## anonymous 2 years ago PLEASE, HELP been waiting nearly and hour... – Drawing a vector displacement problem, need help! After being kidnapped and placed in the trunk of a car, you discover you are able to pry open the trunk. The GPS tells you that you're moving 35 mph in a direction 66 north of east at a location 1.33 mi west of the police station. 5 min later, you take another reading that tells you are now 2.1 mi from the police station in a direction 20 degrees east of north and are moving E at a speed of 10 mph. Find the avg. velocity and acceleration during the trip.

1. anonymous

|dw:1381583505376:dw| The second set of points was meaner to find because it gave them to you in polar form. Remember that polar coordinates in cartesian can be found by x = r cos (theta) y = r sin (theta) But only when theta is measured from quadrant 1. If you used the 20 degrees they gave you, the cos and the sin would be switched around. Remember that the average velocity is the change in position - that's the vector from point A to point B - over time 5 min. The average acceleration is the change in velocities. Whereas you can calculate displacement vector just by making a right triangle using your two points, the change in velocity is most easily done component-wise. Hope this helps :)

2. anonymous

* from the origin in quadrant 1 anticlockwise

3. anonymous

Hey @AllTehMaffs I'm having trouble finding average velocity. By the way, I switches cos and sin around because I'm using 20 degrees. Could you help me out?

4. anonymous

How would I use components to find the avg. velocity @AllTehMaffs

5. anonymous

Kidnapped? What a horrifying question.

6. anonymous

Hi. Sure. For the average velocity you need to find the magnitude of the vector that goes from from the first GPS point to the second. So |dw:1381698737663:dw|

7. anonymous

Okay... I used 20° though

8. anonymous

Does that matter?

9. anonymous

That won't change the x,y position, just make sure that where I have a sin 70, you have a cos 20 and vice versa

10. anonymous

So how would I use components?

11. anonymous

whoops, I definitely forgot my trig identities. scratch what I just said about the swapping. But we'll get back to that in a sec. It's not immediately important.

12. anonymous

Can you see how to get the lengths of the sides of that triangle?

13. anonymous

Using sin and cos?

14. anonymous

|dw:1381699218098:dw| You shouln't need to use sin and cos for the displacement vector. You technically don't know the angles right off the bat :P

15. anonymous

So to find the deltas I would...?

16. anonymous

:) Not yet. you already have the points. What's your change in x?

17. anonymous

2.1sin20-(-1.33)

18. anonymous

right, and you find your y the same way, then use Pythagoras to find C

19. anonymous

I think I contradicted myself by saying we don't need angles, but you're good so far

20. anonymous

This gives the displacement so I would have to divide by time?

21. anonymous

yeah, and then that's your average velocity between those points

22. anonymous

So then as far as acceleration goes?

23. anonymous

That's where you need those other angles decided to overheat

24. anonymous

wow, don't know how those messages got mashed up :P Sorry, my computer decided to overheat

25. anonymous

So! Acceleration. You have to divide your velocities up into components, too. The first one, anyway; the second is only in 1 direction.

26. anonymous

|dw:1381700640583:dw|

27. anonymous

Sorry 35 mph?

28. anonymous

that's the angle. V1 was 35 mph

29. anonymous

yeah

30. anonymous

your acceleration in the x direction is $a_{x}=\frac{ v _{2x}-v_{1x} }{ \Delta t }$ and in the y it's $a_{y}=\frac{ v _{2y}-v_{1y} }{ \Delta t }\ 31. anonymous For v2 I would use 10sin20? 32. anonymous :/ \[a_{y}=\frac{ v _{2y}-v_{1y} }{ \Delta t }]$

33. anonymous

Not quite. Remember that when the person made the measurement, the car was going straight east.

34. anonymous

So all of V2 is in the x direction

35. anonymous

So what would it be?

36. anonymous

my computer is being crashy. So calculate the change of velocity in the x direction, and then calculate it in the y

37. anonymous

What's v2x?

38. anonymous

I'm not sure

39. anonymous

How much of V2 is moving in the x direction?

40. anonymous

2.1sin20?

41. anonymous

|dw:1381701681185:dw| |dw:1381701807535:dw|

42. anonymous

Does that help?

43. anonymous

I can't see the v2 stuff on the dawning

44. anonymous

nooooo!! bummer. k |dw:1381702127697:dw|

45. anonymous

Does it make sense why we're not using the 20 degrees for this part of the calculation?

46. anonymous

Yes

47. anonymous

So 2.1cos0

48. anonymous

?

49. anonymous

Not that either. In the problem, what does it say about his velocity when he measures it a second time?

50. anonymous

2.1 miles is the distance away from the police station

51. anonymous

??

52. anonymous

"tells you are now 2.1 mi from the police station in a direction 20 degrees east of north and are moving E at a speed of 10 mph"

53. anonymous

2.1?

54. anonymous

moving E at a speed of 10 mph"

55. anonymous

V2 = ?

56. anonymous

10

57. anonymous

V2x - V1x = ?

58. anonymous

Where did you get 1.33?

59. anonymous

10-35cos66°

60. anonymous

yes!! Now V2y - V1y = ?

61. anonymous

0-35sin66

62. anonymous

yes!!

63. anonymous

now (V2x - V1x) / t = ax (V2y - V1y)/t = ay a^2 = (ax)^2 + (ay)^2

64. anonymous

and a is your average acceleration

65. anonymous

Should I use | |?

66. anonymous

??

67. anonymous

Abs value

68. anonymous

for acceleration? It has to be positive anyway because you're squaring the term

69. anonymous

Why is a squared?

70. anonymous

You're finding the magnitude of vector a, which is kinda like a right triangle. It's defined as a = sqrt ((ax)^2 + (ay)^2)

71. anonymous

When I divide by t it's 5/60 right?

72. anonymous

ummm , yeah, I think so

73. anonymous

5 min (1 hour/60 min)

74. anonymous

yeah

75. anonymous

I gotta run to work. This problem was mean!! :/

76. anonymous

Ok so I get 387. What units should it be?

77. anonymous

that'd be miles per hour, so something is really wrong there. Because that's insane. I'll look at it again when I get back

78. anonymous

Ok!

79. anonymous

Or, maybe I way over complicated it. It's going 35 miles per hour to start, and it ends up going ten miles per hour

80. anonymous

the magnitude to the deceleration is (10-35)/5min

81. anonymous

it might just be that, I don't know. Good mathin'!

82. anonymous

When you have a chance just double check for me! :)

83. anonymous

First, units on acceleration here should be miles/(hour^2) (acceleration always has units distance/time^2), so sorry I mistyped that before. Second, I think that the answer is right, and the just 10-35 / t is definitely wrong. Not used to working in mph, so the numbers are bigger than I had thought they would be... I got 387miles/h^2 too. It would be negative, since both x and y components are getting smaller, and has a direction. But I don't think you need that for your answer. Hope this was the right method they were looking for! :)

84. anonymous

|dw:1381728262451:dw|

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