Question

A jetliner can fly 4.9 hours on a full load of fuel. Without any wind it flies at a speed of 1.88 x 102 m/s. The plane is to make a round-trip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a 78.2-m/s wind from the jet stream, which blows from west to east. What is the maximum distance (in kilometers) that the plane can travel due west and just be able to return home

Answer 1

1371.26 Km

Step-by-step explanation:

First of all, we need to find the velocity of the plane relative to the ground since the air has a velocity of 78.2 m/s due east and without any wind, it flies at a velocity of 188 m/s.

Thus, during the west trip, the velocity will be;

Vw = Vp - Va

Vp is velocity of plane while Va is velocity of air

and since distance/time = velocity ;

Time = velocity/distance and thus;

Time during this west period ;Tw = X/(Vp - Va)

Now during the east trip,

Ve = Vp + Va

And Te = X/(Vp + Va)

From the question, the plane can fly 4.9 hours on a full load of fuel. Let's convert this to seconds because velocity is in m/s

Thus, 4.9 hours = 4.9 x 60 x 60 = 17640 seconds

So, this time will be equal to the sum of that in the west and east directions.

Thus; T = Tw + Te

From above we know Tw and Te.

Let's substitute them into the equation;

T = [X/(Vp - Va)] + [X/(Vp + Va)]

T = X[(Vp + Va + Vp - Va)/((Vp)² — (Va)²)

T = X[(2Vp)/((Vp)² — (Va)²)

Making X the subject to obtain;

X = [T((Vp)² — (Va)²)]/(2Vp)

X = [17640((188)² — (78.2)²)]/(2 x 188)

X = 515595326.4/376 = 1371264.17m = 1371.26 Km