One Coordinate Describes Motion Along a Line
When an object moves along a fixed straight line, its position changes in space, but its motion can be described by a single scalar. Choose an orientation along the line and measure position from a fixed point using a signed coordinate. The coordinate is positive in the chosen direction and negative in the opposite direction.
This reduces the main questions about the motion to questions about one function of time. Its derivative gives signed velocity: the rate of change of position along the chosen orientation. Speed is the magnitude of that rate, so it is never negative. A second derivative gives signed acceleration along the line.
The choice of orientation is a convention. Replacing \(\mathbf{e}\) by \(-\mathbf{e}\) and \(q\) by \(-q\) describes the same positions, while reversing the signs of signed velocity and signed acceleration. Speed and the physical acceleration vector do not depend on that choice.
Why the Derivative Formulas Work
Because \(\mathbf{a}\) and \(\mathbf{e}\) are fixed, differentiating the position function differentiates only the scalar coordinate. Since \(\mathbf{e}\) has length one, taking the magnitude of the velocity leaves the magnitude of the scalar derivative.
This also explains why signed velocity and speed are not interchangeable. If \(q'(t)<0\), the point moves opposite the chosen orientation, but its speed is the positive quantity \(-q'(t)\). The signed acceleration \(q''(t)\) describes how signed velocity changes; it need not be positive whenever the object is speeding up.
Worked Example: A Reversal Along a Fixed Line
Worked Example: A Reversal Along a Fixed Line
Suppose, for \(t\geq 0\) seconds, the position in metres is \(\mathbf{r}(t)=(1,2)+(t^2-4t+1)(3/5,4/5)\). Find the signed velocity, speed, acceleration, and stopping time, and determine whether the object reverses.
The direction vector is a unit vector because \(\sqrt{(3/5)^2+(4/5)^2}=\sqrt{9/25+16/25}=1\). The scalar coordinate is \(q(t)=t^2-4t+1\) metres. Differentiate it:
Thus the signed velocity is \((2t-4)\) metres per second along the chosen orientation, the speed is \(|2t-4|\) metres per second, and the acceleration vector is \(2(3/5,4/5)=(6/5,8/5)\) metres per second squared.
A stop occurs when the velocity is zero: \(2t-4=0\), so \(t=2\) seconds. For \(0\leq t<2\), \(2t-4<0\); for \(t>2\), \(2t-4>0\). The signed velocity changes sign, so the object reverses at \(t=2\).
At the stopping time, \(q(2)=2^2-4(2)+1=4-8+1=-3\) metres. Its position is therefore \((1,2)-3(3/5,4/5)=(-4/5,-2/5)\) metres. As a check, differentiating the stated position gives \((2t-4)(3/5,4/5)\), whose magnitude is \(|2t-4|\) because the direction vector has unit length. Differentiating once more gives \((6/5,8/5)\) metres per second squared, as found above.
Stopping Times and Reversals
A stopping time is a time at which the velocity vector is zero. In straight-line motion with a fixed unit direction, this is exactly when \(q'(t)=0\). Finding such times identifies instants of zero speed, but it does not by itself determine whether the object changes direction.
For example, a zero of \(q'\) may have even multiplicity, so the sign may stay the same through it. This is why solving \(q'(t)=0\) and checking the sign change are separate steps. In contrast, if \(q'\) changes from negative to positive, the coordinate first decreases and then increases: the point changes direction along the line.
Worked Example: A Stop Without a Reversal
Worked Example: A Stop Without a Reversal
Let \(q(t)=t^3-3t^2+3t\) metres for \(t\geq 0\). Find the signed velocity, speed, and acceleration, and analyze the time \(t=1\) second.
Differentiate the coordinate: \(q'(t)=3t^2-6t+3\). Factoring gives \(q'(t)=3(t-1)^2\) metres per second. Therefore the speed is \(3(t-1)^2\) metres per second as well, because this expression is nonnegative. Differentiating again gives \(q''(t)=6t-6\) metres per second squared, so the acceleration vector is \((6t-6)\mathbf{e}\) for the chosen unit direction \(\mathbf{e}\).
The signed velocity is zero at \(t=1\). Its value is positive both before and after that time, since \(3(t-1)^2>0\) for \(t\ne 1\). Thus the point is instantaneously at rest at \(t=1\), but it does not reverse. Its position then is \(q(1)=1-3+3=1\) metre.
