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Study23 Group Title

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.

  • one year ago
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  1. AllTehMaffs Group Title
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    |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 :)

    • one year ago
  2. AllTehMaffs Group Title
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    * from the origin in quadrant 1 anticlockwise

    • one year ago
  3. Study23 Group Title
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    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?

    • one year ago
  4. Study23 Group Title
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    How would I use components to find the avg. velocity @AllTehMaffs

    • one year ago
  5. loremipsum Group Title
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    Kidnapped? What a horrifying question.

    • one year ago
  6. AllTehMaffs Group Title
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    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|

    • one year ago
  7. Study23 Group Title
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    Okay... I used 20° though

    • one year ago
  8. Study23 Group Title
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    Does that matter?

    • one year ago
  9. AllTehMaffs Group Title
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    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

    • one year ago
  10. Study23 Group Title
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    So how would I use components?

    • one year ago
  11. AllTehMaffs Group Title
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    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.

    • one year ago
  12. AllTehMaffs Group Title
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    Can you see how to get the lengths of the sides of that triangle?

    • one year ago
  13. Study23 Group Title
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    Using sin and cos?

    • one year ago
  14. AllTehMaffs Group Title
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    |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

    • one year ago
  15. Study23 Group Title
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    So to find the deltas I would...?

    • one year ago
  16. AllTehMaffs Group Title
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    :) Not yet. you already have the points. What's your change in x?

    • one year ago
  17. Study23 Group Title
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    2.1sin20-(-1.33)

    • one year ago
  18. AllTehMaffs Group Title
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    right, and you find your y the same way, then use Pythagoras to find C

    • one year ago
  19. AllTehMaffs Group Title
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    I think I contradicted myself by saying we don't need angles, but you're good so far

    • one year ago
  20. Study23 Group Title
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    This gives the displacement so I would have to divide by time?

    • one year ago
  21. AllTehMaffs Group Title
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    yeah, and then that's your average velocity between those points

    • one year ago
  22. Study23 Group Title
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    So then as far as acceleration goes?

    • one year ago
  23. AllTehMaffs Group Title
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    That's where you need those other angles decided to overheat

    • one year ago
  24. AllTehMaffs Group Title
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    wow, don't know how those messages got mashed up :P Sorry, my computer decided to overheat

    • one year ago
  25. AllTehMaffs Group Title
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    So! Acceleration. You have to divide your velocities up into components, too. The first one, anyway; the second is only in 1 direction.

    • one year ago
  26. AllTehMaffs Group Title
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    |dw:1381700640583:dw|

    • one year ago
  27. Study23 Group Title
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    Sorry 35 mph?

    • one year ago
  28. AllTehMaffs Group Title
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    that's the angle. V1 was 35 mph

    • one year ago
  29. AllTehMaffs Group Title
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    yeah

    • one year ago
  30. AllTehMaffs Group Title
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    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 }\

    • one year ago
  31. Study23 Group Title
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    For v2 I would use 10sin20?

    • one year ago
  32. AllTehMaffs Group Title
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    :/ \[a_{y}=\frac{ v _{2y}-v_{1y} }{ \Delta t }]\]

    • one year ago
  33. AllTehMaffs Group Title
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    Not quite. Remember that when the person made the measurement, the car was going straight east.

    • one year ago
  34. AllTehMaffs Group Title
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    So all of V2 is in the x direction

    • one year ago
  35. Study23 Group Title
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    So what would it be?

    • one year ago
  36. AllTehMaffs Group Title
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    my computer is being crashy. So calculate the change of velocity in the x direction, and then calculate it in the y

    • one year ago
  37. AllTehMaffs Group Title
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    What's v2x?

    • one year ago
  38. Study23 Group Title
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    I'm not sure

    • one year ago
  39. AllTehMaffs Group Title
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    How much of V2 is moving in the x direction?

    • one year ago
  40. Study23 Group Title
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    2.1sin20?

