Question

A jet airliner moving initially at 3.50 ✕ 102 mi/h due east enters a region where the wind is blowing at 1.00 ✕ 102 mi/h in a direction 35.0° north of east. (Let the x-direction be eastward and the y-direction be northward.)

A. Find the components of the velocity of the jet airliner relative to the air, vJA.

vJA,x = _____mi/h

vJA,y = _____mi/h

B. Find the components of the velocity of the air relative to Earth, vAE.

vAE,x = ____mi/h

vAE,y = ____mi/h

C. Write an equation analogous to vAB = vAE − vBE for the velocities vJA, vAE, and vJE . (In the given equation, vBE is the velocity of observer B relative to observer E. Object A travels with velocity vAB relative to observer B, and velocity vAE relative to observer E.)

(option a) vJA = vJE + vAE

(option B) vJA = vAE − vJE

(option C) vJA = vAE + 2vJE

(option d) vJA = vJE − vAE

D. What are the speed and direction of the aircraft relative to the ground?

magnitude _____mi/h

direction _______° north of east

Answer 1

(A) The components of the velocity of jet airliner relative to an air which is given as :

On x-axis, we have

v_JA,x = 350 mph

On y-axis, we have

v_JA,y = 0 mph

(b) The components of the velocity of an air relative to Earth which is given as :

On x-axis, we have

v_AE,x = v_JA,x + v_JE,x

v_AE,x = (350 mph) + (100 mph) cos 35⁰

v_AE,x = [(350 mph) + (81.9 mph)]

v_AE,x = 431.9 mph

On y-axis, we have

v_AE,y = v_JA,y + v_JE,y

v_AE,y = (0 mph) + (100 mph) sin 35⁰

v_AE,y = [(0 mph) + (57.3 mph)]

v_AE,y = 57.3 mph

(c) Write an equation analogous to v_AB = v_AE − v_BE for the velocities v_JA, v_AE, and v_JE .

where, v_BE = velocity of observer B relative to observer E

v_AB = object A travels with velocity relative to observer B

v_AE = object A travels with velocity relative to observer E

(option B) : v_JA = v_AE − v_JE

(d) The speed and direction of an aircraft relative to the ground which will be given as :

We know that, v_AE = v_AB + v_BE

v_AE = _$\sqrt{}$v_AE,x² + v_AE,y²

v_AE = _$\sqrt{}$(431.9 mph)²$\hat{}$+ (57.3 mph)²

v_AE = 435.6 mph

Using a trigonometric identity, we have

$\theta$ = tan^-1 [(57.3 mph) / (431.9 mph)]

$\theta$ = 7.55 degree

{ north of east }