## 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.

1. calculusxy

@Michele_Laino

2. calculusxy

I will post the attachment.

3. calculusxy

4. 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: $\Large s\left( t \right) = \frac{1}{2} + \frac{t}{4}$

5. calculusxy

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

6. calculusxy

So the distance would be 1 meters, right?

7. Michele_Laino

yes!

8. calculusxy

I do that for every number?

9. Michele_Laino

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

10. calculusxy

But the graph is only 6-by-6.

11. Michele_Laino

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

12. calculusxy

What do you mean?

13. Michele_Laino

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

|dw:1436127507778:dw|

15. 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.

16. 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|

17. calculusxy

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

18. Michele_Laino

yes!

19. 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

20. 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?

21. 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

22. calculusxy

$0.5(1 + 2) = 0.5(1) + 0.5(2) = 0.5 + 1 = 1.5 m$

23. calculusxy

It seems like there is difference of 0.5m.

24. Michele_Laino

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

25. calculusxy

I think your one is correct right?

26. Michele_Laino

if we use the second equation, of course!

27. 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

28. Michele_Laino

particle*

29. calculusxy

Thank you so much!

30. Michele_Laino

:)