# Why does a perpendicular force not change speed?

I have been taught that a perpendicular force acting on an object will only change the direction of its velocity, not its magnitude. The explanation that was provided to me is that because there no force in the direction of its velocity, there must be no work done and thus change in the object's kinetic energy. However, I have also studied the vectorial form of the equations of motion

$$\vec v_f = \vec v_i + \vec a\cdot t$$

So I assume that the velocity after a constant force has acted on it for a period of time is simply the vector sum of $\vec v_i$ and $\vec a \cdot t$, which I just assume is the acceleration vector (determined by the force) scaled up by a factor of the time the force acts.Diagrammatically, I took the sum as

It is obviously clear from the diagram that the new speed (which is the length of the $\vec v_f$ vector) is bigger than the previous speed. This can be confirmed by Pythagoras's theorem, the magnitude of the $\vec v_f$ vector is

$$|\vec v_f|=\sqrt{|\vec v_i|^2 + |\vec a|^2 \cdot t^2}$$

which is very clearly not equal to $|\vec v_i|$

So, I conclude that the speed of the object has changed even though the force acting on it was perpendicular to its velocity. So my question is, what have I missed?

Is my assumption that the final velocity is the vector sum of the initial velocity and the product of acceleration and time incorrect? Or have I made a mistake somewhere else?

However, if $\vec{F}$ is not of the same magnitude and direction everywhere, one can produce a force that remains perpendicular. The canonical example is that of a body exerting gravity. The direction of the force is always towards the particle (i.e. not the same everywhere). It also decreases with distance, but that's really not an essential point here. Now, if a test particle at a distance $d$ moves past the other body with the right velocity, this force will always remain perpendicular. Can you guess what kind of shape the trajectory of the test particle will trace out? Hint: think about the symmetry of the problem.