Intro Physics Distance Travelved Question

Click For Summary

Homework Help Overview

The original poster presents a problem involving the stopping distance of a car under uniform acceleration, comparing two initial velocities: 90.0 mph and 45.0 mph. The context is rooted in kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to determine the acceleration and stopping distance. Questions arise regarding the equivalence of different forms of the equations and the validity of the original poster's approach.

Discussion Status

Some participants have provided guidance on using a specific kinematic equation to calculate acceleration. The original poster acknowledges a mistake in their reasoning and expresses intent to try the suggested formula. There is an ongoing exploration of the problem without a clear consensus on the solution.

Contextual Notes

Participants note the assumption of uniform conditions for both scenarios and the importance of correctly applying the kinematic equations. There is mention of unit conversion and the implications of time as a variable in the calculations.

maff is tuff
Messages
65
Reaction score
1

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 :)

Homework Equations





The Attempt at a Solution



I have attached my attempt at the solution.

 

Attachments

Physics news on Phys.org
Have you come across this equation?

[itex]v^{2} = u^{2} + 2as[/itex]
 
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;

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

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

The equation you were using;

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

You then changed it to;

[itex]x - x_{0} = x + \frac{1}{2}at^{2}[/itex] (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.
 

Attachments

  • IMG_0164.jpg
    IMG_0164.jpg
    34.2 KB · Views: 511
Last edited:
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!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K