Post a New Question

AP Physics

posted by .

While exploring a cave, a spelunker starts at the entrance and moves the following distances: 75.0 m north, 250 m east, 210 m at an angle 30.0° north of east, and 150 m south. Find the resultant displacement from the cave entrance.

I know I have to use the formulas Ax = A cos è, Ay = A sin è, A = sq rt((Ax)^2 + (Ay)^2), and è = tan^(-1) (Ay/Ax). But I'm not sure whether I should calculate 125*sin(30) on the calculator because I get a negative number.

  • AP Physics -

    è is theta

  • AP Physics -

    125sin(30) is 125x0.5 = 62.5

    For this sort of problem you need to draw it out on paper and so hard to show on here.

    210 m sin(30) + 75 m= 105 m gives the total displacement north. After this he moves south by 150 m so his displacement north is

    105m+-150m +75 m = 30 m

    [North is the +ve direction)

    His displacement east is

    250 m + 210 cos 30 = 250 m +182 m = 432 m

    so the displacement angle is

    tan^-1(30/432) = 3.97 deg (North of east)

    and the displacement is

    Sqrt(30^2 + 432^2) = 433 m

    But check my maths!!

  • AP Physics -


Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question