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Study23

  • one year 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.

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  1. AllTehMaffs
    • one year ago
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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 :)

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

  3. Study23
    • one year ago
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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?

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

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

  6. AllTehMaffs
    • one year ago
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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|

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

  8. Study23
    • one year ago
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    Does that matter?

  9. AllTehMaffs
    • one year ago
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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

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

  11. AllTehMaffs
    • one year ago
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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.

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

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

  14. AllTehMaffs
    • one year ago
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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

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

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

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

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

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

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

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

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

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

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

  25. AllTehMaffs
    • one year ago
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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.

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

  27. Study23
    • one year ago
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    Sorry 35 mph?

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

  29. AllTehMaffs
    • one year ago
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    yeah

  30. AllTehMaffs
    • one year ago
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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 }\

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

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

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

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

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

  36. AllTehMaffs
    • one year ago
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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

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

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

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

  40. Study23
    • one year ago
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    2.1sin20?

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

  42. AllTehMaffs
    • one year ago
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    Does that help?

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

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

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

  46. Study23
    • one year ago
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    Yes

  47. Study23
    • one year ago
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    So 2.1cos0

  48. Study23
    • one year ago
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    ?

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

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

  51. Study23
    • one year ago
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    ??

  52. AllTehMaffs
    • one year ago
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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"

  53. Study23
    • one year ago
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    2.1?

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

  55. AllTehMaffs
    • one year ago
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    V2 = ?

  56. Study23
    • one year ago
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    10

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

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

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

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

  61. Study23
    • one year ago
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    0-35sin66

  62. AllTehMaffs
    • one year ago
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    yes!!

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

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

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

  66. AllTehMaffs
    • one year ago
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    ??

  67. Study23
    • one year ago
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    Abs value

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

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

  70. AllTehMaffs
    • one year ago
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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)

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

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

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

  74. AllTehMaffs
    • one year ago
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    yeah

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

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

  77. AllTehMaffs
    • one year ago
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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

  78. Study23
    • one year ago
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    Ok!

  79. AllTehMaffs
    • one year ago
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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

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

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

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

  83. AllTehMaffs
    • one year ago
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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! :)

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

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