Algebra Problem?

 Two trains leave a station at 8:00 am. 

 One train travels north at a rate (speed) of 96 mph 

 and another travels east at a rate (speed) of 180 mph.   

 Assuming that the trains do not stop, 

 the time the trains will be 340 miles apart:

 (96x)^2 + (180x)^2 = 340^2

 x = 5/3 hours

 9 : 40 AM 

Recall: s = d/t → where s is the speed, d is the distance, t is the time

The first train travels north at a rate of 96 mph.

s₁ = d₁/t₁

d₁ = s₁.t₁ → where: s₁ = 96 mph

d₁ = 96.t₁

The second train travels east at a rate of 180 mph.

s₂ = d₂/t₂

d₂ = s₂.t₂ → where: s₂ = 180 mph

d₂ = 180.t₂

At what time will the trains be 340 miles apart?

d₁² + d₂² = 340²

(96.t₁)² + (180.t₂)² = 340² → here: t₁ = t₂ = t (because the train leave a station at the same time)

t².(96² + 180²) = 340²

t² = 340²/(96² + 180²)

t² = (340/204)²

t = 85/51 ← this is the result in hours → to go further

t = (51 + 34)/51 h

t = (51/51) h + (34/51) h

t = 1 h + (34/51) h → you know that: 1 h = 60 min

t = 1 h + [(34/51) * 60 min]

t = 1 h + 40 min

The trains leave a station at 8:00 am.

…and they will be 340 miles apart at 9:40 am

The distance between the trains is the hypotenuse of a right angle. The legs are 96h and 180h where h is the time of travel measured in hours.

d = √(96² + 180²) h = 340 miles

h = 340/204 = 1:40 hours

The time is 1:40 after 8:00 = 9:40 am

if you assume a flat plain (and ignore the curvature of the earth) you have a right angle triangle

you have the hypotenuse 340

you can express the opposite and adjacent as 96x  and 180x (where x is the time in hours)

now use pythagoras' theorem to solve for x

use x (remember it is in hours and fractions of hours.... not hours and minutes) to give the final time