A navy destroyer is exactly 68 miles North of an aircraft carrier, which is travelling West 21 degrees North at 18 mph. what speed and bearing must the destroyer maintain if it is to reach the aircraft carrier in exactly 3 hrs?
is it 226mph?
To find the speed and bearing the destroyer must maintain to reach the aircraft carrier in exactly 3 hours, we can break this down into two components: the horizontal (westward) component and the vertical (northward) component.
First, let's calculate the horizontal component. The aircraft carrier is traveling west at a bearing of 21 degrees north. This means the vertical component of its speed is given by 18 mph * sin(21 degrees).
To find the horizontal component, we can use the cosine function: horizontal component = 18 mph * cos(21 degrees).
Next, we need to consider the vertical component. The destroyer needs to cover a distance of 68 miles north in 3 hours. Therefore, the vertical component of the destroyer's speed is given by: vertical component = 68 miles / 3 hours.
Now, we can use the Pythagorean theorem to find the total speed the destroyer needs to maintain. The total speed is the hypotenuse of the right triangle formed by the horizontal and vertical components:
total speed = sqrt((horizontal component)^2 + (vertical component)^2).
Finally, we can find the bearing the destroyer needs to maintain by using the arctan function, which is the inverse tangent function: bearing = atan(vertical component / horizontal component).
Now, let's calculate the speed and bearing:
horizontal component = 18 mph * cos(21 degrees) ≈ 16.68 mph
vertical component = 68 miles / 3 hours ≈ 22.67 mph
total speed ≈ sqrt((16.68 mph)^2 + (22.67 mph)^2) ≈ 28.17 mph
bearing ≈ atan(22.67 mph / 16.68 mph) ≈ 52.25 degrees.
Therefore, the destroyer must maintain a speed of approximately 28.17 mph and a bearing of 52.25 degrees to reach the aircraft carrier in exactly 3 hours.