Intro Physics Distance Travelved Question

AI Thread Summary
A physics problem involves calculating the stopping distance of a car moving at 45.0 mph, given that a car traveling at 90.0 mph stops in 400 ft under uniform acceleration. The relevant equation for this calculation is v² = u² + 2as, which helps determine the required acceleration. The user initially misapplied the formula by incorrectly canceling units, leading to confusion in their calculations. After receiving clarification on the correct use of the formula, they acknowledged their mistake and planned to apply the suggested method. The discussion emphasizes the importance of correctly handling units in physics equations.
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Homework Statement



If a car moving at 90.0 mph takes 400 ft to stop with uniform acceleration after its brakes are applied, how far will it take to stop under the same conditions if its initial velocity is 45.0 mph?

Thanks for the help :)

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The Attempt at a Solution



I have attached my attempt at the solution.

 

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Have you come across this equation?

v^{2} = u^{2} + 2as
 
No but I've come across this one:

v^2=v0^2 +2a(x-x0)

Are those the same?
 
Yes they're the same thing, just using different symbols and such.

Use the above formula to calculate the acceleration required to stop the car (in the first case).

The question tells you to assume the same conditions for the second case (therefore the acceleration is the same).

Once you start it should all work itself out :)
 
Okay thanks I'll try that formula. But why didn't the formula that I used work? I solved for (x-x0) and it should have given me the right answer. And for future reference...when I reply to my own thread do you get some kind of notification or are you just checking back here? Thanks
 
The source of your error appears to be this;

400 ft = (132 \frac{ft}{s})t + \frac{1}{2}at^{2} (Line 1 - Correct)

400 ft = 132 ft + \frac{1}{2}at^{2} (Line 2 - Incorrect)

The equation you were using;

x - x_{0}= v_{0}t + \frac{1}{2}at^{2}

You then changed it to;

x - x_{0} = x + \frac{1}{2}at^{2} (No longer the same formula - you can't cancel the per second and the variable t.)

All working after that will then be invalid.

I just attempted to solve the problem using your method, but am left with a variable of time which we have no way of getting without the acceleration. I attached a photo of my workings using your method in case you're interested. (I'm from Australia so I converted the units to metres and metres per second)

If you need any more help let me know. :)

P.s Whenever somebody replies to a thread I've posted in, I am sent an e-mail. I can check my e-mails from my phone.
 

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Ok I see my dumb mistake now. I don't know why but I canceled the seconds out with the time. But now I see that time is not just seconds because for all we know it can be an interval so I can't just cancel out the "t" with the "s". Ok I'm going to go ahead and try the other formula you suggested. Thanks again for the detailed explanation; I probably would have never seen that horrible mistake.
 
No worries mate! We all make mistakes. It was a good attempt though!
 

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