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Homework Help: Velocity, displacement and position time graph help

  1. Jul 16, 2012 #1
    Velocity, displacement and position time graph help!!

    cyclist road east for 1.0x10^2 m with a constant velocity of 25 m/s she then accelerates down a hill and 5s later reached the bottom of the hill with a velocity of 50m/s

    a) how long did the cyclist ride before reaching the hill?
    B)what is her average acceleration down the hill?
    C)what was the distance she traveled down the hill?
    D)if she continued to ride with a constant velocity for 5s how far beyond the hill would she travel
    E) sketch a position, displacement, and velocity time graph (scaled)

    So here in go

    A) 100/25=4s
    B) (50-25)/5 = 5m/s^2
    C)0.5(50+25)(5)= 187.5 m

    I have no idea how to draw anything. All I knew is time is on the horizantle axis and (m) is on the verticle axis !! Please someone help me
  2. jcsd
  3. Jul 17, 2012 #2
    Re: Velocity, displacement and position time graph help!!

    Hello alicia113, you are free to chose axis when you drawing something. But you're right the most convenient is draw a time axis horizontally.

    Since you know all the variables you can simply write an equations and sketch them, I'l put you on the right track. Lets start velocity, for the first 5 seconds the cyclist was riding with constant velocity of v=5m/s. During the next 5 seconds she accelerated, and we can write such equation, since we know the acceleration - 5 m/s^2, So between 5 and 10 seconds v = 25 + 5t, and after that she was driving with constant v = 50 m/s velocity.

    between 0-5 seconds draw v = 25 m/s
    between 5-10 seconds draw v = 25+ 5t and
    between 5-beyond draw v = 50 m/s
  4. Jul 17, 2012 #3


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    Re: Velocity, displacement and position time graph help!!

    Sorry for the edits I miss read the problem earlier.

    I believe that should be...

    between 0-4 seconds draw v = 25 m/s, position increasing from 0 to 100m

    between 4-9 seconds draw v = 25+ 5t (from v=u+at). Position will increase from 100m according to S=ut=0.5at2
    Last edited: Jul 17, 2012
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