Question

Asked Sep 20, 2019

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Two planes are 1470 mi apart and traveling toward each other. One plane is traveling 80 mph faster than the other plane. The planes pass each other in 1.5 h. Find the speed of each plane.

slower plane | mph |

faster plane | mph |

Step 1

Let *v* be the velocity of the slower plane then the velocity of the faster plane is *v* + 80 and the planes pass each other in 1.5 hours.

That is, the two planes are 1,470 miles apart and one plane relative speed with respective to another is *v* + *v* + 80 = 2*v* +80.

It is known that, the product of time and speed is the value of travelled distance.

Step 2

Substitute *v* = 450 in *v *+ 80 to find the speed the faster plane as follows,

450+80 = 530.

Thus, the speed of the faster plane is 530 miles per hour.

Step 3

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