Constant-Speed Motion And Orthogonal Acceleration

When Motion Changes Direction Without Changing Speed

A moving object can have nonzero acceleration even when its speed stays constant. In that situation, its velocity changes direction while keeping the same magnitude. The precise relationship is that velocity and acceleration have zero dot product: they are perpendicular whenever the velocity is nonzero.

This connection also gives a practical test. From a position function, calculate velocity and acceleration, then take their dot product. If the result is zero throughout an interval, the speed is constant there. A zero dot product at just one instant, however, does not establish constant speed.

Position, Velocity, Acceleration, and Speed

Let \(\mathbf{r}(t)\) be a position vector defined on an interval \(I\). Assume that \(\mathbf{r}\) is twice differentiable. Its velocity is \(\mathbf{v}(t)=\mathbf{r}'(t)\), and its acceleration is \(\mathbf{a}(t)=\mathbf{v}'(t)=\mathbf{r}''(t)\). The speed is the magnitude of the velocity, \(\|\mathbf{v}(t)\|\).

Definition: The motion has constant speed on \(I\) if there is a constant \(c\geq 0\) such that \(\|\mathbf{v}(t)\|=c\) for every \(t\in I\). Velocity and acceleration are perpendicular at a time \(t\) when \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\), provided \(\mathbf{v}(t)\ne\mathbf{0}\). The dot-product equation also makes sense when velocity is zero, although a zero vector has no direction to compare with acceleration.

For vectors \(\mathbf{u}=(u_1,\ldots,u_n)\) and \(\mathbf{w}=(w_1,\ldots,w_n)\), their dot product is \(\mathbf{u}\cdot\mathbf{w}=\sum_{j=1}^{n}u_jw_j\). In particular, \(\|\mathbf{v}\|^2=\mathbf{v}\cdot\mathbf{v}\). These definitions apply to planar and three-dimensional motion.

The Constant-Speed Criterion

Theorem: Suppose \(\mathbf{r}\) is twice differentiable on an interval \(I\), with velocity \(\mathbf{v}\) and acceleration \(\mathbf{a}\). Its speed is constant on \(I\) if and only if \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\) for every \(t\in I\). If the constant speed is positive, velocity and acceleration are perpendicular at every time.

The key is to differentiate the square of the speed, not the speed itself. The square is \(\mathbf{v}\cdot\mathbf{v}\), so the product rule and the symmetry of the dot product give

\frac{d}{dt}\bigl(\|\mathbf{v}(t)\|^2\bigr) = \frac{d}{dt}\bigl(\mathbf{v}(t)\cdot\mathbf{v}(t)\bigr) = \mathbf{a}(t)\cdot\mathbf{v}(t) + \mathbf{v}(t)\cdot\mathbf{a}(t) = 2\mathbf{v}(t)\cdot\mathbf{a}(t).

Here is the full argument in both directions.

Constant speed implies a zero dot product.
If \(\|\mathbf{v}(t)\|=c\) throughout \(I\), then \(\|\mathbf{v}(t)\|^2=c^2\) is constant. Its derivative is zero, so the displayed identity gives \(2\mathbf{v}(t)\cdot\mathbf{a}(t)=0\), and hence \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\).

A zero dot product throughout the interval implies constant speed.
If \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\) for every \(t\in I\), then the derivative of \(g(t)=\|\mathbf{v}(t)\|^2\) is zero throughout \(I\). For any two points \(s,t\in I\), the Mean Value Theorem applied between them gives \(g(t)-g(s)=g'(\xi)(t-s)=0\) for some \(\xi\) between them. Thus \(g\) is constant on \(I\). Since speed is the nonnegative square root of \(g\), speed is constant as well.

This proof does not divide by speed. That matters: if the speed is zero, the formula for the derivative of speed in terms of \(\mathbf{v}\cdot\mathbf{a}\) would involve division by zero. The squared-speed argument still applies. In fact, if speed is constantly zero, then \(\mathbf{v}=\mathbf{0}\) throughout the interval and \(\mathbf{a}=\mathbf{0}\) there too.

Worked Examples

Worked Example: Testing a Planar Circular Path

Let \(\mathbf{r}(t)=(3\cos t,3\sin t)\). Determine its speed and check whether velocity and acceleration are perpendicular.

Differentiate each coordinate to obtain velocity, then differentiate again to obtain acceleration:

\mathbf{v}(t)=(-3\sin t,3\cos t), \qquad \mathbf{a}(t)=(-3\cos t,-3\sin t).

The squared speed and speed are

\|\mathbf{v}(t)\|^2 =(-3\sin t)^2+(3\cos t)^2 =9\sin^2t+9\cos^2t =9, \qquad \|\mathbf{v}(t)\|=3.

The dot product is

\mathbf{v}(t)\cdot\mathbf{a}(t) =(-3\sin t)(-3\cos t)+(3\cos t)(-3\sin t) =9\sin t\cos t-9\sin t\cos t =0.

Thus the speed is constantly \(3\), and the dot-product test confirms perpendicularity at every time. To check the calculation, differentiating the stated velocity gives the stated acceleration, while the squared-speed calculation gives \(3^2=9\) at every \(t\).

Worked Example: Detecting Changing Speed

Let \(\mathbf{r}(t)=(t,t^2/2)\). Determine whether this motion has constant speed, and explain what happens at \(t=0\).

Differentiate the position coordinates:

\mathbf{v}(t)=(1,t), \qquad \mathbf{a}(t)=(0,1).

