Why Acceleration Alone Does Not Tell You Whether Speed Changes
A particle’s acceleration describes how its velocity vector changes. But a changing velocity vector does not necessarily mean a changing speed: velocity can change direction while its magnitude stays constant. To decide whether the particle is speeding up or slowing down at a particular time, compare its velocity and acceleration using a dot product.
For a particle moving in the plane along a parametrized position \(\mathbf{r}(t)=(x(t),y(t))\), its velocity is \(\mathbf{v}(t)=(x'(t),y'(t))\), and its acceleration is \(\mathbf{a}(t)=(x''(t),y''(t))\). Its speed is the magnitude of its velocity. This tutorial develops a test for the instantaneous change in that speed.
Speed and the Dot Product
The speed is the nonnegative magnitude of the velocity vector:
The dot product of two plane vectors \((u_1,u_2)\) and \((w_1,w_2)\) is \(u_1w_1+u_2w_2\). In particular, for velocity and acceleration,
This number combines information from both vectors. Its sign, rather than the size of the acceleration alone, diagnoses the instantaneous speed change.
The nonzero-velocity condition matters. Since the denominator is speed, this formula cannot be used when \(|\mathbf{v}(t)|=0\). At such a time, the dot product is automatically zero, but that fact by itself does not classify how speed behaves. The sign test is for a particle that is moving at the time being examined.
Why the Test Works
The formula follows by differentiating the definition of speed. This derivation also explains why the dot product appears: differentiating the sum of the squared velocity components produces the products of matching velocity and acceleration components.
Because \(s(t)=\sqrt{(x'(t))^2+(y'(t))^2}\), equivalently \(s(t)^2=(x'(t))^2+(y'(t))^2\).
Applying the chain rule to both sides gives \(2s(t)s'(t)=2x'(t)x''(t)+2y'(t)y''(t)\).
When \(s(t)>0\), divide by \(2s(t)\) to obtain \(s'(t)=\dfrac{x'(t)x''(t)+y'(t)y''(t)}{s(t)}\).
The numerator is \(\mathbf{v}(t)\cdot\mathbf{a}(t)\), and \(s(t)=|\mathbf{v}(t)|\), giving the stated formula.
Since the denominator \(|\mathbf{v}(t)|\) is positive whenever the particle is moving, dividing by it does not change the numerator’s sign. Therefore the sign of the dot product determines the sign of \(ds/dt\).
Worked Examples
Worked Example: Identifying Increasing Speed
A particle has position \(\mathbf{r}(t)=(t^2,3t)\). Determine whether it is speeding up or slowing down at \(t=1\), and find its instantaneous rate of change of speed.
Differentiate each coordinate to find velocity and acceleration:
At \(t=1\),
Calculate the dot product component by component:
The dot product is positive, so the particle is speeding up at \(t=1\). Its speed then is \(|\mathbf{v}(1)|=\sqrt{2^2+3^2}=\sqrt{13}\), so the speed-change formula gives
Check by differentiating the speed directly. Here \(s(t)=\sqrt{4t^2+9}\), so \(s'(t)=\dfrac{4t}{\sqrt{4t^2+9}}\). At \(t=1\), this is \(4/\sqrt{13}\), agreeing with the formula and confirming that the instantaneous speed change is positive.
Worked Example: Identifying Decreasing Speed
A particle has position \(\mathbf{r}(t)=(3t-t^2,2t)\). Determine whether it is speeding up or slowing down at \(t=1\), and find the rate of change of speed at that time.
Differentiate the coordinate functions:
At \(t=1\), the vectors are
The dot product is
Because the dot product is negative and \(|\mathbf{v}(1)|=\sqrt{1^2+2^2}=\sqrt{5}>0\), the particle is slowing down. More precisely,
Check directly: \(s(t)=\sqrt{(3-2t)^2+4}\), so \(s'(t)=\dfrac{-2(3-2t)}{\sqrt{(3-2t)^2+4}}\). Substituting \(t=1\) gives \(-2/\sqrt{5}\), matching the dot-product calculation.
Worked Example: Zero Instantaneous Change Does Not Mean Constant Speed
A particle has position \(\mathbf{r}(t)=(t,t^2)\). Determine what the dot-product test says at \(t=0\), and decide whether that implies constant speed.
