Lesson 2 · 30 min

Position, Velocity and Acceleration Vectors

Everything in curvilinear motion is built from three vectors: where the particle is, how fast and which way it moves, and how that motion is changing. This lesson defines each one and shows how they relate to the path.

Learning objectives

Define the position vector \(\rvec\) and the displacement \(\Delta\rvec\), and distinguish displacement from distance travelled.
Explain why the velocity \(\vvec = d\rvec/dt\) is always tangent to the path, with magnitude equal to the speed \(v = ds/dt\).
Compute average velocity, average speed and average acceleration over a time interval.
Define the acceleration \(\avec = d\vvec/dt\), explain why it is generally not tangent to the path, and use the hodograph to see its direction.

Position and displacement

Choose a fixed origin \(O\). The position vector \(\rvec(t)\) is the arrow from \(O\) to the particle at time \(t\). As the particle moves, the tip of \(\rvec\) traces the path.

Between times \(t\) and \(t + \Delta t\) the particle moves from \(P\) to \(P'\). Its displacement is the change in position:

\[ \Delta\rvec = \rvec(t + \Delta t) - \rvec(t) \]

Displacement is a vector: the straight arrow from \(P\) to \(P'\). The distance travelled \(\Delta s\) is the length of the path between \(P\) and \(P'\), a positive scalar. On a curve the path is longer than the chord, so

\[ \Delta s \ge |\Delta\rvec| \]

with equality only for straight-line motion in one direction.

Example 2.1 — Half a lap of a track

A runner goes half a lap of a circular track of radius \(R = 50\ \text{m}\), from the east side to the west side, in \(30\ \text{s}\). Find the distance travelled, the size of the displacement, the average speed and the size of the average velocity.

Show solution

Distance. Half the circumference: \(\Delta s = \pi R = 50\pi \approx 157.1\ \text{m}\).

Displacement. The straight line from east to west is a diameter: \(|\Delta\rvec| = 2R = 100\ \text{m}\), pointing west.

Averages.

\[ \begin{aligned} \text{average speed} &= \frac{\Delta s}{\Delta t} = \frac{157.1}{30} \approx 5.236\ \text{m/s} \\ |\vvec_\text{avg}| &= \frac{|\Delta\rvec|}{\Delta t} = \frac{100}{30} \approx 3.333\ \text{m/s}\ \text{(west)} \end{aligned} \]

Interpret. The two "averages" differ because speed follows the path while velocity only cares about the start and end points. After a full lap the displacement, and so the average velocity, would be zero.

Velocity: always tangent to the path

The average velocity over the interval is the displacement divided by the time taken. It points along the chord \(PP'\):

\[ \vvec_\text{avg} = \frac{\Delta\rvec}{\Delta t} \]

Now shrink \(\Delta t\). \(P'\) slides toward \(P\), and the chord \(PP'\) turns until, in the limit, it lies along the tangent to the path at \(P\). The limit is the instantaneous velocity:

Velocity

\[ \vvec = \lim_{\Delta t \to 0}\frac{\Delta\rvec}{\Delta t} = \frac{d\rvec}{dt}, \qquad |\vvec| = v = \frac{ds}{dt} \]

\(\vvec\) is tangent to the path and points in the direction of motion. Its magnitude is the speed \(v\), the rate at which distance along the path is covered.

Figure 2.1 The chord \(\Delta\rvec\) (black) joins the particle's positions at \(t\) and \(t + \Delta t\). The dashed green arrow is the average velocity \(\Delta\rvec/\Delta t\); the solid green arrow is the true velocity \(\vvec\) at \(P\). Shrink \(\Delta t\): the chord turns onto the tangent, the average velocity closes in on \(\vvec\), and chord length over arc length approaches 1.

Acceleration and the hodograph

Acceleration measures how the velocity vector changes. The average acceleration is \(\avec_\text{avg} = \Delta\vvec/\Delta t\), where \(\Delta\vvec = \vvec(t + \Delta t) - \vvec(t)\), and the instantaneous acceleration is its limit:

Acceleration

\[ \avec = \lim_{\Delta t \to 0}\frac{\Delta\vvec}{\Delta t} = \frac{d\vvec}{dt} = \frac{d^2\rvec}{dt^2} \]

To see which way \(\avec\) points, draw every velocity vector from one common origin. Their tips trace a curve called the hodograph. The hodograph is to \(\vvec\) what the path is to \(\rvec\): just as \(\vvec\) is tangent to the path, \(\avec\) is tangent to the hodograph.

Figure 2.2 A particle goes round the ellipse \(x = 3\cos t,\ y = 2\sin t\) (metres, seconds). The small plot on the right is its hodograph: each velocity vector drawn from the origin \(O_v\). The red acceleration is tangent to the hodograph. On the path it is not tangent: it points into the curve, and here it always points at the center of the ellipse. Only at the ends of the axes is it perpendicular to the path, which is when the speed is at a maximum or minimum.

Two facts follow and hold for every path:

\(\avec\) is generally not tangent to the path. It has a part along \(\vvec\), which changes the speed, and a part across \(\vvec\), which turns the direction. Lesson 6 names these \(a_t\) and \(a_n\).
On a curve, \(\avec\) points to the concave (inner) side. The velocity must be turned toward the inside of the bend, so the acceleration always has a component that way.

Worked example

Example 2.2 — A particle given by its position vector

A particle moves in the plane with \(\rvec = (2t^2)\,\ihat + (8t - t^3)\,\jhat\ \text{m}\), where \(t\) is in seconds. At \(t = 2\ \text{s}\), find its velocity, speed and acceleration, and say whether \(\avec\) is tangent to the path.

Show solution

Differentiate. \(\ihat\) and \(\jhat\) are fixed, so differentiate each part (Lesson 3 does this in detail):

\[ \begin{aligned} \vvec &= \frac{d\rvec}{dt} = (4t)\,\ihat + (8 - 3t^2)\,\jhat \\ \avec &= \frac{d\vvec}{dt} = 4\,\ihat - (6t)\,\jhat \end{aligned} \]

At \(t = 2\ \text{s}\).

\[ \begin{aligned} \vvec &= 8\,\ihat - 4\,\jhat\ \text{m/s}, \qquad v = \sqrt{8^2 + 4^2} = \sqrt{80} \approx 8.944\ \text{m/s} \\ \avec &= 4\,\ihat - 12\,\jhat\ \text{m/s}^2, \qquad |\avec| = \sqrt{16 + 144} \approx 12.65\ \text{m/s}^2 \end{aligned} \]

Tangent? \(\avec\) would be tangent only if it were parallel to \(\vvec\). The directions \(8\ihat - 4\jhat\) (slope \(-\tfrac12\)) and \(4\ihat - 12\jhat\) (slope \(-3\)) differ, so \(\avec\) is not tangent: part of it changes the speed and part of it turns the path.

Check your understanding

Key takeaways

\(\rvec\) runs from a fixed origin to the particle; the displacement \(\Delta\rvec\) is a chord, and the distance \(\Delta s\) along the path is never shorter than it.
\(\vvec = d\rvec/dt\) is tangent to the path in the direction of motion, and \(|\vvec| = v = ds/dt\) is the speed.
\(\avec = d\vvec/dt\) is tangent to the hodograph, not to the path. On a curve it points to the concave side.
Average velocity \(\Delta\rvec/\Delta t\) and average speed \(\Delta s/\Delta t\) are different things.
Next: Lesson 3 writes these vectors in \(x\)–\(y\) components and computes them.