Question

A jet airliner moving initially at 548 mph

Answer 1

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

$v_1 =548 mph$ along the x-direction

$v_2 = 343 mph$ in a direction $67^{\circ}$ north of east.

To find the resultant, we must resolve both vectors on the x- and y- axis:

$v_{1x}= 548 mph$

$v_{1y}=0$

$v_{2x} = (343 mph)( cos 67^{\circ})=134.0 mph$

$v_{2y} = (343 mph)( sin 67^{\circ})=315.7 mph$

So, the components of the resultant velocity in the two directions are

$v_{x}=548 mph+134 mph=682 mph$

$v_{y}=0 mph+315.7 mph=315.7 mph$

So the new speed of the aircraft is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(682 mph)^2+(315.7 mph)^2}=751.5 mph$