- #1

orangegalaxies

- 50

- 16

- Homework Statement
- A pilot in a seaplane flies for a total of 3.0 h with an average velocity of 130 km/h [N 32 E]. In the first part of the trip, the pilot flies for 1.0 h through a displacement of 150 km [E 12 N]. She then flies directly to her final destination. Determine the displacement for the second part of the flight.

- Relevant Equations
- velocity = displacement/time

displacement = d2 - d1

a^2 = b^2 + c^2 - 2bc cos A

I rearranged the displacement formula to d2 = d + d1. I used cosine law to solve for d2 since the triangle is not right-angled but I am not getting the correct answer or angle for d2. The angle I used in cosine law (based on the diagram) was 32+12+90 = 134.

d = v(t) = 130(3) = 390 km/h [N 32 E]

d= d2 - d1, therefore d2 = d + d1

d2^2 = 390^2 + 150^2 - 2(390x150) cos134

d2 = 505.8 km

sin ϑ/150 = sin 134/505.8

ϑ = 12

ϑ = 32 + 12 = [N 44 E]

Therefore d2 = 505.8 km [N 44 E]

d = v(t) = 130(3) = 390 km/h [N 32 E]

d= d2 - d1, therefore d2 = d + d1

d2^2 = 390^2 + 150^2 - 2(390x150) cos134

d2 = 505.8 km

sin ϑ/150 = sin 134/505.8

ϑ = 12

ϑ = 32 + 12 = [N 44 E]

Therefore d2 = 505.8 km [N 44 E]

**This answer is wrong and I don't know why.**