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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.
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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calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0I will post the attachment.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0So let's say if we have 2 seconds, we would have: \[s(2) = \frac{ 1 }{ 2 } + \frac{ 2 }{ 4 } \] or \[1\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0So the distance would be 1 meters, right?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0I do that for every number?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! and after 14 seconds the distance is: \[s\left( {14} \right) = \frac{1}{2} + \frac{{14}}{4} = 4meters\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0But the graph is only 6by6.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in that case, for time t, the subsequent condition holds: \[0 \leqslant t \leqslant 6\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I mean that the graph of your function, is a straight line, which connects these 2 points: (0, 0.5) (6, 2)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436127507778:dw

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.1I think that this equation: \[s\left( t \right) = \frac{1}{2} + \frac{t}{4}\] so I think that you have to consider an xaxis like this: dw:1436127695828:dw

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0Okay. So you're saying I should extend the graph right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since this equation: \[s\left( t \right) = \frac{1}{2} + \frac{t}{4}\] is the right equation, which models the motion of your particle

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0Also, 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
 one year ago
Best ResponseYou've already chosen the best response.1that 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
 one year ago
Best ResponseYou've already chosen the best response.0\[0.5(1 + 2) = 0.5(1) + 0.5(2) = 0.5 + 1 = 1.5 m\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0It seems like there is difference of 0.5m.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! at t= 2 seconds, we have s=1.5 meters

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0I think your one is correct right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1if we use the second equation, of course!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! 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
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