Is Velocity Constant if AxVx + AyVy Equals Zero?

[bob1182006]

Homework Statement

A particle is moving in the xy plane with velocity $\vec{v}(t)=v_x(t)\vec{i}+v_y(t)\vec{j}$ and acceleration $\vec{a}(t)=a_x(t)\vec{i}+a_y(t)\vec{j}$. By taking the appropiate derivative show that the magnitude of v can be constant only if $a_xv_x+a_yv_y = 0$

Homework Equations

The Attempt at a Solution

So I know that in order for velocity to have a constant magnitude then acceleration must = 0.

Since acceleration is the derivative of velocity and dv/dt=0 iff v=some constant.
expanding the LHS:

$$\frac{d\vec{v}(t)}{dt}=0$$

$$\frac{dv_x(t)}{dt}\vec{i}+\frac{dv_y(t)}{dt}\vec{j} = 0$$

$$a_x(t)\vec{i}+a_y(t)\vec{j} = 0$$

since i,j are unit vectors and do not equal 0 the components of acceleration must = 0.

$$a_x(t)=a_y(t)=0$$

But it's also possible for $a_x=-a_y$ in which case the acceleration would still = 0 so is this a proof really correct?

 

[learningphysics]

The velocity need not be constant. The magnitude of the velocity is constant. Get the magnitude of the velocity, and use the fact that it is constant... go from there..

 

[m2003]

Your attempt at a solution is on the right track, but it is not a complete proof. Let's break it down step by step.

First, you correctly state that in order for velocity to have a constant magnitude, acceleration must be zero. This is because acceleration is the rate of change of velocity, so if acceleration is zero, velocity is not changing and therefore has a constant magnitude.

Next, you expand the left side of the equation \frac{d\vec{v}(t)}{dt}=0 to get \frac{dv_x(t)}{dt}\vec{i}+\frac{dv_y(t)}{dt}\vec{j} = 0. This is also correct, but it is important to note that this is not the same as v_x(t)\vec{i}+v_y(t)\vec{j} = 0 which is what you have written in the homework statement. The former is the derivative of velocity, while the latter is the velocity itself.

Then, you state that a_x(t)=a_y(t)=0, which is not necessarily true. It is possible for a_x=-a_y, as you mentioned, which would still result in an acceleration of zero. So this is not a complete proof.

To complete the proof, we need to use the fact that acceleration is the derivative of velocity. This means that we can write a_x(t) = \frac{dv_x(t)}{dt} and a_y(t) = \frac{dv_y(t)}{dt}. Substituting these into the equation a_xv_x+a_yv_y = 0, we get \frac{dv_x(t)}{dt}v_x(t)+\frac{dv_y(t)}{dt}v_y(t) = 0.

Now, if we integrate both sides with respect to time, we get \int \frac{dv_x(t)}{dt}v_x(t)dt+\int \frac{dv_y(t)}{dt}v_y(t)dt = 0. Using the chain rule, we can simplify this to \int v_x(t)dv_x(t)+\int v_y(t)dv_y(t) = 0.

We can then use the power rule to integrate each term, giving us \frac{1}{2}v_x(t)^2+\frac{1}{2}v_y(t)^2 = C, where C is a constant of integration.

This shows that the