Two- and three-dimensional motion

Follow one point in a plane

At the start, a point is 1 m left of a fixed origin and 3 m above it. One second later it is 1 m right and 2 m above. It has changed both coordinates: +2 m horizontally and −1 m vertically. A single position number cannot specify both locations.

Choose x positive to the right and y positive upward, with perpendicular axes measured in metres. Write a position as the ordered pair (x, y): the first slot belongs to x, the second to y. Our two locations are (−1, 3) m and (1, 2) m. The trailing m applies to each entry.

We follow one chosen point, rather than the changing outline of an extended object. Time t is elapsed time from the declared start, measured in seconds. The origin and coordinate directions stay fixed throughout this lesson. The motion will be supplied by an exact teaching formula, not reconstructed from measured points.

Supply the coordinate functions

For 0 ≤ t ≤ 2 s, supply x(t) = −1 m + (2 m/s)t and y(t) = 3 m − (1 m/s²)t². The symbol x(t) means the x coordinate at time t. Every added term has units m: a velocity coefficient times time, or an acceleration-sized coefficient times time squared, supplies a length.

To substitute comfortably, let τ = t/(1 s), the dimensionless numerical time in seconds. Then x = (−1 + 2τ)m and y = (3 − τ²)m. At t = 1 s, τ = 1, so (x, y) = (1, 2) m. At t = 2 s the position is (3, −1) m. These formulas prescribe the whole path on the stated interval, not just the three listed locations.

A position vector r points from the origin to the point. In these fixed directions, r = x eₓ + y eᵧ, where eₓ and eᵧ are perpendicular vectors of unit magnitude along the positive axes. They supply directions and have no physical unit. The signed coordinates x and y are scalar components; x eₓ and y eᵧ are component vectors.

Subtract endpoints component by component

Displacement Δr is final position minus initial position. The symbol Δ means a change. From 0 s to 2 s, Δx = 3 m − (−1 m) = 4 m and Δy = −1 m − 3 m = −4 m. Thus Δr = (4, −4) m: rightward and downward. Subtract matching components in the same fixed coordinate system.

The magnitude |Δr| is the nonnegative length of that vector. Because our axes are perpendicular, the Pythagorean rule gives |Δr| = √(Δx² + Δy²) = √(32 m²) = 4√2 m. The square-root symbol selects the nonnegative root. The signed component sum 4 m + (−4 m) = 0 m is not this length.

Displacement uses the endpoints. Distance travelled uses the length along the actual route. A curved route between the endpoints has a greater length than their straight connecting segment. We have not calculated the distance along this curved path by joining the three displayed locations.

Find rates and speed

Average velocity over a nonzero time interval is displacement divided by elapsed time, component by component. Instantaneous velocity is the limiting rate as the interval shrinks around the time of interest. Its components describe movement along the axes at that instant; the position components describe location.

Here is the local calculation for our curve. Let h be a nonzero time increment with t and t + h inside 0–2 s, and let ε = h/(1 s). The x quotient [x(t + h) − x(t)]/h is exactly 2 m/s. Expanding (τ + ε)² gives the y quotient [y(t + h) − y(t)]/h = −(2τ + ε)m/s. As h tends to zero, ε does too, leaving v = (2, −2τ)m/s. This limiting rate is also called the derivative of position. At an endpoint, use only the one-sided limit from inside the supplied interval.

At 1 s the velocity is (2, −2) m/s: rightward and downward, although the position (1, 2) m is rightward and above the origin. Speed is the magnitude of velocity: |v| = √(vₓ² + vᵧ²). It is 2√2 m/s here, rather than the zero signed sum of the two components. A point is at rest only when every velocity component is zero.

Acceleration is the rate of change of velocity. In this example vₓ stays 2 m/s while vᵧ changes by −2 m/s during each additional second. Thus a = (0, −2) m/s² throughout the supplied interval, with one-sided endpoint rates. This is a prescribed acceleration; no force or gravitational cause has been established.

Retain the third coordinate

Now describe a second trajectory in three-dimensional space. Keep the same x(t) and y(t), and add z(t) = 1 m + (1 m/s)t = (1 + τ)m on the same 0–2 s interval. The positive z direction is perpendicular to both x and y, toward the viewer out of the x–y plane. All three directions and the origin remain fixed.

Write its position as R = x eₓ + y eᵧ + z e_z, or as the ordered triple (x, y, z). A capital R distinguishes this trajectory from the planar one. Its velocity is V = (2, −2τ, 1)m/s because z gains 1 m each second; its acceleration is A = (0, −2, 0)m/s². The component positions, rates and units have the same meanings as before.

For three perpendicular directions, apply Pythagoras once to the x–y pair and then to z: |V| = √(Vₓ² + Vᵧ² + V_z²). At 1 s, |V| = √[(4 + 4 + 1)m²/s²] = 3 m/s. The x–y projection still looks like the planar path, whose speed there is 2√2 m/s. Omitting z has omitted a velocity component.

The table records exact stipulated values. The square roots at the other two times are left exact; these entries do not imply an instrument’s precision. Describing a path with three coordinates does not mean it fills a volume.

Read a projection and coordinate histories

The left panel plots spatial x against spatial y. Equal metre intervals have equal drawing lengths on both axes. Its solid curve is the x–y projection of R(t), and also the planar path r(t). Circles, diamonds and squares locate 0 s, 1 s and 2 s. The internal zero lines meet at the fixed origin; the lower-left plot corner is (−2, −2) m.

The right panel plots x(t), y(t) and z(t) against the same time axis. Horizontal movement across this panel means later time, not a spatial direction. The solid blue line, dashed purple curve and dotted green line identify the three coordinates. Read all three at one time to recover R(t). The adjoining table supplies the same selected values.

