Velocity

This page defines and describes velocity.

Main Idea

Velocity, denoted by the symbol [math]\displaystyle{ \vec{v} }[/math], is a vector quantity defined as the rate of change of position with respect to time. In calculus terms, it is the time derivative of the position vector. The magnitude of a velocity vector is speed, and the direction of the vector is the direction of travel. Velocity is an instantaneous value, so it may change over time. Several properties in physics, such as momentum and kinetic energy, are functions of velocity. The most commonly used metric unit for velocity is the meter per second (m/s).

A Mathematical Model

Instantaneous velocity [math]\displaystyle{ \vec{v} }[/math] is defined as:

[math]\displaystyle{ \vec{v} = \frac{d\vec{r}}{dt} }[/math]

where [math]\displaystyle{ \vec{r} }[/math] is a position vector and [math]\displaystyle{ t }[/math] is time.

Average Velocity

Since velocity is an instantaneous quantity, it can change over time. Over any given time interval, there is an average velocity value denoted [math]\displaystyle{ \vec{v}_{avg} }[/math]. The average acceleration over an interval of time [math]\displaystyle{ \Delta t }[/math] is given by

[math]\displaystyle{ \vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} }[/math].

As the duration of the time interval [math]\displaystyle{ \Delta t }[/math] becomes very small, the value of the average velocity over that time interval approaches the value of the instantaneous velocity at any point in time within that interval.

Derivative Relationships

Velocity is the time derivative of position:

[math]\displaystyle{ \vec{v}(t) = \frac{d\vec{r}(t)}{dt} }[/math].

Acceleration, in turn, is the time derivative of velocity:

[math]\displaystyle{ \vec{a}(t) = \frac{d\vec{v}(t)}{dt} }[/math]

Integral Relationships

Position is the time integral of velocity:

[math]\displaystyle{ \vec{r}(t) = \int \vec{v}(t) \ dt }[/math].

Velocity is, in turn, the time integral of acceleration:

[math]\displaystyle{ \vec{v}(t) = \int \vec{a}(t) \ dt }[/math].

Kinematic Equations

The kinematic equations can be derived from the derivative and integral relationships between acceleration, velocity, and displacement. For the equations and more information, view the Kinematics page.

In Physics

According to Newton's First Law of Motion, the velocity of an object must remain constant (whether that velocity is zero or nonzero) unless a force acts on it. Newton's Second Law: the Momentum Principle describes how velocity changes over time as a result of forces.

A Computational Model

In VPython simulations of physical systems, the Iterative Prediction algorithm is often used to move objects with specified velocities and evolve those velocities over time. The program below is an example of such a simulation. In this program, the ball's velocity is represented by a purple arrow.

Click here for velocity simulation

Click "view this program" in the top left corner to view the source code.

Examples

Simple

A runner runs a lap counterclockwise around a circular track at a constant speed of 2.8m/s.

a) What is the average speed of the runner over the course of the lap?

b) What is the average velocity of the runner over the course of the lap?

c) What is the instantaneous velocity of the runner at the northernmost point on the track?

Solutions:

a) The instantaneous speed of the runner is 2.8m/s for the entire duration of the lap, so the average speed of the runner is also 2.8m/s.

b) The instantaneous velocity vector of the runner has a constant magnitude of 2.8m/s, but its direction is constantly changing depending on the runner's position along the track. To find the average value of the velocity vector over time, let us use the following formula:

[math]\displaystyle{ \vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} }[/math].

Because the runner ran a single lap and ended at the point in space where they began, their displacement [math]\displaystyle{ \Delta \vec{r} }[/math] is <0,0>m! The runner's average velocity was therefore <0,0>m/s. This makes sense because the velocity vector spends equal time pointing in every possible direction in the plane of the track, so each instantaneous velocity at any point in the lap cancels with a different instantaneous velocity at a different point in the lap.

c) At the northernmost point on the track, the runner had a speed of 2.8m/s as always and was running in the westward direction because the problem tells us they were running counterclockwise. This is a sufficient answer, as it describes the magnitude and direction of the velocity vector. Alternatively, the answer can be reported in component form: <-2.8,0>m/s using +x to mean east and +y to mean north.

Middling

A particle's position as a function of time is as follows: [math]\displaystyle{ \vec{r}(t) = \lt 6\sin(2 \pi t), -6t^3 + 10, e^{8t}\gt }[/math]

What is the particle's velocity as a function of time?

Solution:

[math]\displaystyle{ \vec{v} = \frac{d \vec{r}}{dt} }[/math]

[math]\displaystyle{ \vec{v} = \lt 12 \pi \cos(2 \pi t), -18t^2, 8e^{8t}\gt }[/math]

Difficult

a particle moves along the x axis. At time t=0, its position is x=2. Its velocity [math]\displaystyle{ v }[/math] varies over time, obeying the following function:

[math]\displaystyle{ v(t) = \begin{cases} 4-2t & \text{if $t\leq 3$} \\ -2 & \text{if $t\gt 3$} \end{cases} }[/math]m/s

What is its position as a function of time after time t=0?

Solution:

[math]\displaystyle{ x(t) = x(0) + \int_0^t v(t')dt' }[/math] ([math]\displaystyle{ t' }[/math] is a "dummy variable" since [math]\displaystyle{ t }[/math] is already our limit of integration)

for [math]\displaystyle{ t\leq 3 }[/math],

[math]\displaystyle{ x(t) = 2 + \int_0^t 4-2t' dt' }[/math]

[math]\displaystyle{ x(t) = 2 + [4t'-t'^2]_0^t }[/math]

[math]\displaystyle{ x(t) = 2 + 4t - t^2 }[/math]

for [math]\displaystyle{ t\gt 3 }[/math],

[math]\displaystyle{ x(t) = 2 + \int_0^3 4-2t' dt' + \int_3^t -2 dt' }[/math]

[math]\displaystyle{ x(t) = 5 + \int_3^t -2 dt' }[/math]

[math]\displaystyle{ x(t) = 5 + [-2t']_3^t }[/math]

[math]\displaystyle{ x(t) = 5 -2t + 6 }[/math]

[math]\displaystyle{ x(t) = 11 - 2t }[/math]

final answer:

[math]\displaystyle{ x(t) = \begin{cases} -t^2 + 4t + 2 & \text{if $t\leq 3$} \\ 11-2t & \text{if $t\gt 3$} \end{cases} }[/math]m

Connectedness

Since the goal of the field of mechanics is to predict the motion of systems, velocity is critical to this branch of physics. In addition to being a value of interest in and of itself, properties such as Linear Momentum, Angular Momentum, and Kinetic Energy depend on the velocities of objects.

The use of velocity in calculations has a wide range of industrial applications including calculating travel time, determining safe operating speeds of vehicles and other machines, vehicle collision testing, determining the inertial forces experienced by roller coaster riders, and the design of equipment such as radar guns. The velocity of moving parts is important in virtually every type of machinery.

See Also

Acceleration

Relative Velocity

Speed and Velocity

External links

The Physics Classroom: Speed and Velocity

HyperPhysics: Average Velocity

YouTube video explaining average vs instantaneous velocity

References

1. Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. Hoboken, NJ: Wiley, 2011. Print.

2. "Velocity." Def. 2. Dictionary.com. N.p., n.d. Web. 29 Nov. 2015.

3. Velocity Expression. Digital image. Physics-Formulas. N.p., n.d. Web. 29 Nov. 2015.

4. Velocity vs Time Graph. Digital image. https://upload.wikimedia.org/wikipedia/commons/a/ae/Velocity-time_graph_example.png. N.p., n.d. Web. 29 Nov. 2015.