Velocity and acceleration for motion along a curve
Master velocity and acceleration for motion along a curve in Year 12 Specialist Mathematics (QCAA, Queensland). When a particle's position is a vector function of time \(\mathbf{r}(t)\), its velocity is the derivative \(\mathbf{v}=\dot{\mathbf{r}}\) — always tangent to the path — and its acceleration is \(\mathbf{a}=\ddot{\mathbf{r}}\).
You will learn to differentiate a vector function component by component, read the speed as the magnitude of velocity, find the direction of motion, and pin down special instants using \(\mathbf{v}\cdot\mathbf{a}=0\) or a minimum speed — the core skills of vector calculus that lead into projectile and circular motion.
Theory
For a particle moving along a curve, its position is a vector function of time \(\mathbf{r}(t)\). In Year 12 Specialist Mathematics (QCAA, Queensland) the velocity is the derivative \(\mathbf{v}=\dot{\mathbf{r}}\) (always tangent to the path), the acceleration is \(\mathbf{a}=\dot{\mathbf{v}}=\ddot{\mathbf{r}}\), and the speed is the magnitude \(|\mathbf{v}|\) — each found by differentiating component by component.
A particle traces a path in the plane (or in space). Its position vector \(\mathbf{r}(t)\) gives where the particle is at time \(t\); each component is a function of \(t\), e.g. \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\).
The velocity is the rate of change of position, \(\mathbf{v}=\dfrac{d\mathbf{r}}{dt}=\dot{\mathbf{r}}\). You differentiate each component separately. Geometrically, \(\mathbf{v}\) always points tangent to the path, in the direction the particle is travelling.
The acceleration is the rate of change of velocity, \(\mathbf{a}=\dfrac{d\mathbf{v}}{dt}=\ddot{\mathbf{r}}\), again found component by component. Unlike velocity, the acceleration need not be tangent to the path — on a curve it usually points to one side.
The speed is the magnitude of the velocity, \(|\mathbf{v}|=\sqrt{\dot{x}^2+\dot{y}^2}\) (add \(\dot{z}^2\) in three dimensions). Speed is a scalar; velocity is a vector that also carries the direction of motion.
For a particle with position \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\), the velocity is the derivative of position:
The acceleration is the derivative of velocity (the second derivative of position):
The speed is the magnitude of the velocity:
How to analyse motion along a curve
- Differentiate for velocity. From \(\mathbf{r}(t)\), find \(\mathbf{v}=\dot{\mathbf{r}}\) by differentiating each component in \(t\).
- Differentiate again for acceleration. Find \(\mathbf{a}=\dot{\mathbf{v}}=\ddot{\mathbf{r}}\), once more component by component.
- Substitute the time if a value at an instant is needed, e.g. \(\mathbf{v}(2)\) or \(\mathbf{a}(1)\).
- Take the magnitude for speed or the magnitude of acceleration: \(|\mathbf{v}|=\sqrt{\dot{x}^2+\dot{y}^2}\); for a special instant use \(\mathbf{v}\cdot\mathbf{a}=0\) or minimise \(|\mathbf{v}|^2\).
Differentiate each component for \(\mathbf{v}\), then again for \(\mathbf{a}\):
| \(\mathbf{v}\) | \(=\) | \(\dfrac{d}{dt}(t)\,\mathbf{i}+\dfrac{d}{dt}(t^2)\,\mathbf{j}\) |
| \(=\) | \(\mathbf{i}+2t\mathbf{j}\) | |
| \(\mathbf{a}\) | \(=\) | \(\dfrac{d}{dt}(1)\,\mathbf{i}+\dfrac{d}{dt}(2t)\,\mathbf{j}\) |
| \(=\) | \(2\mathbf{j}\) |
Substitute \(t=1\) for the velocity at that instant:
| \(\mathbf{v}(1)\) | \(=\) | \(\mathbf{i}+2(1)\mathbf{j}\) |
| \(=\) | \(\mathbf{i}+2\mathbf{j}\) |
\(\mathbf{v}=\mathbf{i}+2t\mathbf{j}\), \(\mathbf{a}=2\mathbf{j}\); at \(t=1\), \(\mathbf{v}=\mathbf{i}+2\mathbf{j}\) (tangent to the curve).
Differentiate for the velocity, substitute \(t=3\), then take the magnitude:
| \(\mathbf{v}\) | \(=\) | \(8\mathbf{i}+2t\mathbf{j}\) |
| \(\mathbf{v}(3)\) | \(=\) | \(8\mathbf{i}+2(3)\mathbf{j}\) |
| \(=\) | \(8\mathbf{i}+6\mathbf{j}\) | |
| \(|\mathbf{v}(3)|\) | \(=\) | \(\sqrt{8^2+6^2}\) |
| \(=\) | \(\sqrt{64+36}\) | |
| \(=\) | \(\sqrt{100}\) | |
| \(=\) | \(10\) |
The speed at \(t=3\) is \(10\) m/s.
Find \(\mathbf{v}\) and \(\mathbf{a}\); perpendicular vectors have \(\mathbf{v}\cdot\mathbf{a}=0\):
| \(\mathbf{v}\) | \(=\) | \(2t\mathbf{i}+(2t-6)\mathbf{j}\) |
| \(\mathbf{a}\) | \(=\) | \(2\mathbf{i}+2\mathbf{j}\) |
| \(\mathbf{v}\cdot\mathbf{a}\) | \(=\) | \((2t)(2)+(2t-6)(2)\) |
| \(=\) | \(4t+4t-12\) | |
| \(=\) | \(8t-12\) |
Set the dot product to zero and solve for \(t\):
| \(8t-12\) | \(=\) | \(0\) |
| \(8t\) | \(=\) | \(12\) |
| \(t\) | \(=\) | \(1.5\) |
The velocity is perpendicular to the acceleration at \(t=1.5\) s.
Minimise \((\text{speed})^2\); differentiate the velocity, then form \(|\mathbf{v}|^2\):
| \(\mathbf{v}\) | \(=\) | \(2t\mathbf{i}+(2t-8)\mathbf{j}\) |
| \(|\mathbf{v}|^2\) | \(=\) | \((2t)^2+(2t-8)^2\) |
| \(=\) | \(4t^2+4t^2-32t+64\) | |
| \(=\) | \(8t^2-32t+64\) |
Differentiate \(|\mathbf{v}|^2\), set it to zero, then take the root at that time:
| \(\dfrac{d}{dt}|\mathbf{v}|^2\) | \(=\) | \(16t-32\) |
| \(16t-32\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(2\) |
| \(|\mathbf{v}|^2\) | \(=\) | \(8(2)^2-32(2)+64\) |
| \(=\) | \(32\) | |
| \(|\mathbf{v}|\) | \(=\) | \(4\sqrt{2}\) |
The minimum speed is \(4\sqrt{2}\) m/s, at \(t=2\) s.
Common pitfalls
Frequently asked questions
How do you find velocity and acceleration from a position vector?
Differentiate the position vector \(\mathbf{r}(t)\) component by component to get the velocity \(\mathbf{v}=\dot{\mathbf{r}}\); differentiate again to get the acceleration \(\mathbf{a}=\dot{\mathbf{v}}=\ddot{\mathbf{r}}\).
What is the difference between speed and velocity?
Velocity \(\mathbf{v}\) is a vector carrying both magnitude and direction. Speed is the scalar magnitude \(|\mathbf{v}|=\sqrt{\dot{x}^2+\dot{y}^2}\), a single number with no direction.
Why is velocity always tangent to the path?
The velocity is the instantaneous rate of change of position, so it points in the direction the particle is heading at that instant — which is along the curve, i.e. tangent to the path.
How do you find the speed at a particular time?
Differentiate for the velocity, substitute the time to get \(\mathbf{v}\) at that instant, then take its magnitude \(|\mathbf{v}|=\sqrt{\dot{x}^2+\dot{y}^2}\).
When is the velocity perpendicular to the acceleration?
When their dot product is zero, \(\mathbf{v}\cdot\mathbf{a}=0\). This is the instant the speed is momentarily stationary — a maximum or minimum of the speed.
How do you find the minimum speed?
Minimise \(|\mathbf{v}|^2=\dot{x}^2+\dot{y}^2\): differentiate it with respect to \(t\), set the derivative to zero to find the time, then take the square root of \(|\mathbf{v}|^2\) there.