On the outbound leg of a 450 mile trip, the plane flew at a slow speed. On the trip back (also 450 miles), the pilot tripled the speed of the plane. How fast did the plane travel on the trip back if the total traveling time was 8 hours?

Answer 1

Speed = distance / time

Out bound

s = 450 / t

In bounds

3s = 450 / t

Algebraically rearranging

t = 450 /s

t = 450 /3s

Adding

2t = 8hrs = 450/s + 450/3s

2t = (3s(450) + 450s) / 3s^2

2t = (1350s + 450s) 3s^2

2t(3s^2) = 1800s

But 2t = 8

Hence

24s^2 = 1800s

Cancel down by '24'

=>

s^2 = 75s Cancel down by 's'

s = 75

So 3s = 3(75) = 225 mph is the speed on the in bound (return journey).