From Position to Motion
A polynomial position function gives a particle’s coordinate at each time. Differentiating it connects that position model to the particle’s instantaneous velocity and acceleration. The signs of velocity and acceleration then describe whether the particle moves in the positive or negative direction and whether its speed is increasing or decreasing.
The interval matters: a polynomial may have several velocity or acceleration zeros, but only those within the stated time interval are relevant. A reliable analysis identifies those times, divides the interval into sign intervals, and interprets the signs there.
For a polynomial position function, both derivatives exist for every real \(t\). If position is measured in meters and time in seconds, velocity is measured in meters per second and acceleration in meters per second squared.
A Sign Rule for Direction and Speed
Velocity answers two related questions. Its sign gives direction: \(v(t)>0\) means motion in the positive coordinate direction, and \(v(t)<0\) means motion in the negative coordinate direction. If \(v(t)=0\), the particle is momentarily at rest.
Acceleration describes how velocity is changing, but its sign alone does not tell whether speed is increasing. Speed is the nonnegative quantity \(|v(t)|\). When \(v(t)\ne 0\), differentiating speed gives a useful test.
Thus speed increases when velocity and acceleration have the same sign, and decreases when their signs are opposite. Equivalently, speed increases where \(v(t)a(t)>0\) and decreases where \(v(t)a(t)<0\). At a time when \(v=0\), the particle is at rest; this test is applied on the intervals around that time, not by dividing by zero.
Why the Speed Test Works
On any interval where \(v(t)>0\), speed equals \(v(t)\), so its derivative is \(a(t)\). Therefore speed increases there if \(a>0\), and decreases if \(a<0\). On an interval where \(v(t)<0\), speed equals \(-v(t)\), so its derivative is \(-a(t)\). In that case speed increases if \(a<0\), and decreases if \(a>0\). These cases are exactly the same-sign and opposite-sign tests.
Worked Example: Direction and Speed Changes on an Interval
Worked Example: Direction and Speed Changes on an Interval
A particle’s position, in meters, is \(s(t)=t^3-6t^2+9t+2\), where \(t\) is in seconds. Analyze its motion on \(0\le t\le4\): find its velocity and acceleration, its rest times, its direction, and where it speeds up or slows down.
Step 1: Differentiate position.
Differentiate velocity to obtain acceleration:
Step 2: Find the rest times and direction intervals. Solve \(v(t)=0\): \(3(t-1)(t-3)=0\), so \(t=1\) and \(t=3\). Both lie in the stated interval. The velocity signs are positive on \((0,1)\), negative on \((1,3)\), and positive on \((3,4)\). The particle moves in the positive direction on \([0,1)\) and \((3,4]\), and in the negative direction on \((1,3)\). It is at rest at \(t=1\) and \(t=3\).
Step 3: Find where acceleration changes sign. From \(a(t)=6(t-2)\), acceleration is negative for \(t<2\), zero at \(t=2\), and positive for \(t>2\). The sign intervals for both \(v\) and \(a\) are therefore:
| Time interval | Sign of \(v\) | Sign of \(a\) | Speed |
|---|---|---|---|
| \((0,1)\) | Positive | Negative | Decreasing |
| \((1,2)\) | Negative | Negative | Increasing |
| \((2,3)\) | Negative | Positive | Decreasing |
| \((3,4)\) | Positive | Positive | Increasing |
For example, on \((1,2)\) both \(v\) and \(a\) are negative, so their product is positive and speed increases. On \((2,3)\) their signs differ, so speed decreases.
Check: Differentiate the original position again to confirm the acceleration: \(s''(t)=6t-12\), agreeing with \(a(t)\). At the rest times, \(v(1)=3(0)(-2)=0\) and \(v(3)=3(2)(0)=0\), as required.
Finding the Right Sign Intervals
The zeros of velocity and acceleration are the times at which the sign information can change. To make a complete classification on a closed interval, find every such zero inside the interval, include the interval endpoints as boundaries, and test signs on each resulting open interval. A zero of acceleration may split a speed interval even though the particle does not stop there.
Compute \(v=s'\), then \(a=v'\).
Solve \(v(t)=0\) and \(a(t)=0\), retaining only times within the stated interval.
Use factored expressions or a test value from each open interval.
