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Constant Acceleration Car PRoblem

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    You're driving at 70 km/h when you accelerate with constant acceleration to pass another car. Six seconds later, you're doing 80 km/h.

    How far did you go in this time?
    Express your answer using two significant figures.

    2. Relevant equations
    V= V(0)+at
    V^2- V(0)^2 = 2a ( X-X(0) )

    3. The attempt at a solution

    I keep finding some large numbers like 125000 m. I think i'm doing something wrong with converting times. So can someone tell me how to do this problem?
  2. jcsd
  3. Oct 15, 2007 #2


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  4. Oct 15, 2007 #3
    you kinda confused me.

    That's what i did. I found the acceleration from the first equation. V=V(0)+at. I mean we know V=80km/h, V(0)= 70km/h, and t=6. Only thing left is a, right?. If i plug the acceleration i found with this in second equation. The answer isn't correct.
  5. Oct 15, 2007 #4


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    make sure one does not mix hours and seconds in the acceleration.

    The v is given in km/h and the time in seconds.

    One could certainly write the acceleration in km/h/s, but then when using v2 one would have units of (km/h)2.

    I recommend converting km/h to m/s first.

    It is better to show the work and make sure units are consistent.
  6. Oct 15, 2007 #5
    Ok, here is a weird thing.

    The site you gave me. I found the average velocity. Then that times with 6s gave me the displacement. which is the answer.

    But the thing is i can't find the answer with those equations. It's basically the same thing. Still can't find it.
  7. Oct 15, 2007 #6


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    What you did is right, but you should watch the units. I repeated your calculation but put in the units...
    For example (6 seconds = 60/10 seconds = 1/10 hour):
    [tex]a = \frac{V - V_0}{t} = \frac{10\text{ km/h}}{1/10\text{ h}} = 100 \text{ km/h}^2.[/tex]

    Then from the second equation,
    [tex]d \equiv (X - X_0) = \frac{ V^2 - V_0^2 }{ 2 a } = \frac{ \left( 80^2 - 70^2 \right) (\text{ km/h})^2 }{ 2 \times 100 \text{ km/h}^2 } = \frac{1500}{200} \frac{\text{km}^2 /\text{h}^2 }{ \text{km} / \text{h}^2 } = \frac{15}{2} \text{ km}, [/tex]
    which IMO is still a lot (are the formulas right, did I make any calculation errors?) but better than 125 km :smile:

    See how it works? Alternatively, you could have converted everything to m and s.
  8. Oct 15, 2007 #7
    Yeah, your answer is correct too.

    Only thing i did wrong was converting units. Everytime i did this problem, I'd have numerator and denominator in different units for the first equation.

    Thanks both.
  9. Oct 15, 2007 #8


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    6 sec = 0.1 minute = 0.1/60 h or 6/3600 h = 0.00167 h ( 1 sec = 1/60 min = 1/3600 hr)

    The delta-V = 80 km/h - 70 km/h = 10 km/h = 10000m/3600s = 2.78 m/s

    dV/dt = 2.78 m/s / 6 s = 0.463 m/s2


    10 km/h/ (0.00167 h) = 5988 km/h2

    So one has to be careful with units.
  10. Oct 15, 2007 #9


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    Hehe, of course!
    I thought there was a mistake somewhere.
    Seven kilometers in 6 seconds seemed like a bit much :smile:
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