Why Multiple Velocity Zeros Require a Sign Chart
A particle can stop more than once during a time interval. At each time when its velocity is zero, its direction may change, or it may stop momentarily and continue in the same direction. Between those times, its direction is determined by the sign of its velocity. To analyze speeding up and slowing down as well, we also need the sign of acceleration.
The reliable approach is to divide the time interval at every velocity zero, determine the sign of velocity on each resulting open interval, and compare it with the acceleration sign. If acceleration changes sign between velocity zeros, that time must also be included as a boundary for the speed-behavior analysis.
Velocity Zeros and the Sign Chart
Let \(x(t)\) denote a particle’s position, \(v(t)=x'(t)\) its velocity, and \(a(t)=v'(t)\) its acceleration. A velocity zero is a time \(t\) at which \(v(t)=0\). At that instant the particle is at rest. A velocity zero is a possible direction-change time, but it is not automatically a direction change: check the velocity signs on either side.
For a specified interval \([t_0,t_1]\), first find all velocity zeros strictly inside the interval and put them in increasing order. They divide the interval into open subintervals. On each such subinterval, check the sign of velocity. If the velocity is continuous and has no zero inside a subinterval, it cannot change sign there. Test values or a factored expression can establish that sign.
To classify speed behavior, check acceleration on the same intervals. If acceleration changes sign inside one of them, split that interval again at the acceleration zero. This extra split matters because velocity alone gives direction, not whether speed is increasing or decreasing.
Why the Sign Rule Works
Speed is the nonnegative quantity \(|v(t)|\). Wherever \(v(t)>0\), speed equals \(v(t)\), so its derivative is \(a(t)\). Wherever \(v(t)<0\), speed equals \(-v(t)\), so its derivative is \(-a(t)\). Thus, on intervals where \(v\ne 0\), the rate of change of speed is determined by the product \(v(t)a(t)\).
The denominator \(|v(t)|\) is positive on these intervals, so the derivative of speed has the same sign as \(v(t)a(t)\). A positive product means speed is increasing; a negative product means it is decreasing. If either sign changes, the classification may change, which is why the interval must be split at relevant zeros.
Worked Example: A Direction Change and an Acceleration Zero
Worked Example: A Direction Change and an Acceleration Zero
A particle has velocity \(v(t)=(t-1)(t-3)\) for \(0\le t\le4\). Find its direction and state where it speeds up or slows down.
First, the velocity zeros satisfy \((t-1)(t-3)=0\), so they are \(t=1\) and \(t=3\). Differentiate to find acceleration:
Acceleration is zero at \(t=2\). Therefore, the full set of interior times needed for the analysis is \(1,2,3\), giving the intervals \((0,1)\), \((1,2)\), \((2,3)\), and \((3,4)\).
The signs follow from the factors. For \(v\), the sign is positive when both factors have the same sign and negative when they have opposite signs. For \(a=2(t-2)\), the sign is negative before \(2\) and positive after \(2\).
| Time interval | Sign of \(v\) | Sign of \(a\) | Motion |
|---|---|---|---|
| \((0,1)\) | Positive | Negative | Positive direction; slowing down |
| \((1,2)\) | Negative | Negative | Negative direction; speeding up |
| \((2,3)\) | Negative | Positive | Negative direction; slowing down |
| \((3,4)\) | Positive | Positive | Positive direction; speeding up |
For a check, at \(t=\tfrac12\), \(v=(-\tfrac12)(-\tfrac52)=\tfrac54>0\) and \(a=-3<0\), so speed decreases. At \(t=\tfrac32\), \(v=(\tfrac12)(-\tfrac32)=-\tfrac34\) and \(a=-1<0\), so speed increases. At \(t=\tfrac52\), \(v=(\tfrac32)(-\tfrac12)=-\tfrac34\) and \(a=1>0\), so speed decreases. At \(t=\tfrac72\), \(v=(\tfrac52)(\tfrac12)=\tfrac54\) and \(a=3>0\), so speed increases. The particle changes direction at \(t=1\) and \(t=3\), where the velocity sign changes. The acceleration zero at \(t=2\) changes speed behavior but not direction.