The sign of \(v\) gives direction; matching signs of \(v\) and \(a\) mean speeding up, and opposite signs mean slowing down.
At an interval endpoint, report the endpoint’s velocity or acceleration by direct substitution if needed. Statements about increasing or decreasing speed are usually made on the open intervals between sign-change times, where the signs remain constant.
Worked Example: A Stop Without a Direction Change
Worked Example: A Stop Without a Direction Change
Let \(s(t)=t^3-3t^2+3t\), in meters, for \(0\le t\le2\) seconds. Determine when the particle is at rest and whether it changes direction there.
Differentiate:
The equation \(v(t)=0\) gives \(3(t-1)^2=0\), so the particle is at rest at \(t=1\). For \(t<1\) and \(t>1\), \(3(t-1)^2>0\). Thus it moves in the positive direction on both sides of the rest time: it stops momentarily but does not change direction.
For \(0<t<1\), \(v>0\) and \(a<0\), so speed decreases. For \(1<t<2\), \(v>0\) and \(a>0\), so speed increases.
Check: \(s''(t)=6t-6\), which agrees with the stated acceleration. Also \(v(1)=3(0)^2=0\), while \(v(0)=3\) and \(v(2)=3\) are both positive. The particle is moving positively before and after its stop.
Worked Example: Several Sign Intervals
Worked Example: Several Sign Intervals
A particle has position \(s(t)=\frac14t^4-t^2\), in meters, on \(-2\le t\le2\) seconds. Find the times it is at rest and classify direction and speed on the resulting intervals.
Differentiate the polynomial:
Velocity is zero when \(t=0\) or \(t^2=2\), giving rest times \(t=-\sqrt2,0,\sqrt2\). Acceleration is zero when \(3t^2-2=0\), giving \(t=-\sqrt{2/3}\) and \(t=\sqrt{2/3}\). These five times divide the interval into six open intervals.
Use the factors of \(v\) and \(a\) to determine signs on each interval:
| Time interval | Sign of \(v\) | Direction | Sign of \(a\) | Speed |
|---|---|---|---|---|
| \((-2,-\sqrt2)\) | Negative | Negative | Positive | Decreasing |
| \((-\sqrt2,-\sqrt{2/3})\) | Positive | Positive | Positive | Increasing |
| \((-\sqrt{2/3},0)\) | Positive | Positive | Negative | Decreasing |
| \((0,\sqrt{2/3})\) | Negative | Negative | Negative | Increasing |
| \((\sqrt{2/3},\sqrt2)\) | Negative | Negative | Positive | Decreasing |
| \((\sqrt2,2)\) | Positive | Positive | Positive | Increasing |
For instance, on \((-2,-\sqrt2)\), \(t<0\) and \(t^2-2>0\), so \(v<0\); also \(t^2>2>2/3\), so \(a>0\). Their signs are opposite, so speed decreases. On \((0,\sqrt{2/3})\), \(v<0\) and \(a<0\), so speed increases.
Check: Differentiating \(v(t)=t^3-2t\) gives \(a(t)=3t^2-2\). Substituting each velocity zero into \(v=t(t^2-2)\) gives zero, confirming all three rest times. The acceleration zeros satisfy \(3t^2-2=0\), as listed.
A Common Sign Mistake
A common error is to conclude that negative acceleration always means the particle is slowing down. Consider the first example at \(t=1.5\) seconds. Its velocity and acceleration are:
A mistaken conclusion would be: “Acceleration is negative, so speed is decreasing.” But velocity is negative too. Since \(v(1.5)a(1.5)=(-2.25)(-3)=6.75>0\), speed is increasing. The particle moves in the negative direction while becoming faster. Always use both signs for a speed conclusion.
Check Your Understanding
Use derivatives and sign intervals to answer each question. State direction and speed behavior on intervals, not just at sample times.
- For \(s(t)=t^3-3t\) on \([-2,2]\), find \(v(t)\), \(a(t)\), and every rest time in the interval.
- For the same model, determine the direction of motion on each interval between rest times.
- At what times does acceleration equal zero in the model \(s(t)=t^3-3t\)?
- For \(s(t)=t^2-4t\) on \([0,3]\), find the rest time and classify the speed on either side of it.
- Explain why a particle with \(v(t)=3(t-1)^2\) is at rest at \(t=1\) without changing direction there.
