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?
So the answer to this question comes from setting your x-component equal to $60t$ and your y-component to $70$. When we use Pythagoreans theorem we get
$$ z(t)= \sqrt {3600t^2 + 4900} $$
$$ z'(t)= \frac {3600t}{ \sqrt {3600t^2 + 4900}} $$
and at $t=4$ we get that the velocity of the train is $57.6$.
When we try to explain why the velocity is 57.6 does our only explanation consist of arguments relating to relativity or is there something else to this? That is the only explanation that makes sense to me because relative to the train the velocity is always constant.
Best Answer
Your explanation needs to take into account that the instantaneous velocity you are computing is that of the rate at which the train is moving away from the observer (relative to the distance measured along the "hypotenuse" formed by the train and the observer). That rate is changing, whereas the "horizontal" velocity of the train (moving east) is constant at $60$ meters per second. The rate at which the train is moving from the observer will be less than the constant east-west velocity of the train because the vertical distance of the observer from the east-west road , measured from the intersection, is constant (only the horizontal distance between the train and the observer is changing).
As time and the corresponding horizontal distance between the observer and train increases, the rate at which the train is moving from the observer will approach the rate at which the train is moving eastbound.