A train leaves the station at 9 AM heading S60º W (at a speed of 100km/h) another train heads North at a speed of 120km/h and reaches the same station at 11 AM

Question

A train leaves the station at 9 AM heading S60º W (at a speed of 100km/h) another train heads North at a speed of 120km/h and reaches the same station at 11 AM. at what time were the two trains closest together? S60ºW is same as 60ºW of S

Answer 1

Let the station be represented by the origin in the coordinate plane and 9 am is t = 0.

At t = 0, the 2nd train is at (0 , - 240) with dy/dt = 120 (its x-coordinate does not change).

At t = 0, the 1st train is at (0 , 0) with dy/dt = - 50 and dx/dt = -50√3 (by special right triangle 30-60-90)

Let d(t): distance between the 2 trains (in km) ; d(t) = √[(-50√3t)² + (-240 + 70t)²]

d(t) = √[12,400t² - 33500t + k ] (k is immaterial so I haven't bothered to calculate it here)

To minimize d(t) it is sufficient to minimize what is under the radical so 24,800t - 33,600 = 0 and t = 336/248

t ~ 81 minutes so the trains are closest at ~ 9:21 am