Question

MEDAL!!!


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

Answer 1

@Michele_Laino

Answer 2

I will post the attachment.

Answer 3

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

Answer 4

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

Answer 5

So the distance would be 1 meters, right?

Answer 6

yes!

Answer 7

I do that for every number?

Answer 8

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

Answer 9

But the graph is only 6-by-6.

Answer 10

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

Answer 11

What do you mean?

Answer 12

I mean that the graph of your function, is a straight line, which connects these 2 points:
(0, 0.5)
(6, 2)

Answer 13

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.

Answer 14

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:

Answer 15

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

Answer 16

yes!

Answer 17

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

Answer 18

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?

Answer 19

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

Answer 20

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

Answer 21

It seems like there is difference of 0.5m.

Answer 22

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

Answer 23

I think your one is correct right?

Answer 24

if we use the second equation, of course!

Answer 25

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

Answer 26

particle*

Answer 27

Thank you so much!

Answer 28

:)