Diagnosing Motion From Position, Velocity, And Acceleration

Reading Motion from Its Derivatives

Position describes where an object is; velocity describes how its position changes; acceleration describes how its velocity changes. These quantities together let us diagnose motion: not just the path an object follows, but whether it is stopped, changing direction, speeding up, or slowing down.

The central distinction is between velocity and speed. Velocity includes direction, while speed is its magnitude. Acceleration can change velocity by changing its magnitude, its direction, or both. In one-dimensional motion, signs make many classifications straightforward. For motion in more than one dimension, the dot product of velocity and acceleration tells us whether speed is increasing or decreasing.

Definition: If the position of an object at time $t$ is $x(t)$ on a line, its velocity is $v(t)=x'(t)$, and its acceleration is $a(t)=v'(t)=x''(t)$, wherever these derivatives exist. For a position vector $\mathbf{r}(t)$, velocity and acceleration are $\mathbf{v}(t)=\mathbf{r}'(t)$ and $\mathbf{a}(t)=\mathbf{v}'(t)$. Speed is $|v(t)|$ in one dimension and $\|\mathbf{v}(t)\|$ in multiple dimensions.

Stops and Direction Changes on a Line

An object on a line is momentarily stopped at a time $t_0$ when its velocity is zero: $v(t_0)=0$. This condition alone does not tell us whether the object reverses direction. To check for a reversal, examine the sign of velocity just before and just after $t_0$. A change from positive to negative velocity, or from negative to positive velocity, means the direction reverses. If the sign is unchanged, the object may stop momentarily without reversing.

On any interval where velocity is positive, the object moves in the positive direction; where velocity is negative, it moves in the negative direction. Away from a stop, acceleration with the same sign as velocity increases speed, while acceleration with the opposite sign decreases speed. This is about the signs of the two quantities, not simply whether acceleration itself is positive or negative.

Worked Example: A Stop Followed by a Reversal

Suppose the position of an object on a line is $x(t)=t^2$. Find its velocity and acceleration, then classify its motion around $t=0$.

Differentiate the position:

$$ v(t)=x'(t)=2t. $$

Differentiate velocity to find acceleration:

$$ a(t)=v'(t)=2. $$

The object is stopped when $2t=0$, so its only stop is $t=0$. For $t<0$, $v(t)<0$; for $t>0$, $v(t)>0$. The velocity changes sign at the stop, so the object reverses direction there. Before $t=0$, velocity is negative and acceleration is positive, so the object slows down. After $t=0$, both are positive, so it speeds up.

For a numerical check, at $t=-1$, $v=-2$ and $a=2$; at $t=1$, $v=2$ and $a=2$. Substitution into the original position gives $x(-1)=1$, $x(0)=0$, and $x(1)=1$. The object reaches its turning point at the origin and moves away from it in the positive direction after the stop.

Finally, differentiating $x(t)=t^2$ gives $2t$, and differentiating $2t$ gives $2$, confirming the formulas used.

Speed Change for Motion in the Plane

For motion in the plane, velocity and acceleration are vectors. Their signs cannot be compared in the same way as scalar signs. Instead, consider their dot product. When velocity is nonzero, the rate of change of speed is the component of acceleration in the direction of velocity.

Theorem: If $\mathbf{v}(t)\ne\mathbf{0}$, then the speed $s(t)=\|\mathbf{v}(t)\|$ satisfies $$ s'(t)=\frac{\mathbf{v}(t)\cdot\mathbf{a}(t)}{\|\mathbf{v}(t)\|}. $$ Consequently, speed is increasing where $\mathbf{v}\cdot\mathbf{a}>0$, decreasing where $\mathbf{v}\cdot\mathbf{a}<0$, and has derivative zero at an instant where $\mathbf{v}\cdot\mathbf{a}=0$. This test requires nonzero velocity.

When the dot product is positive, acceleration has a component in the direction of motion: we can say it acts with the velocity in the sense relevant to speed. When the dot product is negative, acceleration has a component opposing the motion. A zero dot product means acceleration is perpendicular to velocity, so it does not change speed at that instant. It may still change the direction of velocity.

Why the Dot-Product Test Works

The speed is the square root of the velocity vector dotted with itself. Differentiate that identity, then solve for the rate of change of speed. The nonzero-velocity condition matters because solving requires division by speed.

Write speed squared.
Since $s=\|\mathbf{v}\|$, we have $s^2=\mathbf{v}\cdot\mathbf{v}$.

Differentiate both sides.
For $s>0$, the chain rule and product rule for the dot product give $2ss'=2\mathbf{v}\cdot\mathbf{v}'$.

