Is \( V_{\text{average}} = \dfrac{V_{\text{final}}+V_{\text{initial}}}{2} \) ?

This is part of a series on common misconceptions.

Is this true or false?

The average velocity of a body over a course of time is the arithmetic mean of the initial and final velocities of the body.

Why some people say it's true: Average velocity means the average value of the initial and final velocities of the body.

Why some people say it's false: Average velocity is equal to the ratio of displacement to the time required for the displacement to occur, and not the average of the initial and final velocities.

The statement is $\color{#D61F06}{\textbf{false}}$.

Explanation:
In physics, $V_{\text{average}}$ is not simply the average value of the initial and final velocities. Suppose that a body undergoes a change in displacement $\Delta \vec{r}$ in time interval $\Delta t$, then the average velocity of the body is defined as the constant velocity with which the body moves. It will undergo the same change in displacement in the same time interval: $$\vec{V}_{\text{average}} = \dfrac{\Delta \vec{r} }{\Delta t}.$$ If the body has position vectors $\vec{r_{1}}$ and $\vec{r_{2}}$ at time instants $t_{1}$ and $t_{2}$, $$\vec{V}_{\text{average}} = \dfrac{\vec{r_{2}}-\vec{r_{1}}}{t_{2}-t_{1}}.$$ The direction of $\vec{V}_{\text{average}}$ is in the direction of $\Delta \vec{r}$: $$\left| \vec{V}_{\text{average}} \right| = \dfrac{1}{t_{2}-t_{1}} \cdot \left | \vec{r_{2}} - \vec{r_{1}} \right |.$$

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Reply: (Explain why the argument is not vald)

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See Also

• List of Common Misconceptions