A Super Cat left Batangas port on a course 230°20' for 1.5 hours at 210 mph and then at 140°20' for two hours at 180 mph. How is she far from the port and in what direction must she travel to return to Batangas port?
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To determine the Super Cat's distance from the port and the direction to return, we can use vector addition.
First, let's convert the given courses from degrees, minutes, and seconds to decimal degrees for easier calculations.
Course 1: 230°20' = 230 + 20/60 = 230.3333°
Course 2: 140°20' = 140 + 20/60 = 140.3333°
Now, we can calculate the distance traveled in each leg of the journey:
Distance for Course 1: Speed * Time = 210 mph * 1.5 hours = 315 miles
Distance for Course 2: Speed * Time = 180 mph * 2 hours = 360 miles
To find the total distance from the port, we need to determine the resultant of these two distances. This can be done by treating the distances as vectors and adding them together.
To add the vectors, we first find their components:
Distance 1: 315 miles * cos(230.3333°), 315 miles * sin(230.3333°)
Distance 2: 360 miles * cos(140.3333°), 360 miles * sin(140.3333°)
Now, sum the components:
X-component = Distance1 X-component + Distance2 X-component
Y-component = Distance1 Y-component + Distance2 Y-component
Finally, to calculate the total distance from the port:
Total distance = square root of (X-component^2 + Y-component^2)
To find the direction the Super Cat must travel to return to the port, we can use the inverse tangent function:
Direction = inverse tangent (Y-component / X-component)
By applying these calculations, we can determine both the distance and direction.