Use acceleration and divide.
Because $\mathbf{v}'=\mathbf{a}$ and $s=\|\mathbf{v}\|>0$, division by $2s$ gives $s'=(\mathbf{v}\cdot\mathbf{a})/\|\mathbf{v}\|$. The sign of $s'$ is therefore the sign of the dot product.

This derivation also explains why zero velocity needs separate attention. At a stop, the formula for $s'$ above has a zero denominator, and velocity has no direction at that instant. To understand motion through a stop, examine the velocity on either side or compute speed directly.

Worked Example: A Stop with No Reversal

Let $x(t)=(t-2)^3$. Find the stop and decide whether the object reverses direction or changes its speed around that time.

Differentiate position and then velocity:

v(t)=3(t-2)^2,\qquad a(t)=6(t-2).

Velocity is zero only at $t=2$. For every $t\ne2$, $3(t-2)^2>0$, so velocity is positive on both sides of the stop. The object does not reverse direction: it moves in the positive direction before and after $t=2$.

Its speed is $|v(t)|=3(t-2)^2$, since velocity is nonnegative. This speed decreases for $t<2$ and increases for $t>2$, as is also seen from its derivative $6(t-2)$. At $t=2$, both velocity and acceleration are zero. In particular, we do not use the nonzero-velocity dot-product formula at that exact instant.

Check the position and velocity at sample times: $x(1)=-1$, $x(2)=0$, and $x(3)=1$, while $v(1)=3$, $v(2)=0$, and $v(3)=3$. The position increases through the stop, consistent with the positive velocity on both sides. Differentiating $x(t)$ gives $3(t-2)^2$, and differentiating again gives $6(t-2)$, verifying the derivatives.

Using the Test on a Vector Path

Worked Example: Acceleration Opposing Motion

Consider the position vector $\mathbf{r}(t)=\langle t,t^2\rangle$. Determine whether the object speeds up or slows down at $t=-1$, and identify what happens at $t=0$.

Differentiate each coordinate to obtain velocity and acceleration:

\mathbf{v}(t)=\langle 1,2t\rangle, \qquad \mathbf{a}(t)=\langle 0,2\rangle.

At $t=-1$, $\mathbf{v}(-1)=\langle1,-2\rangle$ and $\mathbf{a}(-1)=\langle0,2\rangle$. Their dot product is $1(0)+(-2)(2)=-4$, while speed is $\sqrt{1^2+(-2)^2}=\sqrt{5}$. Therefore the rate of change of speed is $-4/\sqrt{5}<0$: the object is slowing down.

At $t=0$, velocity is $\langle1,0\rangle$, acceleration is $\langle0,2\rangle$, and their dot product is zero. Acceleration is perpendicular to velocity at that instant, so the speed derivative is zero. Indeed, the speed at any time is $\sqrt{1+4t^2}$, which has its minimum value $1$ at $t=0$.

There is no stop, because the first component of velocity is always $1$, so $\mathbf{v}(t)$ is never the zero vector. To check the derivatives, differentiating $\langle t,t^2\rangle$ gives $\langle1,2t\rangle$, and differentiating that gives $\langle0,2\rangle$.

A Common Mistake: Looking Only at Acceleration

A positive acceleration does not always mean an object is speeding up, and a negative acceleration does not always mean it is slowing down. In one dimension, compare the signs of velocity and acceleration. In several dimensions, use their dot product.

For instance, in the vector example above at $t=-1$, the acceleration has magnitude $2$, which is positive. A mistaken conclusion would be, “Acceleration is positive, so speed is increasing.” But the relevant calculation is $\mathbf{v}\cdot\mathbf{a}=-4<0$, so speed is decreasing. The correction is to compare acceleration with the direction of velocity, not to inspect acceleration alone.

Takeaway: First find velocity and acceleration from position. Test for stops by setting velocity equal to zero. For motion on a line, inspect velocity signs on either side of a stop and compare the signs of velocity and acceleration away from stops. For vector motion with nonzero velocity, use $\mathbf{v}\cdot\mathbf{a}$ to classify the change in speed.

Check Your Understanding

For each question, state what calculation or sign test supports your conclusion.

If a one-dimensional velocity changes from negative to positive at a time when it is zero, what happens to the direction of motion?
Can an object have zero velocity at an instant without reversing direction? Give a reason.
For nonzero vector velocity, what does $\mathbf{v}\cdot\mathbf{a}<0$ tell you about speed?
What does $\mathbf{v}\cdot\mathbf{a}=0$ imply about the instantaneous rate of change of speed, and what does it not imply about acceleration?
Why can the dot-product formula for the derivative of speed not be applied directly at a stop?