Constant Velocity and Uniform Motion

From Constant Velocity to Position

A velocity vector describes both how quickly position changes and the direction of that change. If the velocity stays constant, the particle makes the same displacement during every time interval of the same length. Its position therefore changes in a particularly simple way: start at the known position and add the velocity multiplied by the elapsed time.

This rule applies in two or three dimensions, and more generally in any setting where position and velocity are represented by vectors. Here we focus on motion through three-dimensional space, where a position is given by three coordinates.

Definition: A particle is in constant-velocity motion on a time interval if its velocity vector has the same value at every time in that interval. The motion is also called uniform motion. Its speed, the magnitude of its velocity, is constant as well.

Let $\mathbf{r}(t)$ denote the particle’s position at time $t$, and let $\mathbf{v}$ be its constant velocity. If its position at time $t_0$ is $\mathbf{r}_0$, the position at a later time follows from $\mathbf{v}$ and the elapsed time $t-t_0$.

The Constant-Velocity Position Formula

Velocity is the derivative of position, so constant velocity means $\mathbf{r}'(t)=\mathbf{v}$. Integrate this vector equation from the initial time $t_0$ to a time $t$. Vector integration acts on each coordinate, and the fundamental theorem of calculus applies coordinate by coordinate.

\mathbf{r}(t)-\mathbf{r}(t_0) = \int_{t_0}^{t}\mathbf{r}'(s)\,ds = \int_{t_0}^{t}\mathbf{v}\,ds = (t-t_0)\mathbf{v}.

Since $\mathbf{r}(t_0)=\mathbf{r}_0$, rearranging gives the position formula. The variable $s$ in the integral is a time variable of integration; the resulting position is evaluated at $t$.

Constant-velocity position formula: If a particle has constant velocity $\mathbf{v}$ and position $\mathbf{r}_0$ at time $t_0$, then throughout the interval of motion, $$ \mathbf{r}(t)=\mathbf{r}_0+(t-t_0)\mathbf{v}. $$

This derivation also explains why the initial time matters. The factor multiplying velocity is the elapsed time, not necessarily the time $t$ itself. The shortcut $\mathbf{r}(t)=\mathbf{r}_0+t\mathbf{v}$ is valid when the initial time is $t_0=0$, but not in general.

In coordinates, write $\mathbf{r}_0=(x_0,y_0,z_0)$ and $\mathbf{v}=(v_x,v_y,v_z)$. The vector formula means that each coordinate changes linearly:

\mathbf{r}(t) = \bigl(x_0+(t-t_0)v_x,\; y_0+(t-t_0)v_y,\; z_0+(t-t_0)v_z\bigr).

The same elapsed-time factor appears in all three coordinates. A coordinate with zero velocity component does not change, while a negative component means that coordinate decreases as time increases.

Worked Examples

Worked Example: Position After Five Seconds

A particle is at $\mathbf{r}_0=(1,-2,4)$ metres when $t_0=0$ seconds. Its constant velocity is $\mathbf{v}=(3,0,-2)$ metres per second. Find its position at $t=5$ seconds.

The elapsed time is $5-0=5$ seconds. Substitute the data into the formula:

\begin{aligned} \mathbf{r}(5) &=(1,-2,4)+5(3,0,-2)\\ &=(1,-2,4)+(15,0,-10)\\ &=(16,-2,-6)\ \text{metres}. \end{aligned}

Check the initial position by setting $t=0$ in $\mathbf{r}(t)=(1,-2,4)+t(3,0,-2)$: this gives $(1,-2,4)$ metres. Differentiate the position: $\mathbf{r}'(t)=(3,0,-2)$ metres per second, which is the specified velocity. Both checks agree with the model.

Worked Example: Starting at a Nonzero Time

At time $t_0=2$ seconds, a particle is at $(-1,3,2)$ metres and has constant velocity $(2,-1,4)$ metres per second. Find its position at $t=5$ seconds.

Here the elapsed time is $5-2=3$ seconds. Therefore,

\begin{aligned} \mathbf{r}(5) &=(-1,3,2)+(5-2)(2,-1,4)\\ &=(-1,3,2)+3(2,-1,4)\\ &=(-1,3,2)+(6,-3,12)\\ &=(5,0,14)\ \text{metres}. \end{aligned}

The full position function is $\mathbf{r}(t)=(-1,3,2)+(t-2)(2,-1,4)$. At $t=2$, the elapsed-time term is zero, so $\mathbf{r}(2)=(-1,3,2)$ metres. Differentiating gives $\mathbf{r}'(t)=(2,-1,4)$ metres per second. The initial position and constant velocity are both recovered.

