# Help? Vector math problem??

A car travels east at 100 km/h for 3 h, and then N45E at 80 km/h for 2 h.

a) Determine the magnitude of the car's displacement.

b) Determine the bearing of the car.

### 2 Answers

- la consoleLv 75 months agoBest answer
Recall: s = d/t → where s is the speed, d is the distance, t is the time

First step: the car travels EAST at 100 km/h for 3 h

s₁ = d₁/t₁

d₁ = s₁.t₁ → given that: s₁ = 100 km/h

d₁ = 100.t₁ → given that: t₁ = 3 h

d₁ = 300 km

Second step: the car travels N45E at 80 km/h for 2 h

s₂ = d₂/t₂

d₂ = s₂.t₂ → given that: s₂ = 80 km/h

d₂ = 80.t₂ → given that: t₂ = 2 hours

d₂ = 160 km

By using the Pythagorean's theorem, you can see that:

OC² = OB² + BC²

OC² = [OA + AB]² + BC²

OC² = [d₁ + d₂.cos(45)]² + [d₂.sin(45)]²

OC² = d₁² + 2.d₁.d₂.cos(45) + d₂².cos²(45) + d₂².sin²(45)

OC² = d₁² + 2.d₁.d₂.cos(45) + d₂².[cos²(45) + sin²(45)] → recall: cos²(x) + sin²(x) = 1

OC² = d₁² + 2.d₁.d₂.cos(45) + d₂²

OC² = 300² + (2 * 300 * 160).[(√2)/2] + 160²

OC² = 90000 + 48000√2 + 25600

OC² = 115600 + 48000√2

OC² = 400.(289 + 120√2)

OC = √[400.(289 + 120√2)]

OC = 20√(289 + 120√2)

OC ≈ 428.35 km

Recall:

s = d/t

s = (d₁ + d₂) / (t₁ + t₂)

s = (300 + 160) / (3 + 2)

s = 460/5

s = 92 km/h

- TomVLv 75 months ago
a) Use the law of cosines to calculate the displacement

Distance traveled East = 100*3 = 300 km

Distance traveled NorthEast = 80*2 = 160 km

If the displacement of the car, c, is the base of a triangle, the legs are 160 km and 300 km and the apex angle of the triangle is 90 + 45 = 135°

c² = 300² + 160² - 2(300)(160)cos(135°)

Ans: c = 428.3 km

b) Use the law of sines to calculate the displacement angle relative to E

428.3/sin135° = 160/sinΘ

sinΘ = (160/428.3)sin(135) = 0.26415

Θ = arcsin(0.26415) = 15.3° North of East or 74.7° East of North

Ans: N74.7E