Intro Physics Distance Travelved Question

In summary, the equation you were using is incorrect because you incorrectly cancelled out the "t" with the "s". The equation you were using;x-x_{0}= v_{0}t+\frac{1}{2}at^{2}will work correctly.
  • #1
maff is tuff
65
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

  • Problem 2.6.pdf
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  • #2
Have you come across this equation?

[itex]v^{2} = u^{2} + 2as[/itex]
 
  • #3
No but I've come across this one:

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

Are those the same?
 
  • #4
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 :)
 
  • #5
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
 
  • #6
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
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Last edited:
  • #7
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.
 
  • #8
No worries mate! We all make mistakes. It was a good attempt though!
 

Related to Intro Physics Distance Travelved Question

1. What is distance traveled in physics?

Distance traveled in physics is defined as the total length covered by an object in motion. It is a scalar quantity and is usually represented by the symbol 'd'.

2. How is distance traveled calculated?

Distance traveled can be calculated by multiplying the average speed of the object by the time it took to cover that distance. In mathematical terms, it can be expressed as d = v * t, where d is distance, v is average speed, and t is time.

3. What is the difference between distance traveled and displacement?

Distance traveled refers to the total length covered by an object, whereas displacement refers to the shortest distance between the initial and final position of the object. Displacement is a vector quantity, while distance traveled is a scalar quantity.

4. Can distance traveled be negative?

Yes, distance traveled can be negative if the object is moving in the opposite direction of the reference point. This means that the object is moving in the negative direction of the axis and the distance traveled will be represented by a negative value.

5. How does distance traveled relate to velocity?

Distance traveled and velocity are related in the sense that velocity is the rate of change of distance traveled over time. In other words, velocity is the speed and direction at which an object is moving, while distance traveled is the total length covered by the object in motion.

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