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An acceleration problem

  1. Aug 22, 2005 #1
    This problem request the instanteous velocity at a given point which I have not yet discovered how to do, the problem is as follows...

    The engine of a model rocket accelerates the rocket vertically upward for 2.0 s as follows: at t = 0, the rocket's speed is zero; at t = 1.0 s, its speed is 4.0 m/s; and at t = 2.0 s, its speed is 20 m/s. Plot a velocity vs. time graph for this motion.

    I have found that the average acceleration during the 2.0 s interval is 10 m/s2

    I am trying to find the instanteous velocity at 1.5. I know that i must find the slope of the line tangent to the point 1.5 on the graph but am having problms doing so. Any tips?
  2. jcsd
  3. Aug 22, 2005 #2


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    try representing the accln by a polynomial, in time.

    Get this polynomial to fit the known data - i.e velocity at particular time invervals.
  4. Aug 22, 2005 #3
    Since the velocity rate of change varies at the two intervals, and we are dealing with constant acceleration, then I assume you mean try to find two different lines for the two intervals and not one polynomial.

    You need to find the slope of the velocity time surve between 0 and 1 second, the do the same for the interval of 1 to 2 seconds. Once you have this, read the graph and enjoy.


    Last edited: Aug 22, 2005
  5. Aug 22, 2005 #4


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    How do we know that the accln is constant ?
  6. Aug 23, 2005 #5
    Im assuming it is, but you might be right, since it is a rocket, acceleration might vary, but I doubt it since no mass is lost. Its not a momentum problem where they give you the mass flow rate out, I think its a simple kinematics problem, thats why I assumed constant acceleration.


  7. Aug 23, 2005 #6
    what about the instantaneous aceleration at 1.5?
  8. Aug 23, 2005 #7
    What about it, he needs to find instantaneous velocity at 1.5 seconds, not acceleration.


  9. Aug 23, 2005 #8


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    If it's constant acceleration, there are a couple of ways.

    You plotted the velocity-time graph - no ?
    If so, then just read off the graph at t = 1.5s. You should get v = 12 m/s

    Alternatively, you mentioned getting the slope of the graph between 1 and 2 secs.
    let m be the slope. Then m = (v2 - v1)/(t2 - t1), where v1 is the velocity at t1 and v2 is the velocity at t2, and t1 = 1 sec and t2 = 2 sec.
    This should give you m - the acceleration during the 2nd second of movement. Now simply use v = u + at over this movement.

    If the accelerations aren't constant, and it is a rocket after all, then you can approximate its motion over the 2-second interval by a polynomial.

    Let the velocity, v = at² + bt.
    we have two data points, (1,4) and (2,20), where (x,y) == (time,velocity)
    substituting for these data points in the eqn for velocity,

    4 = a + b
    20 = 4a + 2b


    a = 6, b = -2

    .: v = 6t² - 2t
    v = 2t(3t - 1)

    at t = 1.5
    v = 3(4.5 - 1)
    v = 10.5 m/s
    Last edited: Aug 24, 2005
  10. Aug 24, 2005 #9
    Im assuming you meant to write "constant acceleration". The rest looks correct.


  11. Aug 24, 2005 #10


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    Thanks for the note.

    Edited my post
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