physics homework

Question

an aircraft has a cruising speed of 100 m/s. On this particular day or wind is blowing towards the east at 75.0 m/s. If the plane pointed do north what would be the magnitude and direction of the velocity relative to the ground?

Answer 1

We have two perpendicular forces at work here: the planes thrust moving it at 100 m/s NORTH and the wind moving the plane 75 m/s EAST.  We can use these to create a right triangle.  The magnitude of the planes velocity will the length of the hypotenuse, so we use the Pythagorean theorum:

a²+b²=c²

100²+75² = c²

10000 + 5625 = c²

c²= 15625

c = 125 m/s

 

Next to find the direction:

 

We want to find the elevation above due east (to the right on your paper), so we use the same right triangle, with a legs of 75 and 100:

 

           /!

         /   !

       /     !

    /        ! 100

  /          !

/ x         !

----------!

    75

 

 

The angle we need to measure is marked by the x.  We can use our tangent ratio to figure out the angle

Tan = Opposite/Adjacent

 

Tan (x) = 100/75

Tan (x) = 4/3

Using a calculator you can use the inverse tan button to determine x is approx = 53.1 degrees

 

SO

the velocity is 125m/s at an and of 53.13 degrees above due East.