How fast are the trains traveling away from one another at 3:00 PM?

Question

Two trains leave a station at 12:00 noon. One travels north on a track at 50 mph. The second travels east on a track at 70 miles per hour. At 1:00 PM the northbound train stops for one-half hour at a station while the eastbound train continues at 70 miles per hour without stopping. At 1:30 PM the northbound train continues north at 50 mph. How fast are the trains traveling away from one another at 3:00 PM?

 

The distance between the trains is increasing at     ?     miles per hour.

Answer 1

Remember that distance = rate × time.

 

The distance the northbound train has traveled at a given time t is d_n = (50)(t - 0.5) = 50t - 25.

The distance the eastbound train has traveled at a given time t is d_e = 70t.

 

The distance d between them can be found using the Pythagorean Theorem, where the triangle's legs are the distance each train has traveled:

 

d² = (50t - 25)² + (70t)²

d² = 2500t² - 2500t + 625 + 4900t²

d² = 7400t² - 2500t + 625

d = √(7400t² - 2500t + 625)

 

 

 

 

The derivative of that expression (which you can find using the Chain Rule) will tell you the rate at which the distance is changing at a given time t:

 

d' = [ 1/2(7400t² - 2500t + 625)^-1/2 ][ 14800t - 2500 ]

d' = [ 1 / 2√(7400t² - 2500t + 625) ][ 100(148t - 25) ]

d' = [ 1 / 10√(296t² - 100t + 25) ][ 100(148t - 25) ]

d' = (100(148t - 25)) / (10√(296t² - 100t + 25))

d' = (10(148t - 25)) / √(296t² - 100t + 25)

 

Plug in t = 3 to find out how fast the distance between the trains is changing at 3:00 PM:

 

d' = 10(148(3) - 25)) / √(296(9) - 100(3) + 25)

d' = 4190 / √(2389)

d' = 85.7 MPH