A motorboat is headed [W35ºN] with its engine set to move the boat at 25 km/h in still water. The current is flowing at 18 km/h [S]. What is the velocity of the boat relative to the shore?
My work:
cos(35º)=x/25
25cos35=x
x=20km/h
I have an issue with finding the direction. It is supposed to be [W11ºS]
If you draw the diagram, since 25 sin35º = 14, then the angle θ south of W is
tanθ = (18-14)/20 = 4/20
θ = 11.3º
boat components
... W ... 25 cos(35º) = 20
... N ... 25 sin(35º) = 14
add the boat components to the current
(14 N + 18 S) + 20 W = 20
... the tan of the angle is ... S/W
... arctan(4/20) = 11º ... [W11ºS]
To find the velocity of the boat relative to the shore, you need to consider both the speed and direction. Let's break down the steps to solve this problem.
1. Determine the speed of the boat in still water:
The speed of the boat in still water is given as 25 km/h. This means that if there was no current, the boat would be moving at a speed of 25 km/h.
2. Determine the speed and direction of the current:
The current is flowing at a speed of 18 km/h to the south ([S]).
3. Combine the speed and direction of the boat with the speed and direction of the current:
To find the velocity of the boat relative to the shore, you need to add the boat's velocity vector (speed and direction) with the current's velocity vector.
Let's calculate the speed component first:
The component of the boat's velocity in the northern direction is given by sin(35º) * 25 km/h.
sin(35º) ≈ 0.5736
So, the speed component in the northern direction is 0.5736 * 25 km/h ≈ 14.34 km/h.
Now let's calculate the speed component of the current in the opposite (northern) direction:
Since the current is flowing to the south, the speed component in the northern direction will be negative.
The component of the current's velocity in the northern direction is given by -sin(180º) * 18 km/h = -18 km/h.
Adding the boat's speed component and the current's speed component, we get:
14.34 km/h - 18 km/h = -3.66 km/h
This means that the boat's velocity relative to the shore has a speed of 3.66 km/h in the opposite direction of the current (north).
Now let's determine the direction:
To find the direction, we need to break down the boat's velocity vector and determine the angle it makes with the positive x-axis.
The angle of the boat's velocity vector with the positive x-axis is given by:
arctan(cos(35º) / sin(35º))
Plugging in the values:
arctan(cos(35º) / sin(35º)) ≈ arctan(0.8192 / 0.5736) ≈ arctan(1.4294) ≈ 56.1º
Since the boat is moving in the northwest direction (which is between west and north), this angle needs to be added to 270º to get the angle relative to the positive x-axis:
270º + 56.1º ≈ 326.1º
To convert this angle to compass direction, subtract it from 360º:
360º - 326.1º ≈ 33.9º
Therefore, the direction of the boat relative to the shore is approximately [W33.9ºN].