Question

A hiker starts by walking 1400 m in the direction 25 degrees N of W. He then walks and additional distance in the direction 15 degrees W of S, before finally walking 40 degrees N of E back to where he started. How far did he walk on the second and third legs of his trip?

Answer 1

Here ,

let's suppose he walks a and b in second and third leg

Now, let the north is +y direction and east is the positive x direction

as the net displacement is Zero

1400 * (- cos(25 degree) i + j * sin(25 degree)) + a * (- sin(15 degree) i - j * cos(15 degree)) + b * (cos(40 degree) i + sin(40 degree) j ) = 0

hence,

1400 * (- cos(25 degree) + a * (- sin(15 degree)) + b * (cos(40 degree) = 0

also

1400 * (sin(25 degree)) + a * (- cos(15 degree)) + b * ( sin(40 degree) ) = 0

solving both equations

a = 2212.1 m

b = 2404 m

the second leg is 2212.1 m

third leg is 2404 m