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Simple Kinematics Question

  1. Mar 5, 2014 #1
    Hi guys, having some trouble with this question as I cannot see where I am going wrong. Very simple question but I must be misunderstanding something.
    1. The problem statement, all variables and given/known data
    The following table shows how the velocity of a car varied with time:

    Velocity (km/h) 0 19 35 48 48 27 0
    Time (s) 0 3 6 9 12 15 18

    Assuming these speed values to be joined by straight lines, plot a graph of speed against time and determine the total distance travelled.

    2. Relevant equations
    s = dxt
    Area under graph = distance travelled

    3. The attempt at a solution
    I assumed this question would be as simple as finding the area under the graph to get the total distance travelled. So I went about drawing my graph along with converting all the values for speeds into (m/s). I have attached a picture of what my graph looks like in case anyone would like to see it (its a bit messy). I'm sure there is a very simple explanation to where I am going wrong but I'm just not seeing it myself.

    Thanks in advanced guys!
     

    Attached Files:

  2. jcsd
  3. Mar 5, 2014 #2

    BvU

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    So what did you get and why do you think it is wrong ?
     
  4. Mar 5, 2014 #3
    Sorry forgot to put that info in my original post. I keep getting 139.97m by using the area under the graph rule. I do (13.33x9)/2 + (13.33x3) + (13.33x6)/2 = 139.97. Yet my questions sheet gives me an answer of 147m.
     
  5. Mar 5, 2014 #4

    BvU

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    Problem I have with your plot is not that it's messy (it really isn't). I have the same numbers but my plot looks quite diferent!

    Problem I have is that you use different step sizes on the time axis! the 3 sec from 9-12 looks a lot longer than from 12-15.

    This also suggests a linear constant acceleration from 0-9, which is not correct (otherwise the speeds would look x, 2x, 3x for t = 0,6,9 and they don't).

    What do you do to determine the area under the graph ? 9 x 13.3 /2 + 3 x 13.3 + 6 x 13.3 /2 ?

    [edit] well this guess of mine crossed your addendum.

    Draw it with equidistant time intervals and you'll be OK.
     
  6. Mar 5, 2014 #5
    My plot was just a quick sketch as I was unsure how the graph would look, would you recommend drawing it out properly on graph paper?

    I used that to get that area under graph based on splitting the area into two triangles and a rectangle. I then used 1/2 x base x height to get the area of the two triangles and added these onto the area of the rectangle.
     
  7. Mar 5, 2014 #6

    BvU

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    Either that, or you go to ##s=s_0 + v_0 \ t + {1\over 2} a \ t^2##, where ## a_{\ t-3\ \rightarrow\ t} = {v(t) - v(t-3) \over t - (t-3)} ## in steps of 3 sec

    But: did you pick up that 0-9 sec cannot be lumped together because the accelerations in these three sections are not the same ?
    Idem 12-18 sec.
     
  8. Mar 5, 2014 #7
    I think that is where I have gone wrong yes. I have lumped together 0-9 seconds, thank you for pointing that out! I am currently working through it again but this time I have drawn the graph out properly and I am splitting the area beneath the graph into triangles, trapeziums and a rectangle, then adding all these areas together, rather than lumping parts together. Fingers crossed it works this time!
     
  9. Mar 5, 2014 #8

    BvU

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    Let us know if you don't get 147.5 m !

    By the way, do you realize you are doing exactly the same thing as this s = s0 + v0 t + 1/2 a t^2 , when you do it by adding triangles and rectangles ?
     
  10. Mar 5, 2014 #9
    I did get it this time! But if that equation is a faster way to get the answer then I'll try that also. What is that equation? Is it a SUVAT equation because I don't think I've seen that one. what does s0 and v0 represent?
     
  11. Mar 5, 2014 #10

    BvU

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    s0 and v0 are distance and speed at start.
    So first interval: 0 and 0. acceleration a is (5.28 - 0 )/3 = 1.76 m/s2. s = 0 + 0 x 3 + 1/2 x 1.76 x 32 = 7.92 m .
    Then second interval: 7.92 m and 5.28 m/s , a = (9.72-5.28)/3 = 1.48 m/s2.
    s = 7.92 m + 5.28 * 3 + 1/2 * 1.48 * 32 = 30.42 m
    the 5.28 * 3 is a rectangle area and the 1/2 * 1.48 * 32 is a triangle area.
    etc etc
     
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