A boat pilot wishes to go in a straight line across a river from the east dock to the west dock in a ferryboat that has a still water speed 8.0 knots. If the river has a Southward current of 2.5 knots, what compass heading should be maintaned while crossing the river? What is the speed of the boat relative to the land?

Pls help me.. I cant understand

draw a picture, as always. You will have a right triangle representing the speeds involved.

The hypotenuse will be the boat's actual speed (8) and direction (θ).

The vertical leg will be 2.5

So, the angle θ, measured upriver from the east-to-west direction, will be

sin θ = 2.5/8
θ = 18.2°

So, the heading is W18.2°N or 341.8°

To solve this problem, we need to break it down into two components: the boat's heading and the boat's speed relative to the land.

1. Boat's Heading:
Since the boat is moving in a straight line from the east dock to the west dock, it needs to compensate for the southward current of 2.5 knots. This means the boat should aim slightly north of directly west to counteract the current's effect.

To find the boat's heading, we can use trigonometry. The angle between the boat's heading and the westward direction can be calculated using the tangent function:

tan(theta) = opposite/adjacent
tan(theta) = 2.5/8.0

Using a calculator or the inverse tangent function (atan or tan^(-1)), we can find the value of theta, which represents the angle between the boat's heading and the westward direction.

2. Boat's Speed Relative to the Land:
To find the speed of the boat relative to the land, we can use vector addition. The boat's speed in still water is given as 8.0 knots (this is the boat's speed without the current).

We can think of the current as a vector acting on the boat. The magnitude of this vector is 2.5 knots (southward direction) since the current is flowing southward. We need to add this vector to the boat's speed in still water.

The resulting vector will represent the boat's speed relative to the land. We can find the magnitude of this vector using the Pythagorean theorem:

Speed relative to the land = sqrt((8.0)^2 + (2.5)^2)

This will give us the speed of the boat relative to the land.

Note: The actual value of the heading and speed will depend on specific calculations using the equations provided above.