Integrating a velocity over time graph

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Hint: You have to determine the constants of integration.

For example for the first section $x_{\text{Section 1}}(t=0)=0 \implies C = 0 \implies x_{\text{Section 1}}(t) = 45t^2$.

For the second section, we know $x_{\text{Section 2}}(t=0.5)=45\cdot 0.5 +C =x_{\text{Section 1}}(t=0.5)=45\cdot 0.5^2$.

And so forth.


Another more visual way to do this is to first determine the cumulative area (which is equivalent to the total distance) for the endpoint of each section (constant velocity and accelerated/deaccelerated interval). As you already identified there are 7 sections. Then calculate the total area, which is equivalent to the distance.

Section 1: $A_1 = 0.5 \cdot (0.5 \text{min}) \cdot 45 \text{km}/{h}=0.5 \cdot (0.5\cdot 60 s)\cdot 45\frac{1000m}{3600s}=3.725 \,\text{m}$.

Section 2: $A_2 = A_1+1.5 \cdot 60 s\cdot 45\frac{1000m}{3600s}.$

Section 3: $A_3 = A_2+ 0.5 \cdot (0.5\cdot 60 s)\cdot 45\frac{1000m}{3600s}.$

And so forth.

Now, draw a coordinate system with all these points $(t,A_i)$. The last thing is to connect these points by a curve (quadratic, if the velocity is linearly increasing/decreasing; linear, if the velocity is constant for this section)

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Bunny-chan
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Bunny-chan

Just a passionate about mathematics!

Updated on March 10, 2020

Comments

  • Bunny-chan
    Bunny-chan over 3 years

    Velocity over time graph

    In order to elaborate the position over time graph that corresponds to the graph above, I first obtained all of the equations of the straight lines, which are as follows:$$v(t)= \begin{cases} 90t, \textrm{ if } 0 \leq t \leq 0.5 \\ 45, \textrm{ if } 0.5 \leq t \leq 2 \\ -90t+225, \textrm{ if } 2 \leq t \leq 2.5 \\ 0, \textrm{ if } 2.5 \leq t \leq 3 \\ 150t-450, \textrm{ if } 3 \leq t \leq 3.5 \\ 75, \textrm{ if } 3.5 \leq t \leq 4.5 \\ -150t+750, \textrm{ if } 4.5 \leq t \leq 5 \end{cases}$$I know the position over time graph represents the area below the curves in the velocity over time graph, so by integrating:$$x(t)= \begin{cases} 45t^2 + C, \textrm{ if } 0 \leq t \leq 0.5 \\ 45t + C, \textrm{ if } 0.5 \leq t \leq 2 \\ -45t^2+225t + C, \textrm{ if } 2 \leq t \leq 2.5 \\ 0, \textrm{ if } 2.5 \leq t \leq 3 \\ 75t^2-450t + C, \textrm{ if } 3 \leq t \leq 3.5 \\ 75t + C, \textrm{ if } 3.5 \leq t \leq 4.5 \\ -75t^2 +750t + C, \textrm{ if } 4.5 \leq t \leq 5 \end{cases}$$And now this is where things start to get complicated in my head x__x. I searched through the web for the corresponding position over time graph and I found (from an unknown source, so it could be incorrect) the following:

    Position over time graph

    Which didn't match mine, so perhaps there must be something I am missing. Well, one of the main questions I have is what do I take as the value for C in each of the functions in order to properly sketch the graph?

    • amd
      amd over 6 years
      The constants are determined by the requirement that all of the segments match up.
    • Vedvart1
      Vedvart1 over 6 years
      Any chance we could see what you have graphed?
    • amd
      amd over 6 years
      Also, the constant for the first segment is determined by the initial velocity. As well, you’ve integrated a couple of the segments incorrectly.
    • Bunny-chan
      Bunny-chan over 6 years
      I edited one which was incorrect. Is there any other? D: Also, if the constant of the first segment is determined by initial velocity, then the following segments' are determined by their final velocity?
  • Bunny-chan
    Bunny-chan over 6 years
    I solved it. Thank you very much!