At 1 s, the y and z histories both equal 2 m, so their diamond is shared. At 2 s, x and z both equal 3 m, so their square is shared. A shared symbol keeps both component values; it is not a missing coordinate. All curves stop at 0 s and 2 s. Their smooth form comes from the whole-interval supplied functions, not from treating three markers as proof of a trajectory.

Try covering the right panel’s z line. The spatial projection cannot tell whether z is increasing, decreasing or constant. For example, replacing z(t) by the constant 1 m would leave that projection unchanged while changing the three-dimensional motion.

Compare a route with its endpoints

Use a separate exact route to calculate distance without an arc-length method. Start at (0, 0, 0) m. Move uniformly 3 m along +x in 1 s, then 4 m along +y in the next 2 s, then 12 m along +z in the next 4 s. Each leg is straight and monotone, meaning it does not reverse along that leg. The table gives the endpoints.

The distance is the sum of these positive leg lengths: 3 m + 4 m + 12 m = 19 m. The displacement is (3, 4, 12) m, with magnitude √[(9 + 16 + 144)m²] = 13 m. These answers describe different quantities. The route’s total duration is 7 s.

Average speed is total distance divided by total elapsed time: 19/7 m/s. Average velocity is total displacement divided by that time: (3/7, 4/7, 12/7)m/s, with magnitude 13/7 m/s. Average speed and the magnitude of average velocity differ when the path changes direction; neither is obtained by adding the velocity components.

The sharp corners are idealisations. At 1 s, the left-side velocity is (3, 0, 0)m/s and the right-side velocity is (0, 2, 0)m/s. There is no single ordinary instantaneous velocity at that corner, and no finite ordinary acceleration is asserted there. The finite whole-route averages remain well defined.

Keep the coordinate model in scope

Our smooth trajectories give x, y and z as separate functions of the same time. We can evaluate each function and combine its result at that time. This useful separation is not a claim of statistical independence or a rule that physical causes never couple different directions.

For a supplied constant acceleration vector in fixed Cartesian directions, each component follows the constant-acceleration pattern: coordinate = initial coordinate + initial component velocity × t + ½ × component acceleration × t². Our x, y and z formulas match it with initial position (−1, 3, 1)m, initial velocity (2, 0, 1)m/s and acceleration (0, −2, 0)m/s². Keep the declared time origin and interval when using that pattern.

A trajectory describes how a chosen point moves. Explaining why it follows that trajectory requires a suitable model of interactions and conditions. Newton’s laws and free-body diagrams is the next mapped subject. Here, the original stipulated motions let us practise coordinates and rates without assuming that later explanation.

Worked examples

Try it yourself

1. At t = 0.25 s in the planar model, what is x? Is the point moving left or right along x?

   Show the answer and reasoning

   x = −0.5 m; it is moving right along x at +2 m/s.

   τ = 0.25 gives x = (−1 + 0.5)m. The coordinate is left of the origin, but its rate is positive. Position sign does not determine velocity sign.

2. At the supplied start t = 0 s, the planar model has vᵧ = 0. Is the point at rest?

   Show the answer and reasoning

   No. Its x velocity component is +2 m/s, so its planar speed is 2 m/s.

   Rest requires both velocity components to vanish. The rates at this endpoint are one-sided limits inside the supplied interval; they do not assert an earlier trajectory.

3. At 1 s the planar velocity is (2, −2)m/s. Someone adds the components and reports zero speed. What is the correct speed and why?

   Show the answer and reasoning

   2√2 m/s. The perpendicular components combine through their squared magnitudes, not their signed sum.

   Speed² = (2 m/s)² + (−2 m/s)² = 8 m²/s². Opposite signs on different perpendicular axes do not cancel the vector.

4. Compare R(t) with a trajectory S(t) that keeps the same x(t), y(t) and domain but has z(t) = 1 m throughout. Do their identical x–y projections imply equal speeds at 1 s?

   Show the answer and reasoning

   No. R has speed 3 m/s, while S has speed 2√2 m/s at 1 s.

   R has V_z = 1 m/s; S has a zero z rate. Adding that third squared rate changes the full velocity norm, even though the spatial projection is unchanged.

5. For the planar y(t), calculate the average y rate from 1 s to 1.5 s. Is it the instantaneous y rate at 1 s?

   Show the answer and reasoning

   The average y rate is −2.5 m/s; the instantaneous rate at 1 s is −2 m/s. They differ.

   y(1 s) = 2 m and y(1.5 s) = 0.75 m. Their change −1.25 m divided by 0.5 s gives −2.5 m/s. The limiting component formula vᵧ = −2τ m/s gives −2 m/s at τ = 1.

6. On the three-leg route, can the endpoint length 13 m replace distance 19 m when calculating average speed? Does assigning the incoming velocity to the 1 s corner create an ordinary instantaneous velocity there?

   Show the answer and reasoning

   No to both. Average speed uses 19 m, and the corner’s unequal one-sided velocities prevent an ordinary instantaneous velocity.

   The endpoint length belongs to displacement, giving |average velocity| = 13/7 m/s. Average speed is 19/7 m/s. Choosing one side at the sharp corner does not make its two-sided velocity limit exist.

What to carry forward

• Declare fixed coordinate directions, an origin, units and the time domain before combining components.
• Position, displacement, velocity and acceleration are different vectors; speed is the nonnegative magnitude of velocity.
• For perpendicular coordinates, squared components determine a magnitude. Their signed sum does not.
• An x–y projection omits z. Coordinate histories at one common time retain the full three-coordinate position.
• Distance follows a route; displacement joins its endpoints. Average speed and |average velocity| can therefore differ.
• Separate component calculations do not establish uncoupled physical causes, and ideal sharp corners need not have ordinary instantaneous rates.