I get why it works with displacement because that's what the theorem was originally meant for, lengths.... I find it harder to wrap my head around it when its velocity. If anyone has a good visualization or anything that will help me understand It will really help. I scoured the internet for a while but all of them were just:- put V(y) velocity arrow perpendicular to V(x) head ,for some reason this explanation isn't working out for me.... well looking forward to the responses!
cheers!
On
Suppose you’re on a ship moving with velocity $v_1$ (relative to the shore) and you’re walking with velocity $v_2$ (relative to the ship). In time $\Delta t$ the ship is displaced by $v_1 \Delta t$ (relative to the shore) and you are displaced by $v_2 \Delta t$ (relative to the ship). Your displacement relative to the shore is $(v_1 + v_2) \Delta t$. So your velocity relative to the shore is $v = v_1 + v_2$. If $v_1 \perp v_2$, then the Pythagorean theorem (which holds whenever we add orthogonal vectors) tells us that $\| v \|^2 = \| v_1 \|^2 + \| v_2 \|^2$.
If you’d like, you could apply the Pythagorean theorem to the displacement vectors then divide by $\Delta t^2$ to obtain the Pythagorean theorem for the velocity vectors.
On
In Euclidean space, the Pythagorean theorem tells us for orthogonal vectors $u,v,$
$$\|u+v\|^2=\|u\|^2+\|v\|^2.$$
I get why it works with displacement because that's what the theorem was originally meant for, lengths
Go back to what a vector in Euclidean space is: an object with direction and length. The vector may represent many things, e.g. displacement, velocity, acceleration, force, electric field etc. Since vectors have a notion of length, applying Pythagorean theorem makes perfect sense if the vector quantity obeys the superposition principle.
It works because a right triangle of velocity vectors is just a right triangle of displacement vectors, all divided by the same constant time increment.