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.

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.

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?

To find the direction and speed at which the pilot must fly the airliner, we need to consider the effect of the wind on the plane's motion.

Step 1: Break down the wind vector into its North and East components.
- The wind is blowing from the Northwest (45 degrees). This means it has a component blowing due North and a component blowing due West (perpendicular to the direction of travel).
- The North component of the wind can be found using the sine function: North component = 100 km/hr * sin(45 degrees).
- The East component of the wind can be found using the cosine function: East component = 100 km/hr * cos(45 degrees).

Step 2: Calculate the resultant (combined) velocity of the plane.
- Since the airliner wants to reach point B (which is 1000 km due North of point A) in 2.5 hours, the plane's velocity (ignoring the wind) needs to be: Velocity = Distance / Time = 1000 km / 2.5 hr = 400 km/hr (due North).

Step 3: Find the net (combined) velocity of the plane relative to the ground.
- The North component of the plane's velocity is the same as the desired velocity (400 km/hr).
- The East component of the plane's velocity is the East component of the wind subtracted from the desired velocity: East component = 400 km/hr - East component of wind.

Step 4: Calculate the direction and speed of the plane.
- The direction of the plane can be found using the tangent function: Direction = arctan (East component of plane velocity / North component of plane velocity).
- The speed of the plane can be obtained using the Pythagorean theorem: Speed = sqrt((North component of plane velocity)^2 + (East component of plane velocity)^2).

Plugging in the values:
- East component of the wind = 100 km/hr * cos(45 degrees) ≈ 70.7107 km/hr
- East component of plane velocity = 400 km/hr - 70.7107 km/hr ≈ 329.2893 km/hr
- Direction = arctan(329.2893 km/hr / 400 km/hr) ≈ 41.39 degrees

So, the pilot must fly the airliner at a direction of approximately 41.39 degrees (North of East) and at a speed of approximately 329.2893 km/hr to reach point B.