Tough Math Question?

Question

A plane flies north-east with an airspeed of 400 km/hr. If the wind is blowing at a steady speed of 50 km/hr from the west,

a) calculate the distance covered over the ground in one hour

b) calculate the direction in which the plane has travelled.


Thanks again for those who help.

Answer 1

This is a vector addition problem. To make it a little easier, lets rotate the whole thing 45 deg to the south. Then our plane is flying east and the wind is blowing from the north-west.
(a) To find this, we can decompose the wind vector into a +x vector of 0.707x50= 35.5 kph and a -y vector of 35.5 kph. Then to find distance, we can find sqrt (435.5^2+35.5^2) which would be about 437 kpm or so.

(b) The direction would be slightly south of east with angle theta such that
Tan-1 (theta) = 35.5/435.5. (about 6 degrees)
After you find this, the direction will be 45 deg +theta.

Answer 2

First: Draw a diagram with x-y axis, put the line denoting the plane at 45 deg (this is assumed from North east) from the x-axis in the first quadrant-begin from the origin. Then draw the wind going from the west/right along the x-axis into the origin. Now all you have to do is add these two vectors. To do this you have to break the velocities of each into x and y components.

I don't know if you have used vector notation, but it is very useful. i stands for the component in the x direction and j stands for the component in the y direction

So to get the velocity vector of the plane:
the x component is 400*cos(45) and the y component is 400*sin(45). This is using basic trig.

the vector would be:
Vas=282.84i+282.24j (km/hr)

same thing for the wind. except it is traveling along the x axis so:
Vw=50i (km/hr)
Now just add the components. i's add with i's and j's add with j's. So:
Vplane=332.84i+282.84j (km/hr)

Okay so now how do you answer the questions:
First find the magnitude of the Vplane. Do this using the pythagorean theorm.
mag(Vplane)=sqrt(332.84^2+282.84^2) (km/hr)
mag(Vplane)=436.78 (km/hr)
Now to get the distance - multiply it by the time:
Dtrav=436.78 * 1 (km)
Dtrav=436.78 km

To find the direction of travel, draw out the plot again using the i comp of the Vplane as the bottom of a right triangle and the j comp of the Vplane as the side of the right triangle. Then use the inverse tangent to solve for that angle:

theta=arctan(282/332) deg
theta=40.344 deg North of East

or if they want it on a compass rose
90-40.344 = 49 deg East of North

This process can be used on all problems like this. Learn how to use vector notation because it is extremely important in higher level math. This one is a pretty easy problem but I wanted to take the time to spell out the process exactly. Good luck

Answer 3

If you graph this with origin as a starting point, you will find, there are two vectors with a third one that connects both.

You go 50 units from origin to X plus direction
You go 400 units from origin to 45 degrees up from X plus direction.

You will then use the laws of cosine to find the length of the vector that connects both.

c^2 = 50^2 + 400^2 - 2(50)(400)cos(135deg)

C will come out to be about 436 km. That is the answer for the first question.

Knowing this, you can use the laws of sine to find the angle of this vector.

(sin(135deg)/436) = (sin(x)/400)
X = arcsin(400sin(135deg)/436)
x=40.45deg (from x axis)

Plane is traveling at an angle of 40.45 degree from East. If you need to give an answre in DUE NORTH format, you will have to convert it.