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Integrating velocity

  1. Jan 19, 2014 #1

    Qube

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

    A drag racing car starts from rest at t = 0 and moves along a straight line with velocity given
    by v = bt^2, where b is a constant. The expression for the distance traveled by this car from its
    position at t = 0 is:

    A. bt3
    B. bt^3/3

    2. Relevant equations

    Velocity is change in position divided by change in time.

    3. The attempt at a solution

    Three questions:

    1) Is this a bad question? I don't see how we can get distance traveled from what's given in this problem unless we make the assumption that distance = displacement in this problem.

    2) If velocity is displacement / time, then why isn't the answer according to the problem v * time or bt^2 * t = bt^3?

    3) Why is the integral of the velocity function in the question bt^3/3 - or different from simply velocity * time? Isn't the integral the area under the curve on a velocity time graph, or simply the y-axis * the x-axis (velocity * time)?
     
  2. jcsd
  3. Jan 19, 2014 #2
    You are correct; the answer is ##\frac{bt^3}{3}##. Try integrating ##bt^2##, treating ##b## as the constant, by calculus. Which rule do you need to use to integrate such an expression? This is one of the elementary derivative rules.
     
  4. Jan 19, 2014 #3

    CAF123

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    It does say the motion is in a straight line. The quantity bt2 is ≥0 or ≤0 throughout the motion depending on the sign of b.
    This is only true if the body is moving at constant velocity. Is this the case here?
     
  5. Jan 19, 2014 #4

    haruspex

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    That's average velocity.
     
  6. Jan 19, 2014 #5

    Qube

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    You are correct; I cannot use the usual kinematic equations here since they all assume that acceleration is constant, and taking the derivative of the velocity function (given) results in an expression with a variable.
     
  7. Jan 19, 2014 #6
    If dx/dt = bt2, have you learned how to integrate this equation?
     
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