A plane leaves an airport at 3 PM At 4 PM another plane leav

A plane leaves an airport at 3 P.M. At 4 P.M., another plane leaves the same airport traveling in the same direction at a speed that is 150 mph faster than that of the first plane. Four hours after the first plane takes off, the second plane is 300 mi ahead of the first plane. How far did the second plane travel?

Solution

Let x be the speed of the first plane and x+150 is the speed of the second plane

Let d be the distance 2nd plane has traveled when stopwatch reads 3 hours

So, in this time the first plane has covered a distance of d-s-300

Thus, equation for the first plane is d-x-300=3x

i.e. d=4x+300 ....(1)

Equation for the second plane is d=3(x+150)=3x+450 ...(2)

From equations (1) and (2),we get

x=150 and d=900

Hence the required answer is 900mi