A railroad track and a road cross at a right angle. An observer stands on the road 70 meters south of the crossing and watches an eastbound train traveling at 60 meters per second. At how many meters per second is the train moving away from the observer 4 seconds after it passes through the intersection?
Well, 4 seconds after crossing, the train has traveled 240 m. Draw your figure, lets are 60, 240, and the hypotenuse is L=sqrt (60^2+240^2)
But in general, l^2=x^2+60^2
2L*dL/dt=2x dx/dt
you are solveing for dL/dt, you know x, dx/dt, and L.
To solve this problem, we can use the Pythagorean theorem to find the distance between the observer and the train after 4 seconds.
Let's define the distance between the observer and the train as "d". According to the problem, the observer stands 70 meters south of the crossing, so the distance perpendicular to the track is 70 meters.
After 4 seconds, the train will have traveled a distance of 4 seconds * 60 meters/second = 240 meters along the track.
Using the Pythagorean theorem, we can calculate the distance "d" as follows:
d^2 = (70 meters)^2 + (240 meters)^2
d^2 = 4900 meters^2 + 57600 meters^2
d^2 = 62500 meters^2
Taking the square root of both sides, we find:
d ≈ 250 meters
So, the distance between the observer and the train after 4 seconds is approximately 250 meters.
Now, let's calculate how fast the train is moving away from the observer. Since the train is moving along the track, the rate of change of the distance along the track (speed) is the speed of the train. Therefore, the train is moving away from the observer at a speed of 60 meters/second.
To determine the train's velocity away from the observer after 4 seconds, we need to break down the problem and use the Pythagorean theorem.
First, we know that the observer and the train are moving along perpendicular directions, so we can treat their motions separately.
Let's assume the east direction is positive for simplicity.
1. Determine the horizontal distance traveled by the train in 4 seconds (d₁):
Velocity = Distance / Time
d₁ = Velocity * Time = 60 m/s * 4 s = 240 meters
2. Calculate the distance between the observer and the train 4 seconds after it passes through the intersection (d₂):
Since the observer is 70 meters south of the crossing, the vertical distance remains constant.
d₂ = √((d₁)² + (70 meters)²) = √((240 meters)² + (70 meters)²)
d₂ = √(57600 meters² + 4900 meters²) = √(62500 meters²) = 250 meters
3. Find the train's velocity away from the observer after 4 seconds:
To calculate this, we need to find the horizontal distance traveled by the train relative to the observer.
velocity away from the observer = d₁ / 4 seconds = 240 meters / 4 seconds = 60 meters per second
Therefore, the train is moving away from the observer at a velocity of 60 meters per second 4 seconds after it passes through the intersection.