Question

A jet, in calm air conditions, travels with velocity vector (389, 389). The wind velocity (in mph) at the plane’s cruising altitude is given by (0, 50). How fast is the plane’s true speed?
A. 587 mph
B. 590 mph
C. 600 mph
D. 601 mph

Answer 1

After thorough researching, if a jet in calm air conditions travels with velocity vector (389, 389) and the wind velocity (in mph) at the plane’s cruising altitude is given by (0, 50), then the plane’s true speed is 600 mph. The correct answer to the following given statement above is with the letter C.

Answer 2

Answer:

The correct option is A.

Step-by-step explanation:

The magnitude of a vector V=<a,b> is

$|V|=\sqrt{a^2+b^2}$

It is given that a jet, in calm air conditions, travels with velocity vector (389, 389). The magnitude of this vector is

$|V_1|=\sqrt{389^2+389^2}=389\sqrt{2}$

Since the x and y coordinates are same, therefore the angle with x-axis is 45 degrees or $\frac{\pi}{2}$.

The wind velocity (in mph) at the plane’s cruising altitude is given by (0, 50).

$|V_2|=\sqrt{0^2+50^2}=50$

The formula for resultant vector of vectors having a and b is

$R=\sqrt{a^2+b^2+2ab\cos \theta}$

$R=\sqrt{(389\sqrt{2})^2+(50)^2+2(389\sqrt{2})(50)\cos (\frac{\pi}{4})}$

$R=586.55$

$R=587$

The true speed of the plane is 287 mph.

Therefor the correct option is A.