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Velocity and Acceleration questions

  1. Sep 14, 2015 #1
    • Missing template. Originally posted in different forum.
    x(t)= -4m-(9m/s)t+(2m/s)t^2

    I'm lost on velocity and acceleration. Below are questions on a worksheet and i answered positive and negative. Can someone correct me on them please? Thank you!

    Is the initial acceleration in the positive x or negative x direction?
    positive?
    Is the initial velocity in the positive x or negative x direction?
    negative?
    Is the initial position in the positive x or negative x direction from the origin?
    negative?
    After 3 seconds is the acceleration in the positive x or negative x direction?
    positive?
    After 3 seconds, is the velocity in the positive x or negative x direction?
    negative?
    After 3 seconds is the position in the positive x or negative x direction from the origin?
    negative?
     
  2. jcsd
  3. Sep 14, 2015 #2

    jbriggs444

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    If t is in seconds then that formula is dimensionally incorrect. What would the resulting units be on the acceleration term?
     
  4. Sep 14, 2015 #3
    x(t)= -4m-(9m/s)t+(2m/s^2)t^2

    I forgot to square the acceleration...does that help?
     
  5. Sep 14, 2015 #4

    jbriggs444

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    Yes, that addresses that problem.

    But... this should have been posted in the homework section using the supplied template. And you need to show some work. You have a function for position as a function of time. Start with the third question. Can you answer that?

     
  6. Sep 14, 2015 #5
    I want to say negative because of the -4m...but I have a feeling that is wrong
     
  7. Sep 14, 2015 #6

    jbriggs444

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    That would be correct.

    If you evaluate the expression ##-4{meters} -9\frac{meters}{sec}t + 2\frac{meters}{{sec}^2}t^2## for ##t## = 0 sec then the result would be -4 meters.

    Velocity is the rate of change of position with respect to time. If t is near zero and is increasing, what can we say about x? Is x increasing or decreasing? Is there a term in the equation that can tell you that easily?

    If you have to, try evaluating the expression for t = 0.01 and see what the result is.
     
  8. Sep 14, 2015 #7
    so when answering all the "initial" questions, I just have to look at the sign in the equation.


    When it asks for "After 3 seconds", I would still have to look at the signs for acceleration and velocity but not position, correct? For position, I could just plug in 3 seconds into t and solve.
     
  9. Sep 14, 2015 #8

    jbriggs444

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    By this, I assume that you mean "look at the sign of the corresponding term". That works if t is zero.
    Yes, for position, you can just plug in 3 seconds and evaluate the formula.

    But for velocity... You can read off ##-9 \frac{meters}{sec}## for the velocity at t=0. But by t=3, acceleration has occurred. The new velocity is not the same as the initial velocity.
     
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