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Two passengers leave the airport at Kansas City, Missouri. One flies to Los Angeles, California, in 3.5 hr and the other flies in the opposite direction to New York City in 3.5 hr. With prevailing westerly winds, the speed of the plane to New York City is 56mph faster than the speed of the plane to Los Angeles. If the total distance traveled by both planes is 2464mi, determine the average speed of each plane.

Respuesta :

Answer:

324mph los Angeles, 380mph New York

Explanation:

V1 = speed to California

V2 = speed to New York

V2 = V1 + 56

Speed = distance in mi / time

V1 = d1/ 3.5hr

d1 = distance for the first plane = 3.5V1

V2 = d2 (distance for the second plane)/ 3.5hr

d2 = 3.5V2 = 3.5(V1 + 56)

But d1+d2 = 2464 mi = 3.5V1+3.5(V1 +56)

2464 = 3.5V1 + 3.5V1 + 3.5*56

2464 = 7V1 + 196

2464-196 = 7V1

2268 = 7V1

V1 = 2268/7 = 324mph

V2 = V1 + 56 = 324+56 = 380mph

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