1. Limited time only! Sign up for a free 30min personal 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!

Velocity, Acceleration and Displacement time graphs.

  1. Oct 23, 2016 #1
    • Poster has been reminded that showing an attempt is not optional at the PF
    1. The problem statement, all variables and given/known data

    Sketch velocity vs time, acceleration vs time and displacement vs time graphs describing the following motions.


    a) An object begins at rest and accelerates at a constant rate to a velocity of 5.0 m/s [E]. The object continues along at this velocity.

    b) A car accelerates from 0 to 60 km/hr at a constant rate and then hits a wall slowing to a stop almost immediately.


    2. Relevant equations

    Suvat equations

    3. The attempt at a solution


    I am not sure how to solve them at all. How do I figure out the shape of the graph mathematically? I am not sure on how to start.
     
  2. jcsd
  3. Oct 23, 2016 #2

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi unknown physicist and welcome to pF.

    Start with the SUVAT equations that you quoted. Instead of writing the acronym, write down the actual equations and take a good look at them to see how they relate to what is being asked.
     
  4. Oct 23, 2016 #3

    cnh1995

    User Avatar
    Homework Helper

    Welcome to PF!
    You are supposed to make an attempt at a solution when posting in the HW forum.
    Which equation would you use for this?
     
  5. Oct 23, 2016 #4
    v=u+at ?
     
  6. Oct 23, 2016 #5

    cnh1995

    User Avatar
    Homework Helper

    Right. Modify this equation by putting the value of u. What shape will the modified equation give?
     
  7. Oct 23, 2016 #6
    v=u+at
    u=0
    v=at (mx), slope with a 0 y-intercept.

    What about the other part?
     
  8. Oct 23, 2016 #7

    cnh1995

    User Avatar
    Homework Helper

    Yes. Until??
    Again, start with the equation describing the scenario.
     
  9. Oct 23, 2016 #8
    Got it, Thanks. But, what about for the acceleration and displacement time graphs? How do I know what numbers to plug in?
     
  10. Oct 23, 2016 #9

    cnh1995

    User Avatar
    Homework Helper

    The shape of acceleration vs time graph is mentioned in the problem itself. Read again carefully.
    For displacement vs time graph, make use of the known graphs. What is the mathematical relation between displacement and velocity? Which SUVAT equation will you use?
     
  11. Oct 23, 2016 #10
    s=1/2(u+v)t ?
     
  12. Oct 23, 2016 #11
    Mentioned in the problem? I can't seem to understand that.
     
  13. Oct 23, 2016 #12

    cnh1995

    User Avatar
    Homework Helper

    No.
    The equation should be s as a function of t. All the other terms should be known constants.
     
  14. Oct 23, 2016 #13

    cnh1995

    User Avatar
    Homework Helper

     
  15. Oct 23, 2016 #14
    so, I should use s=ut+1/2at^2?

    Also, it accelerates at a constant rate, but why does the graph need to go downward? why doesn't it go up like a slope? (for acceleration)?
     
  16. Oct 23, 2016 #15

    cnh1995

    User Avatar
    Homework Helper

    Yes. It is in the form y=ax+bx2. What is this shape? Put the value of u and see.
    It doesn't. Acceleration is constant w.r.t.time. How would you show this graphically?
     
  17. Oct 23, 2016 #16
    It is a quadratic, so it should be curved.

    w.r.t time?

    But acceleration will be a straight line, then go straight down, to an acceleration of 0, then it will continue to be 0. That is what the answer sheet shows. I can't understand why it is straight, and why it goes down.
     
  18. Oct 23, 2016 #17

    cnh1995

    User Avatar
    Homework Helper

    Yes, after v becomes 5m/s. I was talking about the acceleration period.
    Yes. Which curve is y=bx2?
     
  19. Oct 23, 2016 #18
    yes, I can't seem to understand the acceleration part. Why is it a straight line? then why does it go down?
     
  20. Oct 23, 2016 #19
    the curve will be the s=1/2ut^2
    because (u)(t) = 0
     
  21. Oct 23, 2016 #20

    cnh1995

    User Avatar
    Homework Helper

    It is a straight line (parallel to time axis) because it is "constant" w.r.t. time. Hence, the straight line will have zero slope.
    Yes. What is the name of that curve? Look up "conic sections".
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted