To find the acceleration of the boat, we need to consider the net force acting on it. The net force is the vector sum of the accelerating force of the wind and the drag force of the keel.
Given:
Wind force = 707 N north
Drag force of keel = 500 N west
To find the net force, we need to resolve both forces into their x and y components.
Wind force:
The wind force is acting north, so its y-component is 707 N * sin(90°) = 707 N.
The x-component of the wind force is 0 N since it is acting purely in the vertical direction.
Drag force:
The drag force is acting west, so its x-component is -500 N.
The y-component of the drag force is 0 N since it is acting purely in the horizontal direction.
Now, let's find the net force in the x and y directions:
Net force in the x direction:
Since the wind force and the drag force are the only forces acting in the x direction, the net force in the x direction is the sum of their x-components.
Net force in the x direction = 0 N + (-500 N) = -500 N.
Net force in the y direction:
Since the wind force and the drag force are acting in opposite directions in the y direction, we subtract their y-components to find the net force.
Net force in the y direction = 707 N - 0 N = 707 N.
Now, we can calculate the resultant force:
Resultant force = √((Net force in the x direction)^2 + (Net force in the y direction)^2)
Resultant force = √((-500 N)^2 + (707 N)^2)
Resultant force = √(250000 N^2 + 499849 N^2)
Resultant force ≈ √749849 N^2 ≈ 866 N
The acceleration is given by Newton's second law, F = ma. Rearranging the equation, we have a = F/m.
Acceleration = Resultant force / Mass of the boat = 866 N / 200 kg ≈ 4.33 m/s^2
The direction of acceleration can be determined using the angles provided in the options. Comparing this with the given options:
A) 1.5 m/s^2, 35 degrees north of east
B) 2.5 m/s^2, 55 degrees north of west
C) 3.0 m/s^2, 35 degrees north of east
D) 4.3 m/s^2, 55 degrees north of west
E) 1.5 m/s^2, 55 degrees north of west
The correct answer is D) 4.3 m/s^2, 55 degrees north of west.