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Question

Two passengers leave the airport at Kansas City, Missouri. One flies to Los Angeles, California, in 3.4 hr and the other flies in the opposite direction to New York City in 2.4 hr. With prevailing westerly winds, the speed of the plane to New York City is 60 mph faster than the speed of the plane to Los Angeles. If the total distance traveled by both planes is 2464 mi, determine the average speed of each plane.

Answer 1

(Distance = (Rate)(Time)

 

Let x = speed of the LA bound plane

 

Then, x+60 = speed of NYC bound plane

 

Sum of distances = 2464

 

3.4x + 2.4(x+60) = 2464

 

          5.8x + 144 = 2464

 

                   5.8x = 2320

 

                       x = 400 MPH    x+60 = 460 MPH

 

 

 

Answer 2

Let C = speed of CA plane and N = speed of NY plane

3.4C + 2.4N = 2464 (rate * time = distance)

Multiply by 10 and divide by 2: 17C + 12N = 12320

 

N = C + 60

Substitute into the equation above

17C + 12(C + 60) = 12320

29C + 720 = 12320

Subtract 720 from each side: 29C = 1160

Divide by 29

C = 400 and then N = 460

You should take these values and plug into original problem to be sure they satisfy the conditions.