A railroad track and a road cross at right angles. 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?
The distance z is
z^2 = 70^2 + (60t)^2
at t=4, z=250
z dz/dt = 120t
Now plug in your numbers
To solve this problem, we can use the Pythagorean Theorem to find the distance between the train and the observer at a given time.
Let's denote the distance between the observer and the crossing point as 'd', the distance the train travels in 't' seconds as 's', and the distance the train is from the observer at time 't' as 'x'.
According to the problem, the observer is 70 meters south of the crossing point. Therefore, 'd = 70 meters'.
The distance traveled by the train 's' can be calculated using the formula: 's = velocity × time'.
Given that the velocity of the train is 60 meters per second, and the time is 4 seconds, we can calculate 's' as follows: 's = 60 × 4 = 240 meters'.
Using the Pythagorean Theorem, we can express the relationship between the distances as follows:
x^2 + d^2 = s^2
Substituting the known values, we have:
x^2 + 70^2 = 240^2
Simplifying:
x^2 + 4900 = 57600
x^2 = 57600 - 4900
x^2 = 52700
Taking the square root of both sides:
x = √52700
Now, we can find the distance the train is moving away from the observer by subtracting the distance traveled by the train in the next 4 seconds from the distance traveled by the train in the previous 4 seconds.
At time 't = 4 seconds', the train has traveled 's = 240 meters'. At time 't = 8 seconds', the train would have traveled '2s = 2 × 240 = 480 meters'.
Thus, the distance the train is moving away from the observer is '480 - 240 = 240 meters'.
Therefore, the train is moving away from the observer at 240 meters per second, 4 seconds after it passes through the intersection.
To determine the train's velocity away from the observer, we need to find the component of the train's velocity that is perpendicular to the road.
First, let's consider the train's velocity vector. Since it is traveling eastbound and the observer is standing south of the crossing, we can break down the train's velocity into two components: the component parallel to the road and the component perpendicular to the road.
The component parallel to the road has no effect on the observer's perception of the train moving away. So, we are only interested in the component of the train's velocity that is perpendicular to the road. This is the hypotenuse of the right-angled triangle formed by the train's velocity vector.
Since the observer is 70 meters south of the crossing, we can use this as the adjacent side of the triangle. We can use the formula:
adjacent/hypotenuse = cosine(angle)
To find the hypotenuse, we need to find the angle between the train's velocity vector and the road. Since the train is moving eastbound and the road crosses at a right angle, the angle between the train's velocity vector and the road is 90 degrees.
So, the equation becomes:
70/hypotenuse = cosine(90°)
cosine(90°) is equal to 0, so the equation simplifies to:
70/hypotenuse = 0
This means that the hypotenuse (the component of the train's velocity perpendicular to the road) is 0. In other words, the train is not moving away from the observer in the perpendicular direction at all.
Therefore, the train's velocity away from the observer 4 seconds after it passes through the intersection is 0 meters per second.