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Parametric models of straight-line motion

Model constant-velocity motion with parametric equations. Distinguish position, displacement, distance and speed; find a path equation, valid time interval and direction with original worked practice.

Before you startParametric coordinates, straight-line graphs and Pythagoras

01 / Define the origin, directions, units and time

A position rule needs a coordinate system.

Suppose x measures metres east and y measures metres north of a chosen origin. Let t be seconds after an agreed start. The rules x = 2 + 3t, y = 3 + 4t then give a position for each time.

At t = 0, read the starting position. Coefficients of t give constant velocity components.

The model compares moving east and north, moving west and north, moving south, and remaining still. You control every change of time.

Position and displacementExplore

Velocity (east, north)Elapsed time

x = 2 + 3t; y = 3 + 4t, with metres and seconds.

t = 0 s: position (2, 3) m.

Displacement from the start: (0, 0) m.

Speed 5 m/s; distance travelled 0 m.

Distance from the origin: √13 ≈ 3.606 m.

Grey marks the full six-second path. Blue joins the starting point to the selected position; the dashed triangle separates its signed components. The axis scales differ, so use the numbers to calculate angles and lengths.

02 / Separate position from displacement

Coordinates are measured from the origin; displacement is measured from the earlier position.

For x = 2 + 3t, y = 3 + 4t, find the position and displacement after 2 s.Worked example

Start: (2, 3) m

Set t = 0.

At t = 2: (8, 11) m

Substitute into both position rules.

Displacement: (8 − 2, 11 − 3) = (6, 8) m

Subtract the starting position.

Displacement magnitude: √(6² + 8²) = 10 m

This is not the distance from the origin.

The distance from the origin at that time is √(8² + 11²) = √185 m. Changing the chosen origin changes the coordinates, but not the physical displacement between two positions.

Watch the displacement triangle grow

Pause, replay or seek freely. The notes explain the same idea and stay in view.

03 / Read constant velocity components

A signed component includes its direction.

In x = x₀ + ut, y = y₀ + vt, the constants u and v give horizontal and vertical velocity components. Their units are coordinate units divided by time units.

A robot has x = 12 − 4t, y = −1 + 3t, with metres and seconds.Worked example

Starting position: (12, −1) m

It begins east and south of the origin.

Velocity components: (−4, 3) m/s

4 m/s west and 3 m/s north.

After 5 s: (−8, 14) m

The negative x-position means west of the origin; the negative x-velocity means moving west.

A negative position and a negative velocity describe different facts. A point east of the origin can still be moving west.

04 / Calculate speed and distance travelled

Combine perpendicular velocity components using Pythagoras.

For constant components (−4, 3) m/s, find the speed and distance travelled in 5 s.Worked example

Speed = √((−4)² + 3²) = 5 m/s

Speed is non-negative.

Distance travelled = 5 × 5 = 25 m

Constant velocity gives a straight path with no reversal.

For constant velocity: speed = √(u² + v²), distance over Δt = speed × Δt.

For a journey that changes direction, the length of the total displacement can be smaller than the distance travelled. Do not extend the straight-path equality to a trip with a turn.

05 / Measure displacement between any two times

The interval length is t₂ − t₁.

Use x = 2 + 3t, y = 3 + 4t between t = 1 and t = 4.Worked example

Positions: (5, 7) and (14, 19) m

Evaluate both times.

Displacement: (9, 12) m

Later minus earlier.

Time interval = 3 s; distance = 5 × 3 = 15 m

Do not use 4 s for this interval.

Because the velocity is constant, the starting offset cancels when subtracting positions: Δx = uΔt, Δy = vΔt.

06 / Find a direction with the correct quadrant

An angle alone is ambiguous unless its reference direction is stated.

Find the direction of velocity (−4, 3) m/s.Worked example

4 west and 3 north

Identify the quadrant before using an inverse tangent.

α = arctan(3/4) ≈ 36.87°

The acute angle is north of west.

Direction: 36.87° north of west

Equivalently 143.13° anticlockwise from east.

Bearing: 306.87°

This is measured clockwise from north, written with three digits before the decimal.

Stationary motion has speed zero and no direction of travel. For a vertical velocity, use north or south directly rather than dividing by a zero horizontal component.

07 / Eliminate time to find the path

The path equation does not retain the speed or timing.

Find the path of x = 2 + 3t, y = 3 + 4t for 0 ≤ t ≤ 6.Worked example

t = (x − 2)/3

Solve the x-rule for time.

y = 3 + 4(x − 2)/3 = (4x + 1)/3

Eliminate t.

2 ≤ x ≤ 20, and 3 ≤ y ≤ 27

Retain the time-window restrictions.

Trace from (2, 3) towards (20, 27)

The physical path is this segment, not the entire line.

The rules x = 2 + 6s, y = 3 + 8s, 0 ≤ s ≤ 3, trace the same segment at twice the speed if s also measures seconds. A Cartesian line by itself cannot tell you how quickly it is traversed.

08 / Handle vertical and stationary motion

Do not divide by a component that is zero.

Find the path for x = 7, y = 10 − 2t, 0 ≤ t ≤ 4.Worked example

x = 7 throughout

The horizontal velocity component is zero.

2 ≤ y ≤ 10

The point moves south from (7, 10) to (7, 2).

Speed = 2 m/s; distance = 8 m

The path is a vertical segment.

If both components are zero, both coordinates stay fixed. The path is a single point and the distance travelled is zero.

09 / Solve an event using its coordinate condition

Then check the other coordinate and the allowed time window.

For x = 2 + 3t, y = 3 + 4t, 0 ≤ t ≤ 6, when is the point at height y = 15 m?Worked example

3 + 4t = 15 gives t = 3 s

Solve the height condition.

x = 2 + 3(3) = 11 m

The position is (11, 15) m.

3 lies in [0, 6]

The event occurs during the modelled journey.

The same equations would reach y = 31 at t = 7, but that lies outside the stated journey. Algebraic extrapolation is not evidence that the real motion continues.

10 / Build a model from two timed positions

Use a constant-velocity assumption explicitly.

A point moves at constant velocity. At t = 2 s it is at (5, 7) m; at t = 6 s it is at (17, −1) m. Find its rules.Worked example

Velocity components = ((17 − 5)/4, (−1 − 7)/4) = (3, −2) m/s

The elapsed time is 4 s.

x = 5 + 3(t − 2), y = 7 − 2(t − 2)

Anchor the rules at the known time.

x = 3t − 1, y = 11 − 2t

Equivalent expanded rules.

A point travels 3 m east and then 4 m north. Is its distance travelled 5 m?

Hint

Add the lengths of the two parts.

Worked solution

No. Distance travelled is 7 m. Displacement is (3, 4) m with magnitude 5 m. A single constant-velocity straight-line model does not describe this turn.

12 / Recap

A path, a position and a motion are different descriptions.

Define origin, axis directions, units and time zero.
Read constant velocity components from the coefficients of time.
Subtract positions for displacement; use speed and elapsed time for distance.
Elimination gives a path, so preserve its interval and direction separately.
Check whether an event occurs inside the modelled time window.
State the constant-velocity assumption and avoid unjustified extrapolation.

Section 1 of 12 · Define the origin, directions, units and time