Displacement

Question

Sam went 5 m east, then he turned 30 degrees north and went 8m more. Afterwards, he turned 25 degrees east and went 4 m. How far is Sam from his original position?

Possible Answers:

15.8 m 30.0 degrees W


17.8 m 12.9 degrees W


15.6 m 29.2 degrees N


16.3 m 28.3 degrees N

Answer 1

For this problem, it is easiest to consider the angles in terms of degrees away from East which is set as the positive x direction.  North will be the positive y direction.

 

Based on that, the translation vectors described in the problem are:

 

1:  [ 5, 0]

2:  [ 8 x cos(30), 8 x sin(30)] = [6.93, 4]

3:  [ 4 x cos(5), 4 x sin(5)] = [3.98, 0.349]  (Why is the angle 5? Because if you turn 30 degrees towards north from due east, then 25 degrees back towards east, then you end up 5 degrees towards north from due east.)

 

Using vector addition (summing the x's and y's) the total translation is then:

 [ 5 + 6.93 + 3.98, 0 + 4 + 0.349] = [15.91m north, 4.349m east]

 

The magnitude of that vector is equal to the total distance from the original position, and the magnitude is:

SQRT(15.91^2 + 4.349^2) = 16.494[m] (which should be later rounded to the required number of significant figures)

 

The angle of that vector is equal to the overall direction that the subject traveled, this is obtained with any of the following:

 

Theta = sin^-1(4.349/16.494) = cos^-1(15.91/16.494) = tan^-1(4.349/15.91) = 15.288 Degs north of due east.