Worked Example: A Velocity Zero Without a Direction Change
Worked Example: A Velocity Zero Without a Direction Change
Suppose \(v(t)=(t-1)^2(t-4)\) for \(0\le t\le5\). Analyze direction and speed behavior. The velocity zeros are \(t=1\) and \(t=4\), with \(t=1\) a repeated zero. Differentiate using the product rule:
Thus acceleration is zero at \(t=1\) and \(t=3\). Use the ordered times \(1,3,4\) to divide the interval. For \(v\), the squared factor is positive except at \(t=1\), so away from that zero the sign is the sign of \(t-4\). For \(a\), use the factors \(3(t-1)(t-3)\).
| Time interval | Sign of \(v\) | Sign of \(a\) | Motion |
|---|---|---|---|
| \((0,1)\) | Negative | Positive | Negative direction; slowing down |
| \((1,3)\) | Negative | Negative | Negative direction; speeding up |
| \((3,4)\) | Negative | Positive | Negative direction; slowing down |
| \((4,5)\) | Positive | Positive | Positive direction; speeding up |
For example, at \(t=\tfrac12\), \(v=( -\tfrac12)^2(-\tfrac72)=-\tfrac78<0\) and \(a=3(-\tfrac12)(-\tfrac52)=\tfrac{15}{4}>0\), confirming slowing down. At \(t=2\), \(v=1(-2)=-2\) and \(a=3(1)(-1)=-3\), confirming speeding up. At \(t=\tfrac72\), \(v=(\tfrac52)^2(-\tfrac12)=-\tfrac{25}{8}\) and \(a=3(\tfrac52)(\tfrac12)=\tfrac{15}{4}\), confirming slowing down. At \(t=\tfrac92\), \(v=(\tfrac72)^2(\tfrac12)=\tfrac{49}{8}\) and \(a=3(\tfrac72)(\tfrac32)=\tfrac{63}{4}\), confirming speeding up. The velocity is negative on both sides of \(t=1\), so the particle stops there but does not reverse direction. At \(t=4\), the sign changes from negative to positive, so the particle reverses direction.
Worked Example: Several Splits Within One Time Interval
Worked Example: Several Splits Within One Time Interval
Let \(v(t)=t(t-2)(t-4)\) for \(0\le t\le5\). Analyze the motion, including any times when acceleration changes sign. The velocity zeros are \(t=0,2,4\); only \(2\) and \(4\) are interior times. Differentiate:
Solving \(3t^2-12t+8=0\) gives \(t=2\pm\frac{2\sqrt{3}}{3}\). Define \(\alpha=2-\frac{2\sqrt{3}}{3}\) and \(\beta=2+\frac{2\sqrt{3}}{3}\); both lie between \(0\) and \(4\), with \(0<\alpha<2<\beta<4\). Since the quadratic has positive leading coefficient, acceleration is positive outside its roots and negative between them.
On \((0,2)\), the factors of \(v\) have signs \(+,-,-\), so \(v>0\). On \((2,4)\), they have signs \(+,+,-\), so \(v<0\). On \((4,5)\), all three factors are positive, so \(v>0\). Incorporating the acceleration zeros gives:
| Time interval | Sign of \(v\) | Sign of \(a\) | Motion |
|---|---|---|---|
| \((0,\alpha)\) | Positive | Positive | Positive direction; speeding up |
| \((\alpha,2)\) | Positive | Negative | Positive direction; slowing down |
| \((2,\beta)\) | Negative | Negative | Negative direction; speeding up |
| \((\beta,4)\) | Negative | Positive | Negative direction; slowing down |
| \((4,5)\) | Positive | Positive | Positive direction; speeding up |
As a numerical check, \(\alpha\approx0.845\) and \(\beta\approx3.155\). At \(t=1\), \(v=1(-1)(-3)=3>0\), while \(a=3-12+8=-1<0\), so the particle slows down. At \(t=3\), \(v=3(1)(-1)=-3<0\) and \(a=27-36+8=-1<0\), so it speeds up. At \(