Two planes take off at the same time from an airport. The first plane is flying at 240 miles per hour on a bearing of S 45.0° E. The second plane is flying in the direction S 45.0°W at 290 miles per hour. What is the bearing of the second plane from the first after 5 hours? (Round your answer to one decimal place.)

Question

Two planes take off at the same time from an airport. The first plane is flying at 240 miles per hour on a bearing of S 45.0° E. The second plane is flying in the direction S 45.0°W at 290 miles per hour. What is the bearing of the second plane from the first after 5 hours? (Round your answer to one decimal place.)

Answer 1

Can someone help me?

Answer 2

The first plane ends up at (169.7,-169.7)

The other plane ends up at (-205.1,-205.1)

So, relative to the first, the second plane is at

(-374.8,-35.4)

arctan(-35.4/-374.8) = 185.4°, or W5.4°S

Answer 3

Mr.Steve can you help the Algebra questions that were asked. My class happen to be doing the same thing.

Answer 4

To find the bearing of the second plane from the first after 5 hours, we can use vector addition.

First, let's calculate the displacement vector for each plane after 5 hours.

For the first plane, we know its speed is 240 mph and its bearing is S 45.0° E. To calculate its displacement vector, we can break down the speed into horizontal and vertical components.

The horizontal component (x-direction) can be calculated using the cosine function:
horizontal component = speed * cos(bearing)

So, for the first plane:
horizontal component = 240 * cos(45°) = 240 * 0.7071 = 169.704 miles per hour

The vertical component (y-direction) can be calculated using the sine function:
vertical component = speed * sin(bearing)

So, for the first plane:
vertical component = 240 * sin(45°) = 240 * 0.7071 = 169.704 miles per hour

Therefore, the displacement vector for the first plane after 5 hours is (horizontal: 169.704 miles per hour, vertical: 169.704 miles per hour).

For the second plane, we know its speed is 290 mph and its bearing is S 45.0°W. As before, we can calculate the horizontal and vertical components.

The horizontal component (x-direction) can be calculated using the cosine function:
horizontal component = speed * cos(bearing)

So, for the second plane:
horizontal component = 290 * cos(45°) = 290 * -0.7071 = -205.4845 miles per hour

The vertical component (y-direction) can be calculated using the sine function:
vertical component = speed * sin(bearing)

So, for the second plane:
vertical component = 290 * sin(45°) = 290 * -0.7071 = -205.4845 miles per hour

Therefore, the displacement vector for the second plane after 5 hours is (horizontal: -205.4845 miles per hour, vertical: -205.4845 miles per hour).

Now, to find the bearing of the second plane from the first, we can calculate the angle between the two displacement vectors. We can use the atan2 function in many programming languages or calculators to find the angle.

angle = atan2(vertical2 - vertical1, horizontal2 - horizontal1)

angle = atan2(-205.4845 - 169.704, -205.4845 - 169.704)

angle ≈ atan2(-375.1885, -375.1885)

Using a calculator or a programming language that supports trigonometric functions, we can find the angle to be approximately -45°.

However, since bearing conventionally uses the north direction as the reference (0°), we need to convert the angle to the bearing convention.

To convert the angle to a bearing, we can add 180° to it:

bearing = angle + 180°

bearing = -45° + 180°

bearing = 135°

Therefore, after 5 hours, the bearing of the second plane from the first is approximately S 45.0° E.