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 determine the speed and bearing the destroyer must maintain to reach the aircraft carrier in exactly 3 hours, we need to break down the problem and use some trigonometry.

Let's first calculate the distance between the destroyer and the aircraft carrier using the given information. The destroyer is 68 miles directly north of the aircraft carrier. We also know that the aircraft carrier is traveling west 21 degrees north at 18 mph.

Using trigonometry, we can find the westward component of the aircraft carrier's velocity:

westward velocity = 18 mph * cos(21 degrees)

The westward velocity is therefore:

westward velocity = 18 mph * cos(21 degrees) ≈ 16.59 mph

Since we know the aircraft carrier is moving westward and the destroyer is stationary in that direction, the velocity component of the destroyer relevant to this situation is entirely northward.

Now, to calculate the distance the destroyer needs to cover, we can use the formula:

distance = velocity * time

distance = 16.59 mph * 3 hours ≈ 49.77 miles

Therefore, the initial distance between the destroyer and the aircraft carrier is approximately 49.77 miles.

To reach the aircraft carrier, the destroyer needs to cover this distance in 3 hours.

So, the speed the destroyer must maintain is:

speed = distance / time = 49.77 miles / 3 hours ≈ 16.59 mph

Comparing this result to the given answer of 226 mph, we can see that it is incorrect.

Now, let's calculate the bearing the destroyer must maintain. Since the destroyer needs to travel directly north, the bearing is 0 degrees (or 360 degrees).

In conclusion, the destroyer must maintain a speed of approximately 16.59 mph and a bearing of 0 degrees to reach the aircraft carrier in exactly 3 hours.