and tire companies' marketing figures, which are as far removed from "real world, with non-pro driver, an underinflated left rear tire, 2 kids arguing in the back seat, etc." as the stick-figure physics problem.
 
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