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A step-by-step solution to the distance-rate-time problem, where two airplanes travel in opposite directions with a given difference in speed and distance apart after a certain time. The document uses the distance-rate-time formula and algebraic substitution to find the speeds of both planes.
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Distance-Rate-Time Problem LSC-O 6/2010, Rev. 1/
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Two airplanes depart from an airport simultaneously, one flying 100 km/hr faster than the other. These planes travel in opposite directions, and after 1. hours they are 1275 km apart. Determine the speed of each plane.
relationships that the problem gives you: Distance apart is 1275 km, time traveled is 1. hrs, and one plane is traveling 100 m/hr. faster than the other. Let plane X be the faster plane.
Distance traveled = Rate (or Speed ) times Time.
1275 km is the total of the distances (added together) that each plan travels. Travel time for each plane is the same, 1. hours; however, the planes’ speeds differ by 100 km/hr.
Plane X is traveling X m/hr.
Plane Y Is traveling Y m/hr.
1275 km in 1.5 hours
Speed of Plane Y is 100 m/hr slower than X.
Speed of Plane X is 100 m/hr faster than Y.
Distance-Rate-Time Problem LSC-O 6/2010, Rev. 1/
2 of 2
1275 km = plane X’s distance plus plane Y’s distance: DTotal (or 1275) = Dx + D (^) y Plane X’s distance is its speed x times 1.5, and plane Y’s distance is its speed y times 1.5: 1275 km = (X)(1.5) + (Y)(1.5)
The difference in the planes’ speeds can be expressed as: X – Y = 100
From above, the first equation is: 1275 = 1.5X +1.5Y
The second Equation is: X – Y = 100 On this equation, solve for X by adding Y to both sides: X – Y + Y = 100 + Y X = 100 + Y
Substitute back into the first equation: 1275 = 1.5(100 + Y) + 1.5Y 1275 = 150 + 1.5Y + 1.5Y 1275 = 150 + 3Y 1275 – 150 = 150 – 150 + 3Y 1125 = 3Y
3
1125 = 3
3 Y
And then, back into the second equation: X = 100 + Y X = 100 + 375 X = 475 Answer: The faster plane (plane X) is flying 475 km/hr, and the slower plane (plane Y) is flying 375 km/hr.