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Time taken to travel x meters with constant acceleration

  1. Jul 28, 2010 #1
    Hi, I'm trying to solve something that I'm very sure is simple, I just can't seem to find the right equation.

    A rocket is travelling straight up, with an acceleration of 20m/s^2. I need to know how long it takes to reach a height of 1379.8m (this is the second part of a question).

    I am confused about the acceleration. Does the fact that it's travelling at 20m/s mean that after 2 seconds it's travelling at 400 m/s and at 3 seconds it's travelling at 800 m/s?

    And to work out the time it takes to travel that far the equation would be t = 1379.8 / ?

    Thanks
     
    Last edited: Jul 28, 2010
  2. jcsd
  3. Jul 28, 2010 #2
    It's not "travelling at 20 m/s". The acceleration of 20 m/s^2 means that's the velocity increases by an additional 20 m/s every second. But you need another "given" number to solve this problem, so was it also given that it began at rest, so that the initial velocity can be set to zero? If that's number is also given, you have enough information. Then you need the topic called "kinematics", which provides four formulas. One of those formulas mentions these variables: displacement, initial velocity, acceleration and time.
     
  4. Jul 28, 2010 #3
    [tex]a=\frac{dv}{dt}[/tex]
    [tex]dv={a}*{dt}[/tex]
    [tex]v=at+v_0[/tex]
    [tex]v=\frac{dx}{dt}[/tex]
    [tex]dx={v}*{dt}[/tex]
    [tex]x=x_0+{v_o}{t}+\frac{1}{2}{a}{t}^2[/tex]
     
  5. Jul 28, 2010 #4
    Yep, the starting velocity was 0. I'll look into kinematics and see if I can make sense of those formulas.
     
  6. Jul 28, 2010 #5
    OK, so I've had a go. Can someone please check my answer and see if it's correct? (Sorry for the crappy formatting, I'm not sure how to do sub\superscripts)

    initial velocity = 0;
    acceleration = 20m/s^2;
    displacement = 1379.8;

    d = vi * t + 0.5 * a * t^2
    1379.8 = 0m/s * t + 0.5 * 20m/s^2 * t^2
    1379.8 = (0m) * t + 10m/s^2 * t^2
    1379.8 / -10 = t^2
    SQRT(-137.98) = t //???
    t = 11.746 seconds

    I'm not really sure how I ended up with a negative in my square root, but using the positive I get a number that seems reasonable.
     
  7. Jul 28, 2010 #6

    Doc Al

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    Staff: Mentor

    Your answer is fine.
    When you divided both sides by 10, for some reason you added a negative sign. (Perhaps you were mixing that up with subtracting 10 from both sides?)
     
  8. Jul 28, 2010 #7
    Ah, yep I did. Thanks for everyones help.
     
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