Worked Example: Reading the Three-Dimensional Trajectory

A particle starts at $(1,0,-3)$ metres at time $t=0$, with constant velocity $(2,-1,2)$ metres per second. Find its position function and its position at $t=2$ seconds. Then describe the direction of its path.

Apply the formula with $t_0=0$:

\begin{aligned} \mathbf{r}(t) &=(1,0,-3)+t(2,-1,2)\\ &=(1+2t,-t,-3+2t)\ \text{metres}. \end{aligned}

At $t=2$, the displacement from the starting point is $2(2,-1,2)=(4,-2,4)$ metres, so

\mathbf{r}(2)=(1,0,-3)+(4,-2,4)=(5,-2,1)\ \text{metres}.

The direction of the trajectory is the velocity direction $(2,-1,2)$. To check the result, evaluate the position function at zero to get $(1,0,-3)$, and differentiate to get $\mathbf{r}'(t)=(2,-1,2)$ metres per second. At $t=2$, the coordinate formula gives $(1+4,-2,-3+4)=(5,-2,1)$, as required.

Why the Trajectory Is a Straight Line

The position formula has the form $\mathbf{r}_0+\lambda\mathbf{v}$, where $\lambda=t-t_0$ is a scalar. As time changes, the particle moves from $\mathbf{r}_0$ by a scalar multiple of one fixed vector. This is precisely the vector description of a straight line through $\mathbf{r}_0$ in the direction $\mathbf{v}$.

Geometric interpretation: If $\mathbf{v}\ne\mathbf{0}$, the particle’s trajectory lies on the line through $\mathbf{r}_0$ parallel to $\mathbf{v}$. The specified time interval determines which portion of that line is traced. If $\mathbf{v}=\mathbf{0}$, the position never changes, so the particle remains at $\mathbf{r}_0$.

For increasing time, the particle moves in the direction of $\mathbf{v}$; its speed is $\lVert\mathbf{v}\rVert$. A longer time interval produces a proportionally larger displacement. If time is restricted to $t\ge t_0$, the particle traces the ray beginning at $\mathbf{r}_0$ in the direction of $\mathbf{v}$, rather than necessarily tracing every point on the full line.

Units provide a useful check. Position and displacement have units of length, velocity has units of length per time, and elapsed time has units of time. Thus $(t-t_0)\mathbf{v}$ has units of length, as it must to be added to $\mathbf{r}_0$. The units of $t$ and $t_0$ must also match.

A Common Mistake: Using the Clock Time Instead of Elapsed Time

Suppose a particle is at $(4,1,-2)$ metres at $t_0=3$ seconds and moves with velocity $(1,2,0)$ metres per second. A tempting but incorrect calculation at $t=5$ is to multiply the velocity by $5$, obtaining $(4,1,-2)+5(1,2,0)=(9,11,-2)$. This treats the particle as though it had started at the given position at time zero.

The elapsed time from $3$ seconds to $5$ seconds is only $2$ seconds. The correct calculation is

\begin{aligned} \mathbf{r}(5) &=(4,1,-2)+(5-3)(1,2,0)\\ &=(4,1,-2)+(2,4,0)\\ &=(6,5,-2)\ \text{metres}. \end{aligned}

Substituting $t=3$ into the correct model gives $(4,1,-2)$, while the incorrect expression $(4,1,-2)+t(1,2,0)$ gives $(7,7,-2)$ at $t=3$. It fails the initial-position check. Whenever the initial time is not zero, use $t-t_0$, not $t$.

When This Model Is Useful

Constant-velocity motion is an appropriate model when the velocity vector does not change over the time interval under consideration. It provides a direct connection between an initial position, a fixed direction of travel, and later positions. In applications, it can serve as an idealized model over a short interval or in a situation where changes in velocity can be neglected.

The formula also makes the assumptions visible. The velocity must be constant on the interval, and the stated initial position must correspond to the stated initial time. If velocity changes, the position is not described by this constant-velocity formula over the entire interval. A particle with zero velocity is a special case: its constant position is not a moving straight-line trajectory, even though it satisfies the formula.

Check Your Understanding

Use the constant-velocity formula and check your results against the initial position and velocity.

A particle starts at $(2,-1,0)$ metres at $t=0$ and has velocity $(0,3,-2)$ metres per second. Find its position at $t=4$ seconds.
A particle is at $(1,2,3)$ metres at $t=2$ seconds and has velocity $(-1,0,2)$ metres per second. Find its position at $t=6$ seconds.
Explain why a particle with nonzero constant velocity traces a straight line.
What position does the constant-velocity formula give when $\mathbf{v}=\mathbf{0}$?
Why is it generally incorrect to use $t\mathbf{v}$ as the displacement when the given initial time is $t_0\ne0$?