1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Right.
    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted