A train leaves the 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 2 trains closests?
after t hours,
north train has gone 100t from the station
west train is 120-120t km from the station
so the distance z between the trains is
z^2 = (100t)^2 + (120-120t)^2
= 24400t^2 - 28800t + 14400
= 400(61t^2 - 72t + 36)
z dz/dt = 800(61t-36)
dz/dt = 0 at t = 36/61 = 35.4 minutes
so at 10:35:25 the trains were closest
say time t = hours after 10 am
north train at y = 100 t
west train x= -120 (t-1)
z = d^2 = x^2+ y^2= (100t)^2 + 120^2 (t^2 - 2 t + 1)
z = 10,000 t^2 + 14,400 t^2 - 28,000 t + 14,000
z = 24, 400 t^2 - 28,000 t +14,000
dz / dt = 48,800 t - 28,000
min or max when dz/dt = 0
t = 28.0/48.8 = 0.574 hr = 34.4 minutes after 10am
To determine the time at which the two trains were closest, we need to find the point of intersection of their trajectories. Let's calculate the distance each train has traveled at specific time intervals until they meet.
The first train is traveling at a constant speed of 100 km/h due north. Let's denote the distance traveled by this train as d1 and the time as t.
The second train is traveling at a constant speed of 120 km/h due west. Let's denote the distance traveled by this train as d2 and the time as t.
Since the second train departs one hour later, we can express the distance traveled by the second train as d2 = 120 * (t - 1).
At the time of the intersection, the distance traveled by both trains will be equal, so we can set up the following equation:
d1 = d2
100t = 120(t - 1)
Simplifying the equation:
100t = 120t - 120
120 - 100t = 120
20t = 120
t = 6
Therefore, the time at which the two trains are closest is 6 hours after the first train's departure.
Since the first train departs at 10:00, the two trains are closest at 16:00 (4:00 PM).
To determine the time at which the two trains were closest, we need to find the point at which the distance between them is minimized.
Given that one train is moving due north and the other is moving due west, we can draw a diagram to represent their paths. Let's assume that the trains were closest at time t.
Since the northbound train travels at a constant speed of 100 km/h, the distance it covers from the starting point (at 10:00) in t hours is 100t km.
Similarly, the westbound train also has a constant speed of 120 km/h and has covered a distance of 120 km from its starting point (at an unknown time in the past) in (t-1) hours.
Now, we can use the Pythagorean theorem to find the distance between the two trains at time t. The distance between them is the hypotenuse of a right-angled triangle, with one side being the distance covered by the northbound train (100t km) and the other side being the distance covered by the westbound train (120 km).
Using the Pythagorean theorem:
Distance^2 = (100t)^2 + 120^2
Simplifying the equation:
Distance^2 = 10000t^2 + 14400
To find the minimum distance, we need to differentiate this equation with respect to t and find the value of t that makes the derivative equal to zero. Then we can solve for t to determine the time at which the two trains are closest.
Differentiating the equation:
d(Distance^2)/dt = d(10000t^2 + 14400)/dt
d(Distance^2)/dt = 20000t
Setting the derivative equal to zero:
20000t = 0
t = 0
Now we have the time t = 0, which means the two trains are closest at t = 0 hours. Since t represents the time elapsed from the moment the first train leaves, we can conclude that the two trains are closest when the first train departs from the station at 10:00.
Therefore, the two trains are closest at 10:00.