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Finding the distance in an acceleration problem?

  • #1

Homework Statement


What distance is required to stop a car going 90km/h if the car can stop with an acceleration of -6.0m/s2?


Homework Equations


a = v2-v1/t
d = (v1+v2/2)t


The Attempt at a Solution


I'm thinking first we have to solve for t. Which would be the speed divided by acceleration, which is -0.36 km/h^2 over 80km/h which is 0.0045 hours.

Now to solve for distance, it would be 80km/2 x 0.0045

Which leaves me with 0.18 which is completely wrong because the answer is supposed to be 41 m.

Help please? Thank you.
 

Answers and Replies

  • #2
Oops speed over acceleration is actually 80km/h over -0.36km/h, but that still gives me the wrong answer..
 
  • #3
1,758
57
Check your equation for D again.
 
  • #4
It's right ?

D = (V1 + v2/2) x T
 
  • #5
123
1
1. Is it 90 or 80 km/h?
2. your formula should be [tex]\Delta d=\frac{v_{1}+v_{2}}{2}\Delta t[/tex]
 
  • #6
It says 80km on my sheet.
But yeah, that is the formula I am using.
(I do v1+v2 first, and then divide it by 2)
 
  • #7
123
1
I think I see your mistake. How did you convert [tex]m/s^{2}[/tex] to [tex]km/h^{2}[/tex]
 
  • #8
Oh sorry for writing 90 in the first post, that was a typo. Should be 80km.

Okay so it's 80km/h.
I converted that -5.0m/s into km/^s by doing this:

-5/1000
= - 0.005

Then I multiplied this by 60..

But now I'm getting -0.3. x_x
Ahh, can anyone thoroughly help me? This is so frustrating! I am so lost. =(
 
  • #9
123
1
Ok, I think the only mistake you made was this conversion. It's acceleration, therefore has a unit of either [tex]m/s^{2}[/tex] or [tex]km/h^{2}[/tex], notice the square in the bottom. I tried converting it and got [tex]1m/s^{2}=12960km/h^{2}[/tex]
 
  • #10
How did you get that?
 
  • #11
Now I just confused 2 questions together. Forget this whole thing, it's a flop. =(
I'll ask my teacher.
 
  • #12
123
1
Ok, this is what I did.
[tex]1km/h^{2}=1000m/(3600s)^{2}[/tex]
Try and see what you get.
 
  • #13
Oh, so I have to square the seconds in the speed part?
 
  • #14
123
1
of course!
 
  • #15
Now I got -0.1296..
Kinda on the right track.

I did -0.006/(0.36^2)
 
  • #16
123
1
nope, still wrong. 1/3600 is not equal to 0.36.
 
  • #17
Is this considered partially right?
0.006km/3600 seconds?
 
  • #18
123
1
You really need to work on unit conversion. Let me show you:
[tex]1km/h^{2}=1000m/(3600s)^{2}[/tex] 1km is replaced by 1000m, and 1h is replace by 3600s.
[tex]1km/h^{2}=1000m/(3600^{2}s^{2})[/tex] notice that both the 3600 and s are squared.

[tex]1km/h^{2}=\frac{1000}{3600^{2}}\frac{m}{s^{2}}[/tex] separate the unit from the numbers.

[tex]1km/h^{2}=\frac{1}{12960}\frac{m}{s^{2}}[/tex]

[tex]12960km/h^{2}=1\frac{m}{s^{2}}[/tex]

Right?
 
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