An airliner is at point A and is trying to fly to point B, which is 1000 km due North of point A. The pilot of the airliner wishes to reach point B in 2.5 hr. A wind is blowing from the Northwest (45 degrees) at 100 km/hr. In what direction and at what speed must the pilot fly the airliner? Show your work.

Question

An airliner is at point A and is trying to fly to point B, which is 1000 km due North of point A. The pilot of the airliner wishes to reach point B in 2.5 hr. A wind is blowing from the Northwest (45 degrees) at 100 km/hr. In what direction and at what speed must the pilot fly the airliner? Show your work.

Physics - bobpursley, Wednesday, October 26, 2011 at 11:05am

Resultant=Sum of vectors
1000N=wind+headingspeed
1000N=(100*.707S+100*.707E)2.5+ (XXXXN+ XXXXE)2.5

solve for the N and E components.

I willdo N for you.

1000N=-70.7N*2.5 + XXXXN*2.5 so the north component has to be 471km/hr N

Now do the E component , (it might be negative, or W), then add the vectors to get flight speed, and direction.

Physics - Eliyahna, Wednesday, October 26, 2011 at 11:16am

I do not understand...this is part of my lab questions for a force table lab, and we have not done Sum of Vectors in class at all... Why would you use 2.5hr in the multiplications? Would you not use the 1000km/2.5hr for the speed in some way? Also...I don't understand what you are doing in the second half of the equation... you are adding a NE heading to the SW heading of the wind to correct to go north?

Answer 1

easy anwer. You are looking for in the end speed, but to get there, you start with distance on each side. So XXXXN2.5 is the distance (2.5 is the time, distance=speedtime)

It could have been done your way..

1000N/2.5 (speed)= speed wind+ speed plane

My way was..
1000N (distance)=windspeed*2.5 + speedplane*2.5

There is by the Gift of God, many ways to work these.

Now headings. SW= S + W = S+ (-E)

again,you could have done all of them in N,E, or N,W, orS,E, or S,W allwould have yielded the same direction.

Answer 2

What exactly is your question?

Try the E component as I suggested, post your work, I will critique it this afternoon. I need to see what you are messing up on.

Answer 3

Ummm...I think I'm just going to give up here..you're way over my head. My teacher needs to do a better job explaining this, if he wants to assign it for homework.

Answer 4

He wants a degree for the direction and a speed that would correct for the wind in the right and allow the plane to get to the point at due north in 2.5 hours, the way you are doing things is so completely foreign to anything I've seen in class, which basically has used the Pythagorean theorem, and a force table to explain vector sum resolution, I just don't understand how to work this out...

Answer 5

I understand your confusion, let me explain further. In order to solve this problem, we need to use vector addition.

First, let's break down the problem into components. We have a wind blowing from the Northwest (45 degrees) at a speed of 100 km/hr. This wind can be broken down into two components: one in the north direction and one in the east direction. Since the wind comes from the Northwest, it is at a 45-degree angle. Using trigonometry, we can calculate the values of these components.

N Component of Wind = 100 km/hr * sin(45°) = 100 km/hr * 0.707 = 70.7 km/hr
E Component of Wind = 100 km/hr * cos(45°) = 100 km/hr * 0.707 = 70.7 km/hr

Now, let's consider the speed and direction the pilot needs to fly the airliner. We know that the airliner needs to reach point B, which is 1000 km due North of point A, in 2.5 hours. Therefore, the speed required to cover this distance is 1000 km / 2.5 hr = 400 km/hr.

Let's break down the desired speed of 400 km/hr into components as well. Since the airliner needs to travel due North, all of its desired speed will be in the North direction. Therefore:

N Component of Desired Speed = 400 km/hr
E Component of Desired Speed = 0 km/hr (since the airliner is not flying in the east direction)

Now, let's find out how the pilot needs to fly the airliner to counteract the wind and reach point B.

Resultant N Component = Desired N Component - Wind N Component
= (400 km/hr - 70.7 km/hr) = 329.3 km/hr N

Resultant E Component remains 0 km/hr since the airliner is not flying in the east direction.

Finally, we can find the magnitude and direction of the resultant vector by using the Pythagorean theorem and trigonometry.

Magnitude of Resultant = sqrt((Resultant N Component)² + (Resultant E Component)²)
= sqrt((329.3 km/hr)² + (0 km/hr)²)
= 329.3 km/hr

Direction of Resultant = arctan(Resultant E Component / Resultant N Component)
= arctan(0 km/hr / 329.3 km/hr)
= 0°

Therefore, the pilot must fly the airliner at a speed of 329.3 km/hr due North in order to counteract the wind and reach point B.

easy anwer. You are looking for in the end speed, but to get there, you start with distance on each side. So XXXXN*2.5 is the distance (2.5 is the time, distance=speed*time)

What exactly is your question?

Ummm...I think I'm just going to give up here..you're way over my head. My teacher needs to do a better job explaining this, if he wants to assign it for homework.

I understand your confusion, let me explain further. In order to solve this problem, we need to use vector addition.

easy anwer. You are looking for in the end speed, but to get there, you start with distance on each side. So XXXXN2.5 is the distance (2.5 is the time, distance=speedtime)