A boat needs to travel south at a speed of 20 km/h. However a constant current of 6 km/h is
flowing from the south-east. Use vectors to find:
a.the equivalent speed in still water for the boat to achieve the actual speed of 20 km/h.
b the direction in which the boat must head to compensate for the current.
we need <x,y> such that
<x,y> + <6/√2,6/√2> = <0,-20>
Now it's easy to get x and y, and thence the speed and direction.
Answer
To solve this problem using vectors, we can consider the boat's velocity relative to the water and the current's velocity as separate vectors.
a. The equivalent speed in still water for the boat to achieve the actual speed of 20 km/h can be found by adding the opposite vector of the current's velocity to the boat's velocity relative to the water.
Let's denote the boat's velocity relative to the water as vector Vb and the current's velocity as vector Vc.
Vb = 20 km/h (south)
Vc = 6 km/h (south-east)
To find the equivalent speed in still water, we need to subtract the current's velocity vector from the boat's velocity vector.
Vbw = Vb - Vc
To subtract the vectors, we need to resolve the vector Vc into its components in the south and east directions.
The angle between the south-east direction and the south direction is 45 degrees (halfway between east and south). This means that the east component of Vc is Vc * cos(45°) and the south component is Vc * sin(45°).
Vc_east = 6 km/h * cos(45°)
Vc_south = 6 km/h * sin(45°)
Now we can subtract the components from Vb:
Vbw_south = Vb_south - Vc_south
Vbw_east = -Vc_east (since the current is flowing from the south-east)
Vbw_south = 20 km/h - 6 km/h * sin(45°)
Vbw_east = -6 km/h * cos(45°)
The equivalent speed in still water is the magnitude of the resultant vector Vbw:
|Vbw| = √(Vbw_south^2 + Vbw_east^2)
b. The direction in which the boat must head to compensate for the current can be found by determining the angle between the resultant vector Vbw and the south direction.
θ = arctan(Vbw_east / Vbw_south)
So, to summarize:
a. The equivalent speed in still water for the boat to achieve the actual speed of 20 km/h is |Vbw| km/h.
b. The direction in which the boat must head to compensate for the current is θ degrees east of the south direction.
To solve this problem using vectors, we need to break down the boat's velocity and the current into their respective components.
a. To find the equivalent speed in still water for the boat to achieve the actual speed of 20 km/h, we can assume that the boat is traveling directly south. Let's call this the boat's "undisturbed velocity."
The undisturbed velocity of the boat is 20 km/h to the south.
b. Now let's consider the current. The current is flowing from the south-east, which means it has components in both the south and east directions. We can represent the current's velocity using a vector.
The current's velocity has a magnitude of 6 km/h and is flowing from the south-east direction. To find the components of the current's velocity, we need to consider the right-angled triangle formed by the south, east, and south-east directions.
Given that the current is flowing from the south direction (albeit at an angle), we can determine that the south component of the current's velocity is 6 km/h. By using the right-angled triangle, we can calculate the east component of the current's velocity.
If we assume the east component of the current's velocity is x km/h, we can use Pythagoras' theorem to calculate it:
(6 km/h)^2 = 6 km/h * x km/h + x km/h * x km/h.
Simplifying this equation, we get:
36 km/h = 6x + x^2.
Rearranging the equation, we have:
x^2 + 6x - 36 = 0.
To solve for x, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a.
By substituting the coefficients into the quadratic formula, we find the two possible values for x, in which one will be positive and the other negative. Considering that the current is flowing from south-east, we can discard the negative solution and only consider the positive solution.
After solving the equation using the quadratic formula, we find that x ≈ 3.6 km/h.
Thus, the south-east current has a south component of 6 km/h and an east component of 3.6 km/h.
c. Now, we need to find the boat's actual velocity, taking into account the current. To do this, we can add the vectors representing the boat's undisturbed velocity and the current's velocity.
Adding the south components together, we have:
Boat's south component: 20 km/h + (-6 km/h) = 14 km/h (since the current is flowing in the opposite direction).
Adding the east components together, we have:
Boat's east component: 0 km/h + 3.6 km/h = 3.6 km/h.
Therefore, the boat's actual velocity in the presence of the current is approximately 14 km/h to the south and 3.6 km/h to the east.
d. To find the equivalent speed in still water, we want to determine the magnitude and direction of the boat's velocity in still water.
Using the Pythagorean theorem, we can find the magnitude of the boat's velocity:
Magnitude = sqrt((14 km/h)^2 + (3.6 km/h)^2) ≈ 14.8 km/h.
To find the direction, we can use the inverse tangent function:
Direction = arctan(3.6 km/h / 14 km/h) ≈ 14.5°.
Therefore, the equivalent speed in still water for the boat to achieve the actual speed of 20 km/h is approximately 14.8 km/h, and the direction in which the boat must head to compensate for the current is approximately 14.5° east of south.