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Instantaneous velocity, help me please!

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to find the instantaneous velocity directly before a car collides with another at an intersection. It travels 24 metres with an acceleration of -5.39 m s2. The only time factor given is the light sequence:
    Green for 45 seconds, amber for 3 seconds and red for 5 seconds before the opposing light turns green.
    I don't have the velocity or time, so I don't know how I would draw a graph to figure this out.

    2. Relevant equations
    I know the equation for Instantaneous velocity is Δy/Δx, and that I would have to draw a triangle just touching the point at which I need to find the speed of the vehicle, and then use Pythagoras. But I'm looking for speed, would instantaneous velocity still work? I have also considered the equation vi = vf + at, but again I don't have the time or velocity (neither the initial of final).

    3. The attempt at a solution
    If I used a displacement vs time graph, would it be the same as a velocity vs time graph? Would it give me the same (and correct) answer? Would the slope be negative, and how do I make my graph accurate enough so that my answer is?

    Please, please lend me a hand, I'm so stumped.
    Merci d'avance!
     
  2. jcsd
  3. May 30, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Veloco-raptor! Welcome to PF! :smile:
    I really don't understand the question … there seems a lot missing, including anything about the second car. :confused:


    Speed is just the magnitude of the velocity vector.

    ("v" is usually an abbreviation for "speed" :wink:)
    nooo … the slope of displacement vs time is speed, the slope of speed vs time is acceleration
     
  4. May 30, 2012 #3
    Hi,
    I agree with tiny-tim. There seems to be some missing information here that is preventing us from making a correct calculation.

    Because the acceleration is constant, we know that the velocity will be increasing linearly. However, this means that average velocity ≠ instantaneous velocity.

    So we cannot do delta x/delta t.


    Another note about the graphs.
    1. The slope of x vs. t is velocity. The slope of v vs. t is acceleration.
    2. Similarly, the area UNDER the curve of the a vs. t graph is velocity.
    And the area UNDER the curve of the v vs. t graph is displacement.

    If you decide to continue in physics (I would recommend it), you will learn that these relationships are intuitively connected by the derivative (as in case one) and consequently the integral (as in case two). So as you can see, physics is fun! (no one said physics would be easy, though).
     
  5. Jun 4, 2012 #4

    Simon Bridge

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    So how did you get on?
    You know the acceleration and the distance ...
    You need to know either the initial or final speeds, or the time involved, to draw a v-t graph. The area under the graph is the displacement and the slope of the graph is the acceleration... giving you two equations. The method is also used to complete your more recent homework.

    You should also realize there is a limit to the kind of assistance we can give you when it's homework.
     
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