Question

Description: An airplane is flying from San Francisco to Seattle which is a distance of approximately 1200 km. The cruising speed of the plane (VP ) is 400 km/hr relative to the air, and on a calm day (no wind) the pilot would fly on a path, due north, arriving in Seattle exactly 3.0 hours after leaving San Francisco(ignoring the time spent taxiing, taking off, landing etc.). However, today, the wind is blowing due east with a constant velocity (VW) of 50 km/hr relative to the ground, and the pilot has not taken this into account. Without paying attention, the pilot keeps the plane pointed due north for the entire flight.

1.) (3 points) Relative to the ground, in which direction is the plane flying (use theta for angles: θ)? Derive an algebraic solution and then give a numerical answer (three significant figures is fine) expressed in degrees relative to due north (e.g. ’45.0° west of north’).

2.) (2 pts) After flying for 3.0 hours, how far is the plane from Seattle? Again, derive an algebraic solution, then give the numerical result (3 significant figures). A smarter pilot is flying the next scheduled flight and takes note of the first pilot’s failure to deal with the winds. He takes the wind speed and direction into account so that the path of this plane relative to the ground is due north. This pilot’s plane is identical to the first in every way including the speed, and the wind has not changed in velocity or direction.

3.) (2 pts) How long, in hours, does it take the second pilot to reach Seattle? Again, give the algebraic solution first and then a numerical answer to three significant figures.

4.) (3 pts) Assume the two planes left at the same exact moment. Write an algebraic expression for the distance between two planes as a function of time (∆P(t) = … ). You may only use the variables VP , VW , θ and t as needed.

Hi I just wanted to double check to see if my answers were correct and how you approached the problem. Thanks!

Answer 1

1.

In triangle ABC

tan = BC/AB

tan = V_w/V_p

= tan^-1(V_w/V_p)

inserting the values

= tan^-1(50/400) = 7.125 deg east of north

2.

D = initial distance between San Francisco to Seattle = 1200 km in north direction

After three hours

Y = distance travelled along the north direction = V_p t

D' = distance remaining = D - Y = 1200 - V_p t

x = distance travelled along east direction = V_w t

d = distance from seattle = sqrt(D'² + x²) = sqrt((D - Y)² + (V_w t)²) = sqrt((1200 - V_p t)² + (V_w t)²)

inserting the values

d = sqrt((1200 - (400 x 3))² + (50 x 3)²) = 150 km in east direction

3)

for the plane to travel in north direction

V_p Sin = V_w

= Sin^-1(V_w /V_p) eq-1

Along the north direction

speed = V_p Cos =  V_p Cos(Sin^-1(V_w /V_p) )

time taken = D/V_p Cos = D /(V_p Cos(Sin^-1(V_w /V_p) ))

inserting the values

t = D /(V_p Cos(Sin^-1(V_w /V_p) )) = 1200/(400 Cos(Sin^-1(50/400) )) = 3.02 h