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Acceleration & max displacement from velocity equation

  1. Mar 16, 2017 #1
    1. The problem statement, all variables and given/known data

    The velocity of an object moving along the x-axis is given by Vx = (2.0t-t2)m/s. Initially (at t=0), the object was at the origin.

    a) Determine the object's acceleration at t=3.0s.
    b) Find the object's position at t=3.0s
    c) Calculate the object's maximum positive displacement from the origin.

    2. Relevant equations

    a = dv/dt
    xf = xi + vxit + (1/2)axt2
    vxf2 = vxi2 + 2ax

    3. The attempt at a solution

    a) To get an equation for acceleration,

    ax = dv/dt = d/dt (2.0-2t)
    = 2.0 - 2(3.0s)
    = -4.0 m/s2

    b) xf = xi + vxit + 1/2axt2
    = (0 m) + (0 m/s)(3.0 s) + 1/2 (-4.0 m/s2)(3.0s)2
    = -18.0 m

    c) delta x = [vxf2 - vxi2] / [2ax]
    = [2.0 (3.0s) - (3.0s)2]2 - [0 m/s] / [2 (-4.0 m/s2)]
    = -1.1 m

    First of all, I'm not sure if I've done the derivation for the accleration equation correctly. Also, question c) asks for maximum POSITIVE displacement but I keep getting a negative value. I'm not sure where I'm going wrong!
  2. jcsd
  3. Mar 16, 2017 #2


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    For (b) and (c) you are using "constant acceleration" equations. Is the acceleration of this object constant?
  4. Mar 16, 2017 #3
    The question doesn't say either way... I get the feeling I'm missing something but I don't know what.
  5. Mar 16, 2017 #4


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    Use the given expression for the velocity as a function of time to decide if the acceleration is constant.
  6. Mar 16, 2017 #5
    Ok, so if I plug in different values for t in acceleration equation, the values are not constant... The only solution I can think of is to calculate x separately for each second? Also, am I correct in assuming that my answer for b) is still accurate since I calculated acceleration for the same time (t=3.0s)?

    Thank you for all your help! (I'm a healthcare student who's forgotten all her high school physics...)
  7. Mar 16, 2017 #6


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    Can you derive the position vs time function x(t) from the velocity vs time v(t)? What calculus operation should you use to do this?
  8. Mar 16, 2017 #7
    Alright, so I integrated v(t) to get x(t) = -1/3t3 + t2 + C.
    Input values of t=1,2,3,4 and found that at t=2.0s, the positive displacement is 1.33m, at t=3, it is 0m and at t=4, it is -5.3m.
    So, the maximum positive displacement is 1.33m at t=2.0s.

    Have I finally got it right?
  9. Mar 16, 2017 #8


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    OK. But what is the value of C?
    Yes, at t = 3.0 s, the object is at x = 0.

    For part (c), how do you know that the maximum positive value of x will occur for an integer value of t?
  10. Mar 16, 2017 #9
    When I solve for C:

    From b), I have x(t) for t=3.0s is -18.0m

    x(3) = -18.0 = (3)2 - (1/3)(3)3 + C
    -18.0 = C

    But that doesn't seem right.
  11. Mar 16, 2017 #10


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    In your first post, you didn't get the correct answer for (b) because you assumed that the acceleration is constant.

    You can determine the value of C by using the information given in the problem statement about the location of the object at t = 0.
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