Question

two planes leave an airport at the same time, one flying east, the other flying west. the eastbound plane travels 50mph slower. they are 2550 mi apart after 3 hr. find the speed of each plane

Answer 1

So the westbound planes distance from the start can be described by the following where v is the unknown speed, t is time and d is distance
d_west=v*t
for for the eastbound plane the speed is v-50, so its distance function will look like the following
d_east=(v-50)*t
so the total distance between them is the sum of the two
d_west+d_west=v*t+(v-50)*t
now we can plug in that we know the total distance and time
v*3+(v-50)*3=2550
now we can solve this equation
v=450
so the westbound plane has a speed of 450 and the eastbound a speed 50 mph lower that is 400

Answer 2

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