Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Study23

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.

  • 6 months ago
  • 6 months ago

  • This Question is Closed
  1. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    |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 :)

    • 6 months ago
  2. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    * from the origin in quadrant 1 anticlockwise

    • 6 months ago
  3. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

    • 6 months ago
  4. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  5. loremipsum
    Best Response
    You've already chosen the best response.
    Medals 0

    Kidnapped? What a horrifying question.

    • 6 months ago
  6. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    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|

    • 6 months ago
  7. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay... I used 20° though

    • 6 months ago
  8. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Does that matter?

    • 6 months ago
  9. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • 6 months ago
  10. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    So how would I use components?

    • 6 months ago
  11. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • 6 months ago
  12. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  13. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Using sin and cos?

    • 6 months ago
  14. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    |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

    • 6 months ago
  15. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    So to find the deltas I would...?

    • 6 months ago
  16. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  17. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    2.1sin20-(-1.33)

    • 6 months ago
  18. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  19. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  20. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  21. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  22. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    So then as far as acceleration goes?

    • 6 months ago
  23. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  24. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  25. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  26. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1381700640583:dw|

    • 6 months ago
  27. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry 35 mph?

    • 6 months ago
  28. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    that's the angle. V1 was 35 mph

    • 6 months ago
  29. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah

    • 6 months ago
  30. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    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 }\

    • 6 months ago
  31. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    For v2 I would use 10sin20?

    • 6 months ago
  32. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    :/ \[a_{y}=\frac{ v _{2y}-v_{1y} }{ \Delta t }]\]

    • 6 months ago
  33. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  34. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    So all of V2 is in the x direction

    • 6 months ago
  35. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    So what would it be?

    • 6 months ago
  36. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  37. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    What's v2x?

    • 6 months ago
  38. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not sure

    • 6 months ago
  39. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    How much of V2 is moving in the x direction?

    • 6 months ago
  40. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    2.1sin20?

    • 6 months ago
  41. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  42. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    Does that help?

    • 6 months ago
  43. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    I can't see the v2 stuff on the dawning

    • 6 months ago
  44. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  45. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  46. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

    • 6 months ago
  47. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    So 2.1cos0

    • 6 months ago
  48. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    ?

    • 6 months ago
  49. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  50. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    2.1 miles is the distance away from the police station

    • 6 months ago
  51. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    ??

    • 6 months ago
  52. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    "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"

    • 6 months ago
  53. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    2.1?

    • 6 months ago
  54. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    moving E at a speed of 10 mph"

    • 6 months ago
  55. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    V2 = ?

    • 6 months ago
  56. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    10

    • 6 months ago
  57. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    V2x - V1x = ?

    • 6 months ago
  58. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    Where did you get 1.33?

    • 6 months ago
  59. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    10-35cos66°

    • 6 months ago
  60. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    yes!! Now V2y - V1y = ?

    • 6 months ago
  61. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    0-35sin66

    • 6 months ago
  62. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    yes!!

    • 6 months ago
  63. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  64. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    and a is your average acceleration

    • 6 months ago
  65. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Should I use | |?

    • 6 months ago
  66. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    ??

    • 6 months ago
  67. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Abs value

    • 6 months ago
  68. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  69. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Why is a squared?

    • 6 months ago
  70. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    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)

    • 6 months ago
  71. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  72. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    ummm , yeah, I think so

    • 6 months ago
  73. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    5 min (1 hour/60 min)

    • 6 months ago
  74. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah

    • 6 months ago
  75. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  76. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  77. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • 6 months ago
  78. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok!

    • 6 months ago
  79. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • 6 months ago
  80. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  81. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  82. Study23
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 6 months ago
  83. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    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! :)

    • 6 months ago
  84. AllTehMaffs
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1381728262451:dw|

    • 6 months ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.