Question

Show that the speed of a moving particle over a time interval is constant if and only if its velocity and acceleration vectors are perpendicular over the time interval.

Answer 1

Answer:

$|v|(t)=\sqrt{v_{x}^{2}(t)+v_{y}^{2}(t)+v_{z}^{2}(t)}=C$

$2v(t)\cdot \frac{dv(t)}{dt}=0$

$v(t)\cdot a(t)=0$

Explanation:

Let's start with the definition of a constant velocity.

If the velocity magnitude, in three dimensions,  is a constant value (C) we have a constant velocity, which means.

$|v|(t)=\sqrt{v_{x}^{2}(t)+v_{y}^{2}(t)+v_{z}^{2}(t)}=C$

Now, we know that the dot product between v(t) and v(t) is the |v|².

$v(t)\cdot v(t)=|v|^{2}(t)$

If we take the derivative whit respect to time in both sides of this equation we will have:

$\frac{d}{dt}(v(t)\cdot v(t))=\frac{d}{dt}|v|^{2}(t)$

We apply the product rule on the left side and the right side will zero because the derivative of a constant is 0.

$\frac{dv(t)}{dt}\cdot v(t)+v(t)\cdot \frac{dv(t)}{dt}=0$

$2v(t)\cdot \frac{dv(t)}{dt}=0$

We know that dv(t)/dt = a(t) (using the acceleration definiton)

Therefore, we conclude:

$v(t)\cdot \frac{dv(t)}{dt}=0$

$v(t)\cdot a(t)=0$

If the dot product is 0, it means that v(t) and a(t) are orthogonal.

I hope it helps you!