Find Displacement of Car: Math Homework

[delfam]

Homework Statement

car travels 18.2km due north then 43.2km at 53 degrees west of north. Find magnitude of car's resultant displacement.

b.) calculate the direction of the car's resultant displacement, measured counter-clockwise from the northerly direction.

Homework Equations

The Attempt at a Solution

Ax = 18.2sin(90) =18.2 Ay = 18.2cos(90) = 0
Bx = 43.2sin(143) = 26 By = 43.2cos(143) = -34.501
Cx = 44.2 Cy = -34.501

44.2^2 + -34.501^2 = 3143.82, square root of that = 56.07

arctan(-34.504/44.2) = -37.98 degrees

so part b says counter-clockwise from north, so wouldn't it be from 90 degrees so should the angle be 127.98? I entered that as an asnwer and it's wrong so not sure what I did wrong.

 

[MHrtz]

when you did Bx and By, where did you get 43.2 and 143 degrees?

 

[delfam]

typo, it was due north 43.2km not 432, the 143 degrees is when it says 53 degrees west of north, north is 90, so 90 + 53 = 143. I already got that part right, I just don't understand the wording on part b.

 

[MHrtz]

recheck your math for Bx and By. Bx should be 260 and By should be -345. Also, you're using the Pythagorean theorem when you should be using law of cosines. the angle across from the resultant vector should be 90 + (90 - 53) = 127. In other words, it's 53 degrees from the vertical not the horizontal.

 

[rwisz]

Your calculations for the magnitude of the resultant displacement (56.07) and the direction (37.98 degrees from the northerly direction) are correct. However, the question asks for the direction measured counter-clockwise from the northerly direction, which means the angle should be measured from 0 degrees. Therefore, the correct answer would be 52.02 degrees (90 - 37.98 = 52.02).