Its velocity and acceleration are
At \(t=0\),
The particle is moving because \(|\mathbf{v}(0)|=1\). The formula therefore applies and gives \(s'(0)=0/1=0\). Its speed function is \(s(t)=\sqrt{1+4t^2}\), whose derivative is \(s'(t)=\dfrac{4t}{\sqrt{1+4t^2}}\); substituting \(t=0\) confirms \(s'(0)=0\).
However, for any \(t\ne0\), \(1+4t^2>1\), so \(s(t)=\sqrt{1+4t^2}>1=s(0)\). Thus the speed is not constant around \(t=0\). The zero dot product says only that the instantaneous rate of change of speed at that particular time is zero.
Reading the Dot Product Geometrically
When the particle is moving, define the unit vector in the direction of velocity by \(\mathbf{T}(t)=\mathbf{v}(t)/|\mathbf{v}(t)|\). It points along the particle’s direction of motion. Taking the dot product of acceleration with this direction gives
So the rate of change of speed is the part of acceleration in the direction of motion. Acceleration pointing partly forward increases speed; acceleration pointing partly backward decreases it. Acceleration perpendicular to the velocity has zero dot product with velocity, so it changes direction without producing an instantaneous speed change at that moment.
This interpretation explains why acceleration’s magnitude alone is insufficient. A large acceleration perpendicular to velocity can leave the instantaneous speed rate at zero, while a smaller acceleration with a component opposite the velocity can make the particle slow down.
When Zero Means Constant Speed
A dot product equal to zero at one instant establishes only that \(ds/dt=0\) at that instant, provided velocity is nonzero. It does not establish that speed is constant over an interval, as the third example demonstrated.
There is a stronger conclusion when the dot product is zero throughout an interval on which the particle is always moving. At every time in that interval, the formula gives \(ds/dt=0\). The Mean Value Theorem then implies that \(s(t)\) has the same value at any two times in the interval, so speed is constant there. The conditions matter: the conclusion uses zero dot product throughout the interval and nonzero velocity throughout it, not merely a zero at one time.
Common Pitfall: Multiplying Magnitudes Instead of Taking the Dot Product
Consider the second example at \(t=1\), where \(\mathbf{v}=(1,2)\) and \(\mathbf{a}=(-2,0)\). A mistaken approach is to multiply their magnitudes, obtaining \(|\mathbf{v}||\mathbf{a}|=\sqrt{5}\cdot2=2\sqrt{5}>0\), and conclude that the particle is speeding up. That product is not the dot product: it ignores the angle between the vectors.
The correct calculation uses corresponding components:
The negative result shows that the particle is slowing down. A useful check is to compute the speed rate itself: \(s'=-2/\sqrt{5}<0\), which confirms the classification. Always calculate the signed dot product, not the product of magnitudes.
Units and a Reliable Procedure
If position is measured in meters and time in seconds, velocity has units of meters per second and acceleration has units of meters per second squared. Their dot product has units of meters squared per second cubed. Dividing by speed, which has units of meters per second, gives meters per second squared—the correct units for a rate of change of speed.
For an AP problem asking whether a moving particle is speeding up or slowing down at a time, use this sequence:
- Differentiate the position coordinates to find velocity.
- Differentiate velocity to find acceleration.
- Evaluate both vectors at the specified time.
- Check that velocity is nonzero.
- Compute the dot product and use its sign to classify the speed change.
If the question asks for the numerical rate of change of speed, divide the dot product by the speed at that time. If velocity is zero, do not use the sign test: its denominator would be zero, and the dot product’s value alone does not supply the missing classification.
Check Your Understanding
Use the velocity-acceleration dot product where the particle is moving. For each classification question, distinguish an instantaneous conclusion from a conclusion over an interval.
- If \(\mathbf{v}=(2,-1)\) and \(\mathbf{a}=(3,4)\), is the particle speeding up or slowing down?
- If \(\mathbf{v}=(1,3)\) and \(\mathbf{a}=(-3,1)\), what is \(ds/dt\)?
- What does \(\mathbf{v}\cdot\mathbf{a}=0\) at one instant imply about the speed at that instant?
- Why can the dot-product test not be applied directly when \(\mathbf{v}=\mathbf{0}\)?
- If \(\mathbf{v}\cdot\mathbf{a}=0\) throughout an interval and the velocity never vanishes there, what can be concluded about speed on that interval?
