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Homework Help: Help with this vector problem please

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    1.)A car is driven east for a distance of 52 km, then north for 34 km, and then in a direction 35° east of north for 27 km. Draw the vector diagram and determine the total displacement of the car from its starting point.

    a. Find the magnitude
    b. Find the direction (counter clockwise from east)

    3. The attempt at a solution

    So I got a. using the unit vector method which gave me 49.49 for the y component, and 74.12 for the x component. The answer for part a is 89.1237 km.

    Now, for part b., I'm doing arctan(49.49/74.12) which is equal to 33.73, but it marks me wrong.

    Any idea of what's happening?

    1. The problem statement, all variables and given/known data

    2.) Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.50 m due east, then 0.80 m at 22° north of due east. Beetle 2 also makes two runs and the first is 1.6 m at 41° east of due north.

    (a) What must be the magnitude of its second run if it is to end up at the new location of beetle 1?

    (b) In what direction must it run?

    3. The attempt at a solution

    For part a., I am assuming that Vector A+Vector B= 0.5 i + 0.8(cos(22) i + sin(22) j) and Vector C= 1.6(cos(41) i + sin(41) j). Vector D would be the magnitude of beetle 2's second run.
    So, it'll something like Vector D=Vector A+Vector B-Vector C=(1.24 i +.3 j)-(1.2 i +1.05 j)= .04 i + .75 j

    And by using the Pythagorean theorem: .04^2 + .75^2 = .751 and once again, it marks me wrong.

    What am I doing wrong?
  2. jcsd
  3. Sep 8, 2009 #2
    For the first one, I think you made the wrong assumption. Same for the C vector in second.
    Maybe this could help: http://id.mind.net/~zona/mstm/physics/mechanics/vectors/introduction/introductionVectors.html" [Broken]
    Last edited by a moderator: May 4, 2017
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