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At noon, ship a is 50 km west of ship b. Ship a is sailing south at 25 km/h, and ship b is sailing north at 5 km/h. How fast (in km/h) is the distance between the ships changing at 4:00 p.m.?

A. 20.0 km/h
B. 22.5 km/h
C. 27.5 km/h
D. 30.0 km/h

1 Answer

2 votes

Final answer:

At 4:00 p.m., the rate at which the distance between ship A and ship B is changing is 30.0 km/h, as determined by summing the individual velocities of both ships since they are moving in perpendicular directions.

option d is the correct

Step-by-step explanation:

The question is asking us how fast the distance between two ships is changing at a specific time. To solve this, we'll use the concept of related rates in calculus, which is a mathematical way to determine the rate at which one quantity changes in relation to another. Since the ships are moving in perpendicular directions to each other, we can set up a right triangle where the horizontal side represents the initial distance between the ships, and the vertical sides represent the distances traveled by ships A and B respectively.

At noon, ship A is 50 km to the west of ship B. Since ship A sails south at 25 km/h and ship B sails north at 5 km/h, after four hours, ship A will have sailed 100 km (25 km/h * 4 h) and ship B will have sailed 20 km (5 km/h * 4 h). To find the rate at which the distance between the two ships is changing at 4:00 p.m., we use the Pythagorean theorem to calculate the distance between them, which is the hypotenuse of the right triangle. This distance at 4:00 p.m. is √((50 km)² + (100 km + 20 km)²) = √(50² + 120²) = √(16900) km = 130 km.

To find the rate of change of this distance, we differentiate with respect to time (t). The derivative of this equation gives us the rates at which the sides of the triangle are changing, which we can then combine to find the rate of change of the hypotenuse.

Let x be the distance ship A has traveled from the starting point, y be the distance ship B has traveled from the same point, and z be the distance between ships A and B. At 4:00 p.m., x=100 km, y=20 km, and the ships are moving directly away from each other, with ship A moving at 25 km/h and ship B at 5 km/h. Therefore, their combined rate of movement away from each other is the sum of these individual rates, which is 25 km/h + 5 km/h = 30.0 km/h. This is the rate at which the distance between the two ships is changing at that time. Hence, the correct answer is option D.

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