# 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 the average velocity magnitude.

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

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

Before introducing more problems about velocity itself, let's introduce the distinction between **speed** and **velocity**, as most people generally confuse them:

Speedis a scalar quantity that refers to "how fast/slow an object is moving".

Velocityis a vector that indicates "the rate at which an object changes its position".

In other words, the speed is just a number, \(3 \thinspace m.s^{-1}\), while the velocity is a vector (i.e. it has a direction and an associated value).

Velocity can be thought of as an extension of speed: The speed receives the attribute "direction": The velocity has now two components, its direction, and value (the speed of the object).

It is also worth mentioning that the speed corresponding to a velocity is called the *magnitude* of the velocity.

Now let's get back to velocity and calculations:

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 magnitude 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/