The squared speed, speed, and dot product are

\|\mathbf{v}(t)\|^2=1^2+t^2=1+t^2, \qquad \|\mathbf{v}(t)\|=\sqrt{1+t^2}, \qquad \mathbf{v}(t)\cdot\mathbf{a}(t)=(1)(0)+(t)(1)=t.

Since \(t\) is not zero throughout any interval containing more than one point, the dot product is not identically zero there. The speed is not constant: for instance, it is \(1\) at \(t=0\) and \(\sqrt{2}\) at \(t=1\). At \(t=0\), the dot product happens to be zero, so velocity and acceleration are perpendicular at that instant only. Differentiating \(\sqrt{1+t^2}\) gives \(t/\sqrt{1+t^2}\), which is zero at \(t=0\) but not generally zero. This agrees with the dot-product test wherever the speed is nonzero.

As a check, differentiating \((1,t)\) gives \((0,1)\), and substituting \(t=0\) and \(t=1\) into the speed formula gives the values \(1\) and \(\sqrt{2}\) used above.

Worked Example: Using the Dot Product to Establish Constant Speed

Consider the three-dimensional position function \(\mathbf{r}(t)=(\cos t,\sin t,2t)\). Test whether its speed is constant using the velocity–acceleration dot product, and find the speed.

Differentiate the three coordinates once and then twice:

\mathbf{v}(t)=(-\sin t,\cos t,2), \qquad \mathbf{a}(t)=(-\cos t,-\sin t,0).

The dot product is

\mathbf{v}(t)\cdot\mathbf{a}(t) =(-\sin t)(-\cos t)+(\cos t)(-\sin t)+(2)(0) =\sin t\cos t-\sin t\cos t =0.

Since this holds for every real \(t\), the theorem shows that speed is constant. Compute its value from the velocity:

\|\mathbf{v}(t)\|^2 =(-\sin t)^2+(\cos t)^2+2^2 =\sin^2t+\cos^2t+4 =5, \qquad \|\mathbf{v}(t)\|=\sqrt{5}.

The result checks in both ways: the speed is \(\sqrt{5}\) for every \(t\), and the dot product is zero for every \(t\). Differentiating the stated velocity also recovers the stated acceleration.

What the Dot Product Says About Acceleration

When velocity is nonzero, the equation \(\mathbf{v}\cdot\mathbf{a}=0\) says that acceleration has no component in the direction of motion. Under the tangential and normal decomposition of acceleration, the tangential component measures the change in speed. Constant-speed motion has zero tangential component, though its normal component may be nonzero. The circular-path example illustrates why zero tangential acceleration does not require zero total acceleration: acceleration can still change the direction of velocity.

This gives the test a useful interpretation as well as a calculation. If the dot product vanishes throughout an interval, acceleration does not change the magnitude of velocity there. If it is positive or negative at some time, the squared speed is increasing or decreasing at that time, respectively, because its derivative is \(2\mathbf{v}\cdot\mathbf{a}\).

A Common Pitfall: Checking Only One Instant

A frequent error is to find \(\mathbf{v}(t_0)\cdot\mathbf{a}(t_0)=0\) at one time and conclude that the motion has constant speed. The theorem requires the dot product to vanish at every time in the interval. For \(\mathbf{r}(t)=(t,t^2/2)\), the calculation gives \(\mathbf{v}(0)\cdot\mathbf{a}(0)=0\), but the speed changes from \(1\) at \(t=0\) to \(\sqrt{2}\) at \(t=1\).

Another error is to conclude that constant speed requires zero acceleration. The first worked example disproves that claim: its acceleration is \((-3\cos t,-3\sin t)\), whose magnitude is \(3\), while its speed is also constantly \(3\). The correct requirement is not \(\mathbf{a}=\mathbf{0}\); it is \(\mathbf{v}\cdot\mathbf{a}=0\). Acceleration may be nonzero as long as it is perpendicular to the velocity.

A Reliable Test

For a given position function, keep the logical scope of the test clear: the dot product must vanish throughout the interval to conclude that speed is constant on that interval. A systematic calculation is:

Differentiate position.
Find \(\mathbf{v}(t)=\mathbf{r}'(t)\).

Differentiate velocity.
Find \(\mathbf{a}(t)=\mathbf{v}'(t)\).

Compute the dot product.
Evaluate \(\mathbf{v}(t)\cdot\mathbf{a}(t)\), showing the coordinate products and simplification.

Check the whole interval.
If the result is zero for every time in the interval, the speed is constant. If it is zero only at selected times, the conclusion applies only to the instantaneous perpendicularity at those times.

The method is valuable because it can establish constant speed without first simplifying a square root. It also diagnoses the role of acceleration: a zero dot product throughout an interval means acceleration does not change speed there, even if it continues to change the direction of motion.

Check Your Understanding

Answer each question using the definitions and criterion above.

What regularity of the position function is assumed for the constant-speed criterion, and how are velocity and acceleration defined?
Why does differentiating squared speed give twice the dot product of velocity and acceleration?
If \(\mathbf{v}(t)\cdot\mathbf{a}(t)=0\) only at one time, what can you conclude at that time, and what can you not conclude about the whole interval?
For \(\mathbf{r}(t)=(2\cos t,2\sin t)\), find velocity, acceleration, speed, and their dot product.
Can a motion have constant speed and nonzero acceleration? Explain using the dot-product criterion.