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Long flights at midlatitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane's speed relative to Earth's surface. If a pilot maintains a certain speed relative to the air (the plane's airspeed), the speed relative to the surface (the plane's ground speed) is more when the flight is in the direction of the jet stream and less when the flight is opposite the jet stream. Suppose a round-trip flight is scheduled between two cities separated by 4300 km, with the outgoing flight in the direction of the jet stream and the return flight opposite it. The airline computer advises an airspeed of 930 km/h, for which the difference in flight times for the outgoing and return flights is 61 min. What jet-stream speed is the computer using

Answer :

Answer:

103.52 km/h

Explanation:

We are given;

Distance between two cities; d = 4300 km

airspeed; v_as = 930 km/h

Difference in flight time; Δt = 61 min = 1.0167 h

Now, the equation of motion to find the distance is given as;

d = vt

Where v = v_as + v_js

v_as is the airspeed

v_js is the jet speed

Thus;

d = (v_as + v_js)t

Thus, time(t1) for outgoing flight is;

t1 = d/(v_as + v_js)

Meanwhile, time(t2) for the return flight, the jet stream velocity will be negative and time is;

t2 = d/(v_as - v_js)

Recall that Difference in flight time; Δt = 61 min.

Thus;

Δt = t2 - T1 = [d/(v_as - v_js)] - [d/(v_as + v_js)]

Factorizing out, we have;

Δt = d[1/(v_as - v_js)] - [1/(v_as + v_js)]

Furthermore, it gives;

Δt = d[(v_as + v_js - v_as + v_js)]/((v_as - v_js) × (v_as + v_js))

Δt = d(2v_js)/((v_as)² - (v_js)²)

Cross multiply to get;

(2dv_js)/Δt = ((v_as)² - (v_js)²)

(v_js)² + ((2dv_js)/Δt) - (v_as)² = 0

Plugging in values for d,v_as and Δt gives;

(v_js)² - ((2 × 4300 × v_js)/1.0167) - (930)² = 0

(v_js)² - (8458.7329v_js) - 864900 = 0

Using quadratic formula, we have;

v_js = 103.52 km/h

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