    • one year ago
  41. AllTehMaffs Group Title
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    |dw:1381701681185:dw| |dw:1381701807535:dw|

    • one year ago
  42. AllTehMaffs Group Title
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    Does that help?

    • one year ago
  43. Study23 Group Title
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    I can't see the v2 stuff on the dawning

    • one year ago
  44. AllTehMaffs Group Title
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    nooooo!! bummer. k |dw:1381702127697:dw|

    • one year ago
  45. AllTehMaffs Group Title
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    Does it make sense why we're not using the 20 degrees for this part of the calculation?

    • one year ago
  46. Study23 Group Title
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    Yes

    • one year ago
  47. Study23 Group Title
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    So 2.1cos0

    • one year ago
  48. Study23 Group Title
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    ?

    • one year ago
  49. AllTehMaffs Group Title
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    Not that either. In the problem, what does it say about his velocity when he measures it a second time?

    • one year ago
  50. AllTehMaffs Group Title
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    2.1 miles is the distance away from the police station

    • one year ago
  51. Study23 Group Title
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    ??

    • one year ago
  52. AllTehMaffs Group Title
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    "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"

    • one year ago
  53. Study23 Group Title
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    2.1?

    • one year ago
  54. AllTehMaffs Group Title
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    moving E at a speed of 10 mph"

    • one year ago
  55. AllTehMaffs Group Title
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    V2 = ?

    • one year ago
  56. Study23 Group Title
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    10

    • one year ago
  57. AllTehMaffs Group Title
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    V2x - V1x = ?

    • one year ago
  58. AllTehMaffs Group Title
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    Where did you get 1.33?

    • one year ago
  59. Study23 Group Title
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    10-35cos66°

    • one year ago
  60. AllTehMaffs Group Title
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    yes!! Now V2y - V1y = ?

    • one year ago
  61. Study23 Group Title
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    0-35sin66

    • one year ago
  62. AllTehMaffs Group Title
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    yes!!

    • one year ago
  63. AllTehMaffs Group Title
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    now (V2x - V1x) / t = ax (V2y - V1y)/t = ay a^2 = (ax)^2 + (ay)^2

    • one year ago
  64. AllTehMaffs Group Title
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    and a is your average acceleration

    • one year ago
  65. Study23 Group Title
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    Should I use | |?

    • one year ago
  66. AllTehMaffs Group Title
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    ??

    • one year ago
  67. Study23 Group Title
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    Abs value

    • one year ago
  68. AllTehMaffs Group Title
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    for acceleration? It has to be positive anyway because you're squaring the term

    • one year ago
  69. Study23 Group Title
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    Why is a squared?

    • one year ago
  70. AllTehMaffs Group Title
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    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)

    • one year ago
  71. Study23 Group Title
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    When I divide by t it's 5/60 right?

    • one year ago
  72. AllTehMaffs Group Title
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    ummm , yeah, I think so

    • one year ago
  73. AllTehMaffs Group Title
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    5 min (1 hour/60 min)

    • one year ago
  74. AllTehMaffs Group Title
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    yeah

    • one year ago
  75. AllTehMaffs Group Title
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    I gotta run to work. This problem was mean!! :/

    • one year ago
  76. Study23 Group Title
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    Ok so I get 387. What units should it be?

    • one year ago
  77. AllTehMaffs Group Title
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    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

    • one year ago
  78. Study23 Group Title
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    Ok!

    • one year ago
  79. AllTehMaffs Group Title
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    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

    • one year ago
  80. AllTehMaffs Group Title
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    the magnitude to the deceleration is (10-35)/5min

    • one year ago
  81. AllTehMaffs Group Title
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    it might just be that, I don't know. Good mathin'!

    • one year ago
  82. Study23 Group Title
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    When you have a chance just double check for me! :)

    • one year ago
  83. AllTehMaffs Group Title
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    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! :)

    • one year ago
  84. AllTehMaffs Group Title
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    |dw:1381728262451:dw|

    • one year ago
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