Question

Two planes start flying towards each other simultaneously from two different cities. The distance between the cities is 960 miles and one of the planes' speed exceeds the other by 75 mph. If after 1.5 hours of the flight they were still 75 miles apart, what were the speeds of the planes, in mph?

Answer 1

Answer:

257.5 mph

332.5 mph

Step-by-step explanation:

The initial distance between the two planes is 960 miles, while the final distance (after t=1.5 h) is 75 miles, so the total distance covered by the two planes in 1.5h is

$d=960-75=885$ miles

Calling v1 and v2 the velocities of the two planes, we have the following equations:

(1) $v_1 = v_2 + 75$ --> velocity of plane 1 exceeds velocity of plane 2 by 75 mph

(2) $v_1 t+ v_2 t=885$ --> the total distance covered by the two planes is 885 miles (t=1.5 h is the time, and the products v1 t and v2 t represent the distance covered by each plane)

Substituting t=1.5 h, the second equation becomes:

$1.5 v_1+1.5 v_2=885\\v_1 + v_2 = 590$

By substituting (1) into this last equation, we find:

$(v_2+75)+v_2 = 590\\2v_2 + 75 = 590\\2v_2 = 515\\v_2=257.5$

And substituting this back into eq.(1), we find

$v_1 = 257.5 + 75=332.5$

So, the speeds of the two planes are

257.5 mph

332.5 mph