Question

A ship leaves port and travels 50 miles at a standard position of 45⁰. The ship

then travels for 90 miles in a standard position angle of 80⁰. At that point the ship

drops anchor. A speedboat travelling 30 miles per hour moves towards the ship in a

straight line. At what time will it get there? ​

Answer 1

t = 4.468 h

Step-by-step explanation:

For this exercise let's start by calculating the distance where the boat is

the first trip is 50 miles at 45 °, then 90 miles at 80 °,

to find the total distance let's find the distance of each displacement

cos 45 = x₁ / 50

sin 45 = y₁ / 50

x₁ = 50 cos 45 = 35.35 miles

y₁ = 50 cos 45 = 35.35 miles

cos 80 = x₂ / 90

sin 80 = y₂ / 90

x₂ = 90 cos 80 = 15.63 miles

y₂ = 90 sin 80 = 88.63 miles

let's find the total displacement in each axis

x_total = x₁ + x₂

x_total = 35.35 + 15.63

x_total = 50.98 miles

y_total = y₁ + y₂

y_total = 35.35 + 88.63

y_total = 123.98 miles

Let's use the Pythagorean theorem to find the modulus of the displacement

R = √ (x_total² + y_total²)

R = √ (50.98² + 123.98²)

R = 134.05 miles

The boat goes at a constant speed,

v = R / t

t = R / v

let's calculate

t = 134.05 / 30

t = 4.468 h