# Average Velocity

Average velocity is a concept that measures the displacement of an object over time. This can mathematically be described as the average rate of change in position over time:

\[\bar{v}=\frac{\Delta x}{\Delta t}=\frac{x_f-x_i}{\Delta t},\]

where \(x_i\) and \(x_f\) denote the initial and final positions of the object, respectively. Thus, average velocity is a physical concept consistent with the mathematical concept "average rate of change." Observe that average velocity only concerns the initial and final positions of the object, regardless of the path it takes.

An object on the \(x\)-axis moves from \(x=10\) to \(x=-2\) over \(3\) seconds. Find its average velocity.

Simply plugging in \(x_i=10, x_f=-2,\) and \(\Delta t=3\) in the equation above gives the answer:

\[\bar{v}=\frac{-2-10}{3}=-4.\ _\square\]

An athlete runs a lap on a circular track of radius \(100\text{ m}.\) He runs at a constant speed, and it takes him \(50\) seconds to complete one lap. What is the magnitude of his average velocity when he is halfway through? How about when he has finished one lap?

First let's find his average velocity when he has finished one lap. Since his position has not changed from his initial, the displacement is equal to zero. Therefore his average velocity is also equal to zero.

When he is halfway through, he is \(2\times100\text{ m}=200\text{ m}\) far from his starting point, and it would take him \(\frac{50}{2}=25\) seconds for him to get there. Hence we have \(\Delta x=200\) and \(\Delta t=25,\) so the magnitude of his average velocity is

\[\bar{v}=\frac{200}{25}=8\text{ (m/s)}.\ _\square\]

The same goes for two-dimensional motion. Just think of the displacement and average velocity as vectors, and apply the same methods.

A creature crawls \(4\) meters east, and \(3\) meters north over \(10\) seconds. What is the magnitude of its average velocity?

Consider east as the \(+x\) direction, and north as the \(+y\) direction. Then the creature's displacement vector is \(\vec{x}=(4,3).\) Therefore the magnitude of its average velocity is \[\lvert\vec{v}\rvert=\left\lvert\frac{\vec{x}}{\Delta t}\right\rvert=\frac{\sqrt{4^2+3^2}}{10}=0.5.\ _\square\]

A dot moves on the coordinate plane along the graph of the function \(f(x)=x^2.\) If the dot's \(x\)-coordinates at times \(t=1\) and \(t=4\) are \(x\rvert_{t=3}=-2\) and \(x\rvert_{t=4}=5,\) what is its magnitude of average velocity?

Since \(f(-2)=4\) and \(f(4)=25,\) the displacement vector is

\[\vec{x}=(4,25)-(-2,4)=(6,21).\]

Therefore the magnitude of average velocity is

\[\lvert\vec{v}\rvert=\left\lvert\frac{\vec{x}}{\Delta t}\right\rvert=\lvert(2,7)\rvert=\sqrt{2^2+7^2}=\sqrt{53}.\ _\square\]

**Cite as:**Average Velocity.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/average-velocity/