Letting t stand for time turns a pair of parametric equations into a description of where something actually is, moment by moment.
9 min read2 worked examples
Modelling motion parametrically
Parametric equations are a natural way to model an object's position over time, with $t$ representing time and $(x,y)$ its coordinates. For an object moving with constant velocity between two known points over a fixed time interval, the position at time $t$ can be written directly using the displacement between the points.
MethodConstant-velocity motion between two points
1
Suppose an object moves from point $(x_1,y_1)$ at $t=0$ to point $(x_2,y_2)$ at $t=T$, with constant velocity throughout.
Set up the scenario: known start and end positions, known total time.
2
The total displacement is $(x_2-x_1, y_2-y_1)$, achieved over time $T$, so the velocity components are $\dfrac{x_2-x_1}{T}$ and $\dfrac{y_2-y_1}{T}$.
Constant velocity = total displacement ÷ total time, in each direction separately.
3
Position at time $t$: $x(t) = x_1 + \dfrac{x_2-x_1}{T}t$, $\quad y(t) = y_1 + \dfrac{y_2-y_1}{T}t$.
Starting position plus (constant velocity) × (time elapsed), in each coordinate.
4
These are valid for $0\le t\le T$, matching the interval over which the motion is modelled.
The domain of t is fixed by the physical time interval given in the problem.
▪ derived
Worked example 1
Modelling constant-velocity motion between two points
A particle moves with constant velocity in a straight line from $(2,5)$ at time $t=0$ seconds to $(14,-1)$ at time $t=6$ seconds. Find parametric equations for the particle's position at time $t$, and hence find its Cartesian path.
MethodFind the velocity components from the displacement and time, then eliminate t
1
Displacement: $\Delta x = 14-2=12$, $\Delta y=-1-5=-6$, over $T=6$ seconds.
Total change in each coordinate, and the total time taken.
2
Velocity components: $\dfrac{12}{6}=2$ (in $x$), $\dfrac{-6}{6}=-1$ (in $y$).
Constant velocity = displacement ÷ time, applied to each coordinate.
3
Position at time $t$: $x(t) = 2+2t$, $\quad y(t) = 5-t$, for $0\le t\le6$.
Starting position plus velocity × time, using the formula derived above.
4
Check: at $t=6$, $x=2+12=14$ and $y=5-6=-1$. ✓ Matches the given end point.
Verify the parametric equations against the known end point.
5
Eliminate $t$: from $y=5-t$, $t=5-y$. Substitute into $x=2+2t$: $x=2+2(5-y)=12-2y$.
Use the simpler equation to isolate t, then substitute into the other.
6
So $x=12-2y$, i.e. $x+2y=12$ — a straight line, as expected for constant-velocity motion. Since $0\le t\le6$, $x$ ranges from $2$ to $14$.
Constant velocity always produces straight-line motion; state the restricted range of x from the time domain.
▪ solved
Look deeper
This confirms a general fact: constant-velocity motion always traces a straight line, because both $x(t)$ and $y(t)$ are linear in $t$, and eliminating a linear parameter between two linear equations always produces another linear (Cartesian) equation. The parametric form additionally encodes when the particle is at each point — information the Cartesian equation $x+2y=12$ alone does not capture.
Worked example 2
Deciding whether two moving objects collide · Challenge
Two objects move with constant velocity. Object $A$ has position $(3t,4t)$ at time $t$; object $B$ has position $(20-2t,5+3t)$ at time $t$, both for $t\ge0$. Determine whether the two objects ever occupy the same point at the same time.
MethodMatch x-coordinates to find a candidate time, then check whether the y-coordinates also agree at that same t
1
A collision requires both $x_A(t)=x_B(t)$ and $y_A(t)=y_B(t)$ at the same value of $t$ — matching only one coordinate is not enough, since the two objects' paths could cross without them ever being there at the same time.
Both objects must be at the same place at the same instant, not just on paths that geometrically cross.
2
Set $x_A=x_B$: $3t=20-2t \Rightarrow 5t=20 \Rightarrow t=4$.
Find the only time at which the x-coordinates could possibly match.
3
At $t=4$: $y_A=4(4)=16$, but $y_B=5+3(4)=17$.
Check the y-coordinates at that same candidate time.
4
Since $16\ne17$, the two objects are at different heights when their $x$-coordinates match — they do not collide.
The x-coordinates only match at one instant, and the y-coordinates disagree there.
5
Geometrically, the two straight-line paths do cross at some point (since they aren't parallel) — but the objects pass through that crossing point at different times, so they miss each other.
Distinguishes a geometric path-crossing from an actual collision in time.
▪ solved
Common mistake
Finding a value of $t$ that matches the $x$-coordinates (or the $y$-coordinates) and stopping there. Both coordinates must match at the same $t$ for a genuine collision — this is the parametric analogue of the "points of intersection" idea from the previous lesson, but with the extra requirement that the parameter itself (time) must also agree.
Check yourself
Two objects move with constant velocity. Object $A$ has position $(1+2t,3+t)$ at time $t$; object $B$ has position $(9-2t,1+2t)$ at time $t$, both for $t\ge0$. Determine whether the two objects collide, and if so, find the time and position. (Hint: match the x-coordinates first to find a candidate time, then check the y-coordinates at that same t.)
Method
1
Set $x_A=x_B$: $1+2t=9-2t \Rightarrow 4t=8 \Rightarrow t=2$.
Find the candidate time where the x-coordinates match.
2
At $t=2$: $y_A=3+2=5$, and $y_B=1+2(2)=5$.
Check the y-coordinates at that same t.
3
Since both coordinates agree at $t=2$, the objects do collide, at the point $(1+2(2),\,5)=(5,5)$, at time $t=2$.
Both x and y match at the same t, so this is a genuine collision.