Acceleration Describes Change in Velocity
Velocity records how position changes over time. But velocity itself can also change: an object may speed up, slow down, or move in a new direction. The vector that describes this change is acceleration.
For motion along a space curve, position and velocity are vector-valued functions of time. Differentiating velocity gives acceleration. Because the derivative of a vector-valued function is computed component by component, each coordinate of acceleration measures the rate of change of the corresponding velocity coordinate.
In three dimensions, write the velocity in components as \(\mathbf{v}(t)=\langle v_1(t),v_2(t),v_3(t)\rangle\). Then $$ \mathbf{a}(t)=\langle v_1'(t),v_2'(t),v_3'(t)\rangle. $$ The same rule applies in two dimensions, with two components. Acceleration is a vector, not a single number: its components give the rates of change of the velocity components.
If position is measured in metres and time in seconds, velocity has units of metres per second and acceleration has units of metres per second squared. Each acceleration component describes the change in the corresponding velocity component per second.
Why Differentiation Is Componentwise
The componentwise rule follows from the definition of a vector derivative. A vector difference quotient is formed by subtracting vectors and dividing by a scalar. Both operations act coordinate by coordinate. Taking the limit therefore gives the derivative of each component.
This calculation assumes the component derivatives exist at \(t\). The last equality follows because a limit of a finite-dimensional vector exists exactly when the limits of its components exist, in which case the vector limit consists of those component limits. If velocity is itself the derivative of position, differentiating once more gives \(\mathbf{a}(t)=\mathbf{r}''(t)\), wherever that second derivative exists.
Worked Example: Acceleration as Speed Changes on a Line
Worked Example: Acceleration as Speed Changes on a Line
An object moves along the \(x\)-axis with position \(\mathbf{r}(t)=\langle t^2,0,0\rangle\) metres, where \(t\) is measured in seconds. Find its velocity and acceleration, and compare its speed at \(t=1\) and \(t=2\).
Differentiate each position component to obtain velocity: $$ \mathbf{v}(t)=\mathbf{r}'(t) =\left\langle \frac{d}{dt}(t^2),\frac{d}{dt}(0),\frac{d}{dt}(0)\right\rangle =\langle 2t,0,0\rangle\ \text{m/s}. $$ Differentiate the velocity components to obtain acceleration: $$ \mathbf{a}(t)=\mathbf{v}'(t) =\left\langle \frac{d}{dt}(2t),\frac{d}{dt}(0),\frac{d}{dt}(0)\right\rangle =\langle 2,0,0\rangle\ \text{m/s}^2. $$
At \(t=1\) second, \(\mathbf{v}(1)=\langle 2,0,0\rangle\) m/s, so the speed is \(2\) m/s. At \(t=2\) seconds, \(\mathbf{v}(2)=\langle 4,0,0\rangle\) m/s, so the speed is \(4\) m/s. The velocity points in the positive \(x\)-direction at both times, while its magnitude is larger at the later time. The acceleration also points in the positive \(x\)-direction, consistent with the increasing positive \(x\)-component of velocity.
Check: Differentiating \(\mathbf{v}(t)=\langle 2t,0,0\rangle\) gives \(\langle 2,0,0\rangle\) m/s², exactly the acceleration found above. Differentiating \(\mathbf{r}(t)\) twice gives the same result.
Acceleration Can Change Direction Without Changing Speed
Acceleration does not have to point in the direction of motion. An object can move at steady speed while its velocity direction changes. The next example shows this with circular motion: the velocity continually changes direction, and the acceleration points toward the centre of the circle.
Worked Example: Acceleration in Circular Motion
Suppose an object moves on a circle of radius \(3\) metres according to \(\mathbf{r}(t)=\langle 3\cos(2t),3\sin(2t),0\rangle\) metres, with \(t\) in seconds. Find its velocity and acceleration, then evaluate both at \(t=\pi/4\) seconds.
Differentiate the three position components: $$ \mathbf{v}(t) =\langle -6\sin(2t),6\cos(2t),0\rangle\ \text{m/s}. $$ Now differentiate each velocity component: $$ \mathbf{a}(t) =\langle -12\cos(2t),-12\sin(2t),0\rangle\ \text{m/s}^2. $$ At \(t=\pi/4\), the angle \(2t=\pi/2\), so \(\sin(2t)=1\) and \(\cos(2t)=0\). Therefore, $$ \mathbf{r}(\pi/4)=\langle 0,3,0\rangle\ \text{m},\qquad \mathbf{v}(\pi/4)=\langle -6,0,0\rangle\ \text{m/s},\qquad \mathbf{a}(\pi/4)=\langle 0,-12,0\rangle\ \text{m/s}^2. $$
At this instant, the object is at the top of the circle and moving to the left. Its acceleration points downward, toward the centre at the origin. To check that speed is constant, compute the velocity magnitude: $$ \|\mathbf{v}(t)\| =\sqrt{(-6\sin(2t))^2+(6\cos(2t))^2} =6\sqrt{\sin^2(2t)+\cos^2(2t)} =6\ \text{m/s}. $$ Thus the speed is \(6\) m/s even though the velocity direction changes around the circle.
