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

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

`2v(t)\cdot (dv(t))/(dt)=0`

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

Step-by-step 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:

`(d)/(dt)(v(t)\cdot v(t))=(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.

`(dv(t))/(dt)\cdot v(t)+v(t)\cdot (dv(t))/(dt)=0`

`2v(t)\cdot (dv(t))/(dt)=0`

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

Therefore, we conclude:

`v(t)\cdot (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!