## anonymous one year ago The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = 5t − 4, 0 ≤ t ≤ 3 (a) Find the displacement. (b) Find the total distance that the particle travels over the given interval.

1. anonymous

Hint: $v(t) = \frac{ dx(t) }{ dt }$ You'll have to integrate this over the given interval

2. anonymous

I got all the way through it but not the right answer. I got -3/4 for my displacement

3. anonymous

I did integrate it. I'm just not coming up with the correct answer for some reason

4. anonymous

You may have made an arithmetic mistake, because -3/4 doesn't sound right

5. anonymous

where did you get the 9t? I got

6. anonymous

When you integrate a constant, you get at+C so $\int\limits 9 dt = 9t+C$

7. anonymous

you don't need a constant it's on a definite integral

8. anonymous

I know, I'm just explaining the 9t

9. anonymous

$\int\limits_{0}^{3} 9 dt \implies 9t|_0^3$

10. anonymous

yes! okay awesome I fixed part a with 21/2

11. anonymous

|dw:1443314364425:dw| some handy rules, just in case/ review

12. anonymous

AWESOME that's so helpful thank you

13. anonymous

Yw, so what did you get for your displacement now?

14. anonymous

21/2

15. anonymous

hmmm it told me it was correct

16. anonymous

but I can't get part b

17. anonymous

Oh sorry I kept using 9 when there is a 4 haha :)

18. anonymous

hahaha okay :) now I got 87/10 for part b & they counted it wrong

19. anonymous

Let me just fix that to show it again $\int\limits\limits_{0}^{3} (5t-4) dt \implies (\frac{ 5t^2 }{ 2 }-4t)|_0^3$ ah there we go

20. anonymous

The distance is a bit tricky, because we have to find the total interval so we have $5t-4 = 0 \implies t = \frac{ 4 }{ 5 }$ so our integral should be $\int\limits_{0}^{4/5} (5t-4)dt + \int\limits_{4/5}^{3} (5t-4)dt$ and note that distance is a scalar quantity so you will have to use absolute values for your integrals

21. anonymous

At t =4/5 is really just when it changes the direction

22. anonymous

OH oh my gosh that makes so much more sense

23. anonymous

i got 73/10 this time.. still incorrect any suggestions?

24. anonymous

$|\int\limits_{0}^{4/5} (5t-4)dt| \implies |(\frac{ 5t^2 }{ 2 }-4t)|_0^{4/5} |$ the integral is the same your interval is just changed so here is the first one

25. anonymous

The |...| represent absolute values

26. anonymous

Yes! I did that

27. anonymous

Ok what did you get for this integral?

28. anonymous

$(4t-\frac{ 5t ^{2} }{ 2 }) Integral (0, 4/5) + (\frac{ 5t ^{2} }{ 2 } - 4t) \int\limits (3, 4/5)$

29. anonymous

Why did you write $4t-5t^2/2$

30. anonymous

because that's on the integral (0, 4/5)? maybe not?

31. anonymous

Noo, why are you changing the integrand? It's the same we are just seeing where the velocity is negative and positive

32. anonymous

okay so do I set the integral negative? hahaha I'm so sorry

33. anonymous

No, we are finding the distance, which is the magnitude it cannot be negative

34. anonymous

so we're just finding the velocity which means just plugging in the integral? & keeping the integral the same?

35. anonymous

Nooo, we're finding the distance, and the reason we have split it into two integrals is because the integral 0 to 3 does not account for the negative and positive path of the particle, that is just the displacement, but for part b we want distance, that's total area travelled, so we have to find the change in velocity which is at $t=4/5$ so we go from 0 to 4/5 and we add the positive area 4/5 to 3 for the total distance...I hope that makes sense now

36. anonymous

v(t) = 5t-4 is a straight line, the line does not change, but the path does

37. anonymous

If you meant put the first integral as $- \int\limits_{0}^{4/5} (5t-4)dt$ that would work

38. anonymous

OH okay let me try!

39. anonymous

would my second integral be $\int\limits_{0}^{3} (5t-4)$

40. anonymous

Noooo haha, I have already set up the integrals for you, hmm ok, I think you're not understanding the concept

41. anonymous

I feel like I'm losing it more and more hahaha okay let me look this over again

42. anonymous

$\left| \int\limits_{0}^{4/5}(5t-4)dt \right|+\left| \int\limits_{0}^{4/5} (5t-4)dt\right|$ that's all there is!

43. anonymous

HAHAHAHAHA OH MY GOSH

44. anonymous

any ideas why it's not taking my answer of 3.2? sometimes it's picky

45. anonymous

That's not right

46. anonymous

Lets do this one at a time $\int\limits_{0}^{4/5} (5t-4)dt$ what does this integral =? Remember to include absolute value

47. anonymous

I got -1.6 but of course the absolute value is 1.6

48. anonymous

the integral is $\frac{ 5t ^{2} }{ 2 }-4t \int\limits (0,4/5)$

49. anonymous

Why is your notation like that, $\int\limits\limits_{0}^{4/5} (5t-4)dt = (\frac{ 5t^2 }{ 2 }-4t)|_0^{4/5} = \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right) ^2}{ 2 }-4(\frac{ 4 }{ 5 }) \right)-0 = |\frac{ 5 \times 16}{ 2 \times 25}-\frac{ 16 }{ 5 }| = \frac{ 8 }{ 5 }$ so that's good, that is 1.6

50. anonymous

Now lets do the other integral $\int\limits_{4/5}^{3} (5t-4)dt = \left( \frac{ 5t^2 }{ 2 }-4t \right)|_{4/5}^{3}$

51. anonymous

it's normally not like that I just struggle with plugging it into a computer, my apologies

52. anonymous

$\frac{ 121 }{ 10 }$ this integral gives the above, you can check

53. anonymous

Now we add the two integrals $\frac{ 8 }{ 5 } + \frac{ 121 }{ 10 } = \frac{ 137 }{ 10 }$

54. anonymous

when I plug 3 in, i get $\frac{ 45 }{ 2 } - 12=21/2$ ...what is going on

55. anonymous

$\left( \frac{ 5(3)^2 }{ 2 } -4(3)\right) - \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right)^2 }{ 2 } -4(\frac{ 4 }{ 5 })\right)$ this is the result of the second integral, you're forgetting about $\frac{ 4 }{ 5 }$

56. anonymous

yes so this would be 21/2 - 8/5 = 89/10

57. anonymous

No, it should be +8/5, the second term gives -8/5 remember? And you're already subtracting using FTC so a (-)(-) = +

58. anonymous

We figured this out with the first integral

59. anonymous

ohhhhhhhhhhhhhh