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Average speed over time interval

  1. Sep 14, 2015 #1
    Edit Background: The teacher doesn't teach we just do activities and have whole chapters worth of 20 mini assignments due every few days. Yes I did read the book found nothing.

    1. The problem statement, all variables and given/known data

    The average velocity of a car over a certain time interval is 35 mi/h. If the velocity of the car was 65 mi/h at the end of this interval, what was its initial velocity? Assume that acceleration was constant. 

    2. Relevant equations
    avg velocity=Vo+at

    v start+v end/2=constant accel

    3. The attempt at a solution
    Possibly either 0 or 35 is this a trick question?
     
    Last edited: Sep 14, 2015
  2. jcsd
  3. Sep 14, 2015 #2
    First thing first, v = u + at does not give avg velocity, it gives final velocity . Average velocity is given by: change in displacement / change in time . In case of constant acceleration it's (initial velocity + final velocity)/2

    The data given seems insufficient to get a numerical answer (assuming I'm not missing something) as the time interval is not given...

    Edit : As pointed out by others , it's possible to get a numerical answer as the OP has got. I forgot it's constant acceleration.
     
    Last edited: Sep 15, 2015
  4. Sep 14, 2015 #3

    NascentOxygen

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    You have never seen the letters "avg" in front of velocity like this, because that equation does not give average velocity. It gives final velocity of a body that has been experiencing constant acceleration.

    You've got this saying something like the average of two velocities equals acceleration? You have never seen that equation. One side has units of miles/hr and the other side of miles/hr/hr, and this is a reliable way of recognizing that it can't possibly be correct.

    Physics doesn't allow us to invent equations of convenience. You must use the correct equations of motion you have studied.
     
  5. Sep 14, 2015 #4
    Okay I wrote the equation wrong. I mean the fill out sheet says to say "any relevant equations" (which I don't know because the teacher doesn't teach.) That is word for word the exact question. I said the same thing "how do I solve this?" and "why is this question not already on the internet???"

    I'm not making up equations (I leave that for Calc class tho). I forgot to say thanks for the quick replies.
     
  6. Sep 15, 2015 #5

    haruspex

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    The correct versions of your equations (for constant acceleration only!) are:
    Final velocity = initial velocity + (acceleration)*(elapsed time)
    Average velocity = ((initial velocity)+(final velocity))/2
    For more such equations, and the standard notation, Google SUVAT.
     
  7. Sep 15, 2015 #6

    NascentOxygen

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    The only place you may trip up here involves average velocity. What is average velocity? It is the fixed velocity a body would need in order to travel the specified displacement in the specified time.

    For any journey, average velocity = displacement ÷ time

    (Displacement should not be confused with "distance travelled", e.g., a runner who ends up back where he started covered the course with an average velocity of zero, even though he may have demonstrated record-breaking speed.)
     
  8. Sep 15, 2015 #7
    well 35mph+95mph=130mph/2=65mph so is that correct initial speed is 95mph?
     
  9. Sep 15, 2015 #8

    haruspex

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    You seem to have substituted the numbers into the wrong terms of the equation. What is the final velocity? Where does that go in the equation?
     
  10. Sep 15, 2015 #9
    OH 5mph+65mph-(end of interval which means final velocity)=70mph/2=35mph avg velocity so initial is 5mph.
     
  11. Sep 15, 2015 #10

    haruspex

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    Yes.
     
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