A long distance swimmer is able to swim through still water at 4km/hr. She wishes to swim from Port Angeles, WA due north to Victoria, B.C., a distance of 50km. An ocean current flows through the Strait of Juan de a from west to east at 3km/hr. In what direction should she swim to make the crossing along a straight line between the two cities?
49 degrees west of north
To determine the direction in which the long distance swimmer should swim, you need to consider the velocity of the swimmer and the ocean current.
Let's break down the given information:
- The swimmer's velocity in still water is 4 km/hr, which means she can swim at this speed without the influence of any external factors like tides or currents.
- The ocean current flows from west to east at a velocity of 3 km/hr.
To find the optimal direction for the swimmer, we need to account for the combined effect of her swimming velocity and the ocean current.
Since the ocean current is flowing from west to east, the swimmer should aim to swim slightly west of the straight line between Port Angeles and Victoria.
To calculate the exact angle, we can use the concept of vector addition:
1. Define the velocity of the swimmer as a vector pointing north with a magnitude of 4 km/hr (her swimming speed) and a direction represented by angle θ.
2. Define the velocity of the ocean current as a vector pointing east with a magnitude of 3 km/hr and a direction represented by an angle of 90 degrees (due east).
Now, the resulting velocity (or resultant vector) will be the vector sum of the swimmer's velocity and the current's velocity.
To calculate the magnitude of the resultant velocity:
- Use the Pythagorean theorem: the magnitude of the resultant is the square root of the sum of the squares of the two velocities.
To calculate the direction of the resultant velocity:
- Use the trigonometric inverse tangent function (arctan) to find the angle between the resultant vector and the north direction.
For the magnitudes:
- Swimmer's velocity (Vswimmer) = 4 km/hr
- Ocean current's velocity (Vcurrent) = 3 km/hr
Using the Pythagorean theorem:
Resultant velocity (Vresultant) = sqrt((Vswimmer)^2 + (Vcurrent)^2)
= sqrt((4 km/hr)^2 + (3 km/hr)^2)
= sqrt(16 km^2/hr^2 + 9 km^2/hr^2)
= sqrt(25 km^2/hr^2)
= 5 km/hr
For the direction:
Angle (θ) = arctan((Vcurrent)/(Vswimmer))
= arctan((3 km/hr)/(4 km/hr))
≈ 36.87 degrees
Therefore, the long distance swimmer should aim to swim at an angle of approximately 36.87 degrees west of north to make the crossing in a straight line between Port Angeles, WA, and Victoria, B.C.