A train starting at 11 am travels east at 45 km / hr while another starting at 12 noon from the same point trivels south at 60 km / hr . how fast the train separating at 3 p. m
at time t hr after second train travels from starting point
train 1 , X = 45 * (t+1)
train 2 , Y = 60 * t
distance apart
s^2 = [ 60^2 t^2 + 45^2 ( t+1)^2]
2s ds/dt = 2 * 3600* t + 2025 * 2 (t+1)
2s ds/dt = 7200 t + 4050 t + 4050
2s ds/dt = 11250 t + 4050
s ds/dt = 5625 t + 2025
when t= 3
s2= 32400+32400= 64800
s=254.56
ds/dt = 18900/ 254.56
= 74.245 mph ------answer
train 1, x= 45*3 = 135, y = 0
dx/dt = 45
train 2, x = 0, y= -60*3 = -180
dy/dt = -60
z = hypotenuse
z^2 = x^2+y^2 = 3^2 (45^2+60^2)
= 3^2 (2025 + 3600) = 3^2(5625)
so
z = 3*75 = 225
z^2 = x^2 + y^2
2 z dz/dt = 2 x dx/dt +2 y dy/dt
dz/dt = (xdx/t+ydy/dt)/z
= (135*45-180*-60)/225
=etc
Well, considering that the train started at 11 am and the other one started at 12 noon, they would have been traveling for 4 hours and 3 hours respectively by the time they separate at 3 pm. Now, if we take into account the speed of the eastward train (45 km/hr) and the southward train (60 km/hr), we can use some basic math to find the speed at which they are separating.
To calculate the distance traveled by the eastward train, we can multiply its speed (45 km/hr) by the time it traveled (4 hours), resulting in a distance of 180 km.
Similarly, to calculate the distance traveled by the southward train, we can multiply its speed (60 km/hr) by the time it traveled (3 hours), resulting in a distance of 180 km.
Now, if we draw a right-angled triangle using the eastward and southward distance as the two sides, the train's separation speed is the hypotenuse of this triangle. Since the two sides are equal (180 km), it means we have an isosceles right-angled triangle.
According to the Pythagorean theorem, the length of the hypotenuse can be calculated by multiplying the length of one side by the square root of 2. So, the separation speed of the two trains at 3 pm would be approximately 180 * √2 km/hr.
To find the speed at which the two trains are separating at 3 p.m., we can use the concept of relative velocity.
1. Determine the time difference between the two trains: The second train starts at 12 noon, which is 1 hour after the first train starts at 11 am. So, there is a time difference of 1 hour between the two trains.
2. Determine the distance traveled by each train during this time difference:
- The first train travels for 3 hours from 11 am to 3 pm, so its distance is 45 km/hr x 3 hrs = 135 km.
- The second train travels for 2 hours from 12 noon to 3 pm, so its distance is 60 km/hr x 2 hrs = 120 km.
3. Apply Pythagoras' theorem to find the distance between the two trains at 3 pm:
- Using the distances calculated above, the vertical distance covered by the second train is 120 km (facing south).
- The horizontal distance covered by the first train is 135 km (facing east).
- The distance between the two trains at 3 pm can be found using the Pythagorean theorem: distance = √(120^2 + 135^2) = √(14400 + 18225) = √32625 = 180.4 km.
4. Determine the time taken for the separation to occur:
- The time taken for the separation is the same for both trains, as they started from the same point. Therefore, we can consider the time taken by the first train (3 hours) or the second train (2 hours). In this case, we will consider 3 hours.
5. Calculate the speed at which the trains are separating:
- Speed = distance / time = 180.4 km / 3 hours = 60.1 km/hr.
Therefore, the trains are separating at a speed of approximately 60.1 km/hr at 3 p.m.
To find the speed at which the two trains are separating at 3 p.m., we can use the concept of relative velocity. Since the two trains are traveling in different directions, we can treat their velocities as vectors.
Let's break down the problem step by step:
Step 1: Calculate the time elapsed for the first train (eastbound train) from 11 a.m. to 3 p.m.
The first train travels for 4 hours (11 a.m. to 3 p.m.).
Step 2: Calculate the distance covered by the first train.
Distance = Speed × Time
Distance = 45 km/hr × 4 hr = 180 km
Step 3: Calculate the time elapsed for the second train (southbound train) from 12 noon to 3 p.m.
The second train travels for 3 hours (12 noon to 3 p.m.).
Step 4: Calculate the distance covered by the second train.
Distance = Speed × Time
Distance = 60 km/hr × 3 hr = 180 km
Step 5: Calculate the distance between the two trains at 3 p.m.
Since the two distances are equal (180 km), the trains are initially 180 km apart.
Step 6: Calculate the relative velocity of the two trains.
Since the trains are moving in perpendicular directions, we can use the Pythagorean theorem to find the relative velocity.
Relative Velocity = √(Velocity₁² + Velocity₂²)
Velocity₁ (eastbound train) = 45 km/hr
Velocity₂ (southbound train) = 60 km/hr
Relative Velocity = √(45² + 60²)
Relative Velocity = √(2025 + 3600)
Relative Velocity = √5625
Relative Velocity = 75 km/hr
Therefore, the two trains are separating at a speed of 75 km/hr at 3 p.m.