To check the calculation, differentiating \(q(t)\) gives \(3t^2-6t+3=3(t-1)^2\), and differentiating again gives \(6t-6\). At \(t=1\), the speed is \(3(0)^2=0\), while the signed velocity has the same positive sign on the two sides. The zero velocity is therefore a momentary stop, not a reversal.
Signed Acceleration Is Not the Same as “Speeding Up”
Signed acceleration tells whether signed velocity is increasing or decreasing. Whether speed is increasing depends instead on how the velocity and acceleration signs compare. On an interval where \(q'(t)>0\), the object moves in the chosen direction: positive \(q''(t)\) increases its speed, while negative \(q''(t)\) decreases it. Where \(q'(t)<0\), the object moves in the opposite direction, so the sign interpretation switches: negative \(q''(t)\) increases its speed, while positive \(q''(t)\) decreases it.
At a time when \(q'(t)=0\), that interval-based sign test should not be applied as if the object were moving. Examine the velocity on the neighbouring intervals to decide whether there is a reversal. Acceleration at the instant may help describe the local motion, but the sign test for reversal is based on the signed velocity.
Worked Example: Two Reversals
Worked Example: Two Reversals
For \(t\geq 0\) seconds, let \(q(t)=\frac{1}{3}t^3-2t^2+3t\) metres. Find the signed velocity, speed, acceleration, and reversal times.
Differentiating gives \(q'(t)=t^2-4t+3=(t-1)(t-3)\) metres per second. The speed is \(|(t-1)(t-3)|\) metres per second. Differentiating the signed velocity gives \(q''(t)=2t-4\) metres per second squared, so the acceleration vector is \((2t-4)\mathbf{e}\).
The velocity is zero at \(t=1\) and \(t=3\). Check its sign on the intervals separated by those times. For \(0\leq t<1\), both factors \(t-1\) and \(t-3\) are negative, so their product is positive. For \(1<t<3\), the first factor is positive and the second negative, so their product is negative. For \(t>3\), both are positive, so their product is positive. The velocity changes sign at each zero, and the object reverses at both \(t=1\) and \(t=3\).
The positions at these times are \(q(1)=\frac13-2+3=\frac43\) metres and \(q(3)=\frac13(27)-2(9)+9=9-18+9=0\) metres. For a derivative check, differentiating the original coordinate gives \(t^2-4t+3\), and differentiating that expression gives \(2t-4\). The signed velocity and acceleration therefore match the stated results.
A Common Mistake: Calling Signed Velocity the Speed
In the first example, the signed velocity is \(2t-4\). A common error is to report the speed as \(2t-4\) metres per second for every \(t\). At \(t=1\), that expression equals \(-2\), which cannot be a speed. The correct calculation is \(\text{speed}=|2t-4|\). In particular, at \(t=1\) the speed is \(|-2|=2\) metres per second, and at \(t=3\) it is \(|2|=2\) metres per second.
Another error is to assume that every zero of velocity is a reversal. In the second example, \(q'(t)=3(t-1)^2\) vanishes at \(t=1\), but remains positive on both sides. The correction is to check the sign of the velocity on the intervals before and after each stopping time, rather than infer direction from the zero alone.
A Reliable Analysis
For a position function whose image lies on a fixed line, the following sequence keeps the quantities distinct and makes the direction analysis explicit.
Write the position as a fixed point plus a scalar coordinate times a constant unit direction vector.
The first derivative is signed velocity; its absolute value is speed.
The second derivative is signed acceleration, and multiplying by the unit direction vector gives the acceleration vector.
Solve \(q'(t)=0\), then compare the signs of \(q'\) on the available sides to identify reversals.
This scalar viewpoint is useful because it separates the geometry of the fixed line from the timing of the motion along it. The chosen unit vector describes where the line points; the function \(q(t)\) describes how the point moves along it. A change in speed can occur without changing the path, and a reversal can occur even though the path remains the same line.
Check Your Understanding
Use a scalar coordinate \(q(t)\) along a fixed line and explain each conclusion from its derivatives.
- If \(q'(t)=-5\) metres per second, what are the signed velocity and speed?
- For \(q(t)=t^2-6t\) metres, find the stopping time and determine whether the object reverses.
- For \(q(t)=(t-2)^3\) metres, does the zero of velocity at \(t=2\) represent a reversal? Justify using the sign of \(q'\).
- If \(q'(t)<0\) and \(q''(t)<0\) on an interval, is the object speeding up or slowing down? Explain.
- Why does the formula for speed use the absolute value of signed velocity?