Check: Differentiating \(-6\sin(2t)\) gives \(-12\cos(2t)\), by the chain rule; differentiating \(6\cos(2t)\) gives \(-12\sin(2t)\); and the derivative of \(0\) is \(0\). These are exactly the stated acceleration components.
Worked Example: Components Can Change in Different Ways
Worked Example: Components Can Change in Different Ways
Consider the position \(\mathbf{r}(t)=\langle t^2,t^3,2t\rangle\) metres, where \(t\) is in seconds. Find velocity and acceleration, and evaluate them at \(t=1\) second.
Differentiate each position component: $$ \mathbf{v}(t) =\langle 2t,3t^2,2\rangle\ \text{m/s}. $$ The first velocity component changes linearly, the second changes quadratically, and the third is constant. Differentiate each component in turn: $$ \mathbf{a}(t) =\langle 2,6t,0\rangle\ \text{m/s}^2. $$ At \(t=1\), the values are $$ \mathbf{v}(1)=\langle 2,3,2\rangle\ \text{m/s}, \qquad \mathbf{a}(1)=\langle 2,6,0\rangle\ \text{m/s}^2. $$
The \(x\)- and \(y\)-components of velocity are increasing at this instant, at \(2\) and \(6\) metres per second per second, respectively. The \(z\)-component of velocity is constant, so its acceleration component is zero. Acceleration is not simply “how fast the object is moving”; it describes how the velocity vector changes, one component at a time.
Check: Differentiating \(\langle 2t,3t^2,2\rangle\) gives \(\langle 2,6t,0\rangle\). Substituting \(t=1\) gives \(\langle 2,6,0\rangle\) m/s², as required.
Reading the Acceleration Components
The signs and sizes of the components describe changes in the corresponding velocity coordinates. A positive acceleration component means that velocity component is increasing; a negative one means it is decreasing; and a zero one means it is not changing at that instant. These statements concern velocity components, not automatically the total speed.
| Acceleration component | Effect on the corresponding velocity component |
|---|---|
| \(a_x(t)>0\) | The \(x\)-component of velocity is increasing at that time. |
| \(a_y(t)<0\) | The \(y\)-component of velocity is decreasing at that time. |
| \(a_z(t)=0\) | The \(z\)-component of velocity has zero instantaneous rate of change. |
Acceleration direction also needs to be interpreted in relation to the motion. In the one-dimensional example, acceleration points along the velocity and the speed increases. In the circular example, acceleration is perpendicular to the instantaneous velocity and changes its direction while the speed stays constant. These examples show why the acceleration vector cannot be interpreted from a single component, or assumed always to point along the path of motion.
A Common Differentiation Error
A frequent mistake is to copy a constant velocity component into acceleration instead of differentiating it. For example, if \(\mathbf{v}(t)=\langle 3t^2,\cos t,4\rangle\), it is incorrect to write \(\mathbf{a}(t)=\langle 6t,-\sin t,4\rangle\). That last \(4\) is the velocity component, not its derivative.
The correct calculation differentiates all three components: $$ \mathbf{a}(t) =\left\langle \frac{d}{dt}(3t^2),\frac{d}{dt}(\cos t),\frac{d}{dt}(4)\right\rangle =\langle 6t,-\sin t,0\rangle. $$ For instance, at \(t=0\), the incorrect result would give \(\langle 0,0,4\rangle\), whereas the correct acceleration is \(\langle 0,-0,0\rangle=\langle 0,0,0\rangle\). The constant third velocity component means there is no change in velocity in that coordinate.
Why Acceleration Is Useful
Velocity alone gives the current rate and direction of motion, but not whether those are changing. Acceleration supplies that information in a form that can be calculated directly from a velocity function. Its components help identify which coordinate motions are changing; its vector direction gives the direction of the instantaneous velocity change.
For reliable calculations, write velocity as a component vector, differentiate every component, and retain the units. Then evaluate the resulting functions at the requested time. If position is given, differentiating it twice provides a separate check; if velocity is given, differentiating it once verifies acceleration.
Check Your Understanding
Use componentwise differentiation and state units where appropriate.
- If \(\mathbf{v}(t)=\langle t^2,4t,-3\rangle\) metres per second, find \(\mathbf{a}(t)\) and evaluate it at \(t=2\) seconds.
- For \(\mathbf{r}(t)=\langle 2\cos t,2\sin t,0\rangle\) metres, find \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\).
- What does a zero \(y\)-component of acceleration tell you about the \(y\)-component of velocity at that instant?
- Give a reason acceleration can be nonzero even when speed is constant.
- Identify the error in the claim that the derivative of a constant velocity component \(5\) is \(5\).