Prove that if the speed of a particle is constant, $\vec a$ is perpendicular to $\vec v$

Question

My train of reasoning (well, not even that, more like "translating") so far:

Let $C$ be some constant. $\|\vec {v(t)}\|=C$, so $\frac {d\|\vec {v(t)}\|} {dt}=0$.

But, where to go from here? I don't know how to express $\frac {d\|\vec {v(t)}\|} {dt}=0$ in terms of $\vec a$, aside from just flatly saying "$\mathrm{comp}_va=0$", which I feel doesn't sound very smooth or explanatory.

Answer 1

With some help from Semiclassical, I figured it out:

Let $C$ be some constant; we know $||\vec {v(t)}||=C$ and $||\vec{v(t)}||^2=\vec v • \vec v$ so $\vec v • \vec v=C^2$ and $$\frac {d(\vec v • \vec v)} {dt}=2(\vec v • \frac {d \vec v} {dt})=0$$Therefore, $$\vec v • \vec a=0$$

Answer 2

$$\vec {v(t)}\ = \|\vec {v(t)}\|(sin(\theta) + cos(\theta))$$ $$\therefore \vec a = \frac {d\vec {v(t)}} {dt}=\frac {d\|\vec {v(t)}\|} {dt}(sin(\theta) + cos(\theta)) + \|\vec {v(t)}\|\frac{d(sin(\theta) + cos(\theta))}{dt}$$ $$\therefore \vec a = 0 + \|\vec{v(t)}\|\frac{d\theta}{dt}(cos(\theta)-sin(\theta))$$ $$\therefore \vec a = \|\vec{v(t)}\|\frac{d\theta}{dt}(sin(\pi/2 + \theta)+cos(\pi/2 + \theta))$$

Thus, $\vec a$ is perpendicular to $\vec v.$

Answer 3

a in direction of v = $\frac{d|v|}{dt}=0$
a in perpendicular direction of v = $\frac{|v^2|}{R}\ne0$ where R is radius of curvature