# a lost hiker radios you and tells you...?

...he walked 8.5 km south then turned right 37 degrees and walked 4.2 km. what's his displacement

some physics help?! please i have no idea how to do this!

Relevance

1. Draw a vector digram, 8.5 km south, then 4.2 km S37W

2. Use cosine Law c^2 = a^2+b^2-2ab cos c

c^2 = 8.5^2 + 4.2^2 - 2*8.5*4.2 cos (180-37)

c^2 = 72.25 + 17.64 + 57

c^2 = 146.9

c = 12.1 km

3. Use sine Law to find angle

12.1/sin 143 = 4.2/ sin X

sin X = 0.2

X = 12

Final Displacement is 12.1 km 12 degrees West of South

• Draw a vector diagram of each leg of the trip so that you may

determine the North(+)/South(-) and East(+)/West(-):

[i.e.,y component and x component, respectively,] of his

movement.

For the 8.5 km south leg denote by: -8.5 km S.

For the 37 deg.right turn and his 4.2 km distance covered,

use right triangle trigonometry:

4.2 cos37 north => 4.2*0.7986 = 3.35 km N.

4.2 sin37 east => 4.2*0.6018 = 2.5 km E

Now, collect and algebriacally sum respective components:

y component: => -8.5 + 3.35 = -5.15 km

x component: => 2.5 km

Then calculate the total displacement using:

d = √ x² + y² = √ 2.5² + (-5.15)² = 5.72 km

Hope this helps!