Question

A mountain climber hikes 25.5 km from her base camp, going 20.0º north of east. On the second day, she walks 41.0 km in a direction 65.0º north of east, at which point she has reached a summit. Determine the magnitude and direction of her resultant displacement between the base camp and the second camp(Remember to analyze the two motions separately, and then add and use Pythagorean theorem for the resultant and trigonometry for the angle.)

Answer 1

magnitude: 61.72 km

direction: 48° (north of east)

Explanation:

Two vectors are formed, the first one starts at the origin, has a length of 25.5 km and form an angle of 20° with the positive x-axis; the second one starts when the first vector finish, has a length of 41 km and form an angle of 65° with a line which is parallel to the positive x-axis.

From trigonometry relationships we can decompose the vectors as follows

first hike

25.5*sen(20°) = 8.72 km (north direction)

25.5*cos(20°) = 23.96 km (east direction)

second hike

41*sen(65°) = 37.16 km (north direction)

41*cos(65°) = 17.32 km (east direction)

Then, the resultant has the next components:

8.72 km + 37.16 km = 45.88 km (north direction)

23.96 km + 17.32 km = 41.28 km (east direction)

From Pythagorean theorem

Resultant = sqrt(45.88^2 + 41.28^2) = 61.72 km

From trigonometry

angle = arctan(45.88/41.28) = 48° (north of east)

Answer 2

As we can check from the angles that are in the positive region (0 to 90º)
all terms are positive. decomposing the vectors we have

25.5 sen(20) = 8.72 & 25.5 cos(20) = 23.96 for the first hike

41 sen(65) = 37.16 & 41 cos(65) = 17.32 for the second hike

adding we have

45.88 and 41.28
√( 45.88^{2} + 41.28^{2} ) = 61.72 miles this will be the total displacement

and the angle would be arctan ( 41.28/45.88) ≈ 42º north of east