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

- calculusxy

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

@Michele_Laino

- calculusxy

I will post the attachment.

- calculusxy

##### 1 Attachment

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## More answers

- 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

So let's say if we have 2 seconds, we would have:
\[s(2) = \frac{ 1 }{ 2 } + \frac{ 2 }{ 4 } \]
or \[1\]

- calculusxy

So the distance would be 1 meters, right?

- Michele_Laino

yes!

- calculusxy

I do that for every number?

- Michele_Laino

yes!
and after 14 seconds the distance is:
\[s\left( {14} \right) = \frac{1}{2} + \frac{{14}}{4} = 4meters\]

- calculusxy

But the graph is only 6-by-6.

- Michele_Laino

in that case, for time t, the subsequent condition holds:
\[0 \leqslant t \leqslant 6\]

- calculusxy

What do you mean?

- 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

|dw:1436127507778:dw|

- 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

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

Okay. So you're saying I should extend the graph right?

- Michele_Laino

yes!

- 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

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?

##### 1 Attachment

- 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

\[0.5(1 + 2) = 0.5(1) + 0.5(2) = 0.5 + 1 = 1.5 m\]

- calculusxy

It seems like there is difference of 0.5m.

- Michele_Laino

yes! at t= 2 seconds, we have s=1.5 meters

- calculusxy

I think your one is correct right?

- Michele_Laino

if we use the second equation, of course!

- 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

particle*

- calculusxy

Thank you so much!

- Michele_Laino

:)

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