# Car accelerating up a hill

• nateb27
This car will accelerate at an average of .091 g (2.93 ft/s2).The 12.42 mile long mountain course has over 156 turns. (Not sure what to do with this information. It says we remove all the turns, so the course should no longer be 12.42 miles, right?)GIVEN EQUATIONS:d=v^2/2at=v_f/at=d_remaining/v_fI got that the time is equal to 1 minute and 41 seconds on one attempt, another gave 3 minutes and 7 seconds, and my latest attempt gave 4 minutes and 1 second. If anyone can
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 $2.93 \ \mathrm{{\frac{ft}{s^2}}}$ 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.

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.. ojorojo said:
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.

Ok maybe I am going to be wrong to put it this way but seems that distance traveled 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 traveled to reach the 203mph at the end..if that average acceleration doesn't 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.

The problem statement, as written, has a few ambi
JamesJones said:
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.)

collinsmark said:
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

fyithis is next...im coming up with 43,446 HP...i think i might be missing a proper conversion...

#### Attachments

ojorojo said:
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.

yep..wouldve saved us all a bunch of headaches...

haruspex said:
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.

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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