Solving Vector Problems: Displacement and Magnitude Calculations

[schrodingersequation]

A jogger travels a route that has two parts. The first is a displacement A of 2.50 km due south, and the second involoves a displacement B that points due east. (a) The resultant displacement A+B has a magnitude of 3.75 km. What is the magnitude of B, and what is the direction of A+B relative to due south? (b) suppose that A-B had a magnitude of 3.75 km. What then would be the magnitude of B, and what is the direction of A-B realtive to due south?


I have an answer for this problem I just want to know if it is right.
(a)The magnitude of B is 2.80 km
A+B points 48.188 degrees East of South

(b)the magnitude of B is 2.80 km
A-B points 48.188 degrees West of South

 

[AKG]

Looks pretty good. In finding the direction, I got the angle to be 48.190. Make sure you're not using rounded values, in fact you can just use cosX = 2.50/3.75, and solve for X (which is what I did). However, you should round to 3 significant digits, giving you 48.2 anyways, which would be the same using my number or yours.

 

[M98Ranger]

Your answer is correct! Good job solving this vector problem. It's important to remember that displacement is a vector quantity, meaning it has both magnitude and direction. In part (a), the magnitude of B can be found by using the Pythagorean theorem to find the missing side of a right triangle (3.75^2 = 2.50^2 + B^2). And the direction of A+B can be found by using the inverse tangent function to find the angle between the resultant displacement and the due south direction. Similarly, in part (b), the magnitude of B can be found using the same method, and the direction of A-B can be found by subtracting the angle between A and B from 180 degrees. Keep up the good work!