Two planes just took off from Salt Lake City, UT. The first plane is traveling 33 times as fast as the second plane. After traveling in the same direction for 33 hours, they are 630630 miles apart. What is the average speed of each plane? (Hint: Since they are traveling in the same direction, the distance between them will be the difference of their distances.)

Answer: Hello!

Here we have two planes with V1 and V2 their respective velocities, and both planes are traveling in the same direction.

We know that the first plane is traveling 33 faster as the second one, this means that V1 = 33V2

and we know that after 33 hours, the distance between the planes is of 630630 miles, knowing that velocity multiplicated by time is distance; this means that V1*33h - V2*33h = 630630 mi

so now we have two equations:

1) V1 = 33V2

2) (V1 - V2) * 33h = 630630mi.

then, using the first equation we can replace V1 in the second equation and obtain the value for V2, this is:

(V1 - V2) * 33h = 630630mi

(33V2 - V1) * 33h = 630630mi

32V2*33h = 630630mi

V2 = 630630mi / (32*33h) = 597.2 mi/h

and V1 = 33V2 = 33*597.2mi/h = 19707.2mi/h

Now the numbers are kinda odd, because you may writen wrong the numbers (V1 3 times faster than V2, 3 hours instead of 33, and 630mi instead of 630630mi)

doing these replacements our equations are:

1) V1 = 3V2

2) (V1 - V2) * 3h = 630mi.

and doing the same steps as before:

V2 = 630mi / (2*3h) = 105 mi/h

V1 = 3V2 = 315mi/h