A train leaves a station at 10:00 and travels due North at a speed of 100km/h. Another train has been heading due west at 120km/h and reaches the same station at 11:00. At what time were the two trains closest together?

The time were the two trains are closest together is 10:35:41

Step-by-step explanation:

Here we are given;

Train A: Speed = 100 km/h

Direction = North or y axis

Departure time = 10:00

Train B: = Speed = 120 km/h

Direction = West or x axis

Departure time = 11:00

Therefore, we have;

Distance of separation, d between the two trains is given as follows

d(t)² = (100·t)² + (120 - 120·t)²

d(t) = √((100·t)² + (120 - 120·t)²) = √(24400·t²-28800·t+14400)

Differentiating and equating to zero, we have;

Cross multiplying we have;

61×2×t -72 = 0

Which gives t = 36/61 hours = 35.41 mins after Train A starts the journey or 10:35:41 ≈ 10:36.

69

step-by-step explanation:

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.