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Trava [24]
3 years ago
5

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

re perpendicular over the time interval.
Physics
1 answer:
lawyer [7]3 years ago
8 0

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!

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