Writing the fourth kinematic formula ($v^2 = v_0^2 + 2a \Delta x$) in a different form

Question

I have noticed that if we play with the formula a bit, we get

\begin{align} v^2 &= v_0^2 + 2a \Delta x \\ \frac{(v+v_0)}{2} \cdot (v-v_0) &= a \Delta x \\ \overline{v} \cdot \Delta v &= a \Delta x \end{align}

Meaning the average velocity $\overline{v}$ times the change in velocity $ \Delta v $ is equal to the (constant) acceleration times the change in place

Is this correct? If so, is there a "physical meaning" behind this form of the formula? Meaning, why does multiplying the average velocity with the difference yields acceleration times displacement?

Answer 1

The formula is valid and has a geometric and a physical interpretation.

The hand side corresponds to the kinetic energy of a unit mass, and the right-hand side the work done by a force on the same unit mass. Multiply both sides with $m$ the mass and get

$$ m\, \overline{v}\, \Delta v = m a\,\Delta x \tag{1}$$

Use $F=m a$ and the definition of work $W = F\,\Delta x$ to get

$$ W = m\, \overline{v}\, \Delta v \tag{2}$$

but you can show that the RHS is the change in kinetic energy as

$$ W = m \frac{(v+v_0)}{2} \cdot (v-v_0) = \tfrac{1}{2} m v^2 - \tfrac{1}{2} m v_0^2 = \Delta E \tag{3} $$

which goes down to the fundamental principle of work = change in energy.

But there is a geometric interpretation also.

Take the velocity plot vs. time, which under constant acceleration, it is a straight line between $v_0$ and $v$

The area under the curve is the distance traveled. From the triangle area formula that is

$$ \Delta x = \tfrac{1}{2} \Delta t ( v-v_0) \tag{4} $$

but the distance traveled can also be found by the average speed $\overline{v} = \tfrac{v_0+v}{2}$ which you can see above as the area under the rectangle defined by the dashed line.

$$ \Delta x = \overline{v}\, \Delta t \tag{5} $$

Combine this with the kinematic equation $\Delta v = a \, \Delta t$ which is solve for time $\Delta t = \tfrac{\Delta v}{a}$ to get

$$ \Delta x = \overline{v}\, \tfrac{\Delta v}{a} \tag{6} $$

or

$$ a \Delta x = \overline{v}\, \Delta v \tag{7} $$

Note that if/when you learn calculus, you will deal with infinitesimal changes and the equation $a\, {\rm d}x = v\, {\rm d}v$ is solve by integration (summation of all the small parts) as

$$\int a\, {\rm d}x =\int v\, {\rm d}v =\tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2 \tag{8}$$

which is the solution for non-constant $a$ and matches with equation (3) above as it is interpreted as work = change in kinetic energy.

An example is when you have a spring with $F=-k x$ and thus $ a =-\tfrac{1}{m} k x$, you use it in (8) to get

$$- \int \tfrac{1}{m} k x\,{\rm x} = -\tfrac{1}{2} \tfrac{k}{m} x^2 = \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2$$

which is equivalent to the energy balance between the potential energy of the spring and the kinetic energy of the mass at initial conditions (rhs) and at current conditions (lhs).

$$ \tfrac{1}{2} k x^2 + \tfrac{1}{2} m v^2 = \tfrac{1}{2} m v_0^2 $$

Answer 2

The average acceleration is defined as $\bar{a} = \frac{\Delta v}{\Delta t}$, and the average velocity is $\bar{v} = \frac{\Delta x}{\Delta t}$. For a constant acceleration $a = \bar{a}$, so the right hand side of your expression could be rewritten: $$ a\, \Delta x = \frac{\Delta v}{\Delta t} \Delta x = \Delta v \, \frac{\Delta x}{\Delta t} = \Delta v\, \bar{v}.$$

This shows that your expression is correct, but I can't find any deep physical significance in it.

Answer 3

Average velocity by basic definition is the average distance covered by an object divided by the average time required to do so.

Hence, v avg = s/t

Also, acceleration is the rate of change of velocity, or simply change in velocity divided by the time interval.

Hence, the LHS is the displacement of the body times its acceleration