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calculusxy

  • one year ago

MEDAL!!! Start at the 0.5 meter mark and walk towards the 4 meter mark at 0.25 meters per second.

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  1. calculusxy
    • one year ago
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    @Michele_Laino

  2. calculusxy
    • one year ago
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    I will post the attachment.

  3. calculusxy
    • one year ago
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  4. Michele_Laino
    • one year ago
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    I think that the function s=s(t), namely the space traveled by our particle, at time t, is given by the subsequent formula: \[\Large s\left( t \right) = \frac{1}{2} + \frac{t}{4}\]

  5. calculusxy
    • one year ago
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    So let's say if we have 2 seconds, we would have: \[s(2) = \frac{ 1 }{ 2 } + \frac{ 2 }{ 4 } \] or \[1\]

  6. calculusxy
    • one year ago
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    So the distance would be 1 meters, right?

  7. Michele_Laino
    • one year ago
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    yes!

  8. calculusxy
    • one year ago
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    I do that for every number?

  9. Michele_Laino
    • one year ago
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    yes! and after 14 seconds the distance is: \[s\left( {14} \right) = \frac{1}{2} + \frac{{14}}{4} = 4meters\]

  10. calculusxy
    • one year ago
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    But the graph is only 6-by-6.

  11. Michele_Laino
    • one year ago
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    in that case, for time t, the subsequent condition holds: \[0 \leqslant t \leqslant 6\]

  12. calculusxy
    • one year ago
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    What do you mean?

  13. Michele_Laino
    • one year ago
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    I mean that the graph of your function, is a straight line, which connects these 2 points: (0, 0.5) (6, 2)

  14. Michele_Laino
    • one year ago
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    |dw:1436127507778:dw|

  15. calculusxy
    • one year ago
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    But it said to go to the 4 meter mark. I don't what I should do with that small of a graph because I feel like I need to extend it.

  16. Michele_Laino
    • one year ago
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    I think that this equation: \[s\left( t \right) = \frac{1}{2} + \frac{t}{4}\] so I think that you have to consider an x-axis like this: |dw:1436127695828:dw|

  17. calculusxy
    • one year ago
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    Okay. So you're saying I should extend the graph right?

  18. Michele_Laino
    • one year ago
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    yes!

  19. Michele_Laino
    • one year ago
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    since this equation: \[s\left( t \right) = \frac{1}{2} + \frac{t}{4}\] is the right equation, which models the motion of your particle

  20. calculusxy
    • one year ago
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    Also, I asked this question before. Another user answered it, but I didn't know if it was right or wrong. So can you check his/her graph?

  21. Michele_Laino
    • one year ago
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    that drawing, is the graph of the subsequent equation: \[s\left( t \right) = 0.5\left( {1 + t} \right)\] as you can check, by substitution

  22. calculusxy
    • one year ago
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    \[0.5(1 + 2) = 0.5(1) + 0.5(2) = 0.5 + 1 = 1.5 m\]

  23. calculusxy
    • one year ago
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    It seems like there is difference of 0.5m.

  24. Michele_Laino
    • one year ago
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    yes! at t= 2 seconds, we have s=1.5 meters

  25. calculusxy
    • one year ago
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    I think your one is correct right?

  26. Michele_Laino
    • one year ago
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    if we use the second equation, of course!

  27. Michele_Laino
    • one year ago
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    yes! since the general formula, is: \[s\left( t \right) = {s_0} + vt\] where s_0 is the initial space, namelythe space at t=0, and v is the speed of our particel. Now in our case, we have: s_0=0.5=1/2 and v=0.25=1/4

  28. Michele_Laino
    • one year ago
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    particle*

  29. calculusxy
    • one year ago
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    Thank you so much!

  30. Michele_Laino
    • one year ago
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    :)

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