Aircraft Speed with Wind: Calculating New Velocity

[pringless]

A jet airliner moving initially at 795 mph to the east where there is no wind moves into a region where the wind is blowing at 140 mph in a direction 56 degrees north of east. What is the new speed of the aircraft with respect to the ground? Answer in units of mph.

how can i start this off

 

[HBar]

The first step would be to draw a vector diagram. Draw the normal cartesian coordinate system, assign the axes a direction (i.e. +y is north, -y is south, +x is east, -x is south). Since the plane relative to the air is traveling east at 795 mph draw a vector starting from the origin and going east (+x). Now draw the wind as a vector (starting from origin with an angle of 56 degrees from the +x axes). Now you have a vector diagram. That is the first step. Good Luck!
-HBar

 

[M98Ranger]

To solve this problem, we can use vector addition to find the new velocity of the aircraft with respect to the ground. First, we can draw a diagram to represent the initial velocity of the aircraft (795 mph to the east) and the wind velocity (140 mph at 56 degrees north of east). From this diagram, we can see that the wind velocity can be broken down into two components: a northward component of 140*cos(56) mph and an eastward component of 140*sin(56) mph.

Next, we can add these components to the initial velocity of the aircraft to find the new velocity. We can do this by using the Pythagorean theorem: c^2 = a^2 + b^2, where c represents the magnitude of the new velocity, a represents the eastward component, and b represents the northward component.

Plugging in the values, we get c^2 = (795 + 140*sin(56))^2 + (140*cos(56))^2. Solving for c, we get c = √((795 + 140*sin(56))^2 + (140*cos(56))^2) = 855.14 mph.

Therefore, the new speed of the aircraft with respect to the ground is approximately 855.14 mph.