calculusxy
  • calculusxy
MEDAL!!! Start at the 0.5 meter mark and walk towards the 4 meter mark at 0.25 meters per second.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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calculusxy
  • calculusxy
@Michele_Laino
calculusxy
  • calculusxy
I will post the attachment.
calculusxy
  • calculusxy

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Michele_Laino
  • Michele_Laino
I think that the function s=s(t), namely the space traveled by our particle, at time t, is given by the subsequent formula: \[\Large s\left( t \right) = \frac{1}{2} + \frac{t}{4}\]
calculusxy
  • calculusxy
So let's say if we have 2 seconds, we would have: \[s(2) = \frac{ 1 }{ 2 } + \frac{ 2 }{ 4 } \] or \[1\]
calculusxy
  • calculusxy
So the distance would be 1 meters, right?
Michele_Laino
  • Michele_Laino
yes!
calculusxy
  • calculusxy
I do that for every number?
Michele_Laino
  • Michele_Laino
yes! and after 14 seconds the distance is: \[s\left( {14} \right) = \frac{1}{2} + \frac{{14}}{4} = 4meters\]
calculusxy
  • calculusxy
But the graph is only 6-by-6.
Michele_Laino
  • Michele_Laino
in that case, for time t, the subsequent condition holds: \[0 \leqslant t \leqslant 6\]
calculusxy
  • calculusxy
What do you mean?
Michele_Laino
  • Michele_Laino
I mean that the graph of your function, is a straight line, which connects these 2 points: (0, 0.5) (6, 2)
Michele_Laino
  • Michele_Laino
|dw:1436127507778:dw|
calculusxy
  • calculusxy
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.
Michele_Laino
  • Michele_Laino
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|
calculusxy
  • calculusxy
Okay. So you're saying I should extend the graph right?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
since this equation: \[s\left( t \right) = \frac{1}{2} + \frac{t}{4}\] is the right equation, which models the motion of your particle
calculusxy
  • calculusxy
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?
Michele_Laino
  • Michele_Laino
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
calculusxy
  • calculusxy
\[0.5(1 + 2) = 0.5(1) + 0.5(2) = 0.5 + 1 = 1.5 m\]
calculusxy
  • calculusxy
It seems like there is difference of 0.5m.
Michele_Laino
  • Michele_Laino
yes! at t= 2 seconds, we have s=1.5 meters
calculusxy
  • calculusxy
I think your one is correct right?
Michele_Laino
  • Michele_Laino
if we use the second equation, of course!
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
particle*
calculusxy
  • calculusxy
Thank you so much!
Michele_Laino
  • Michele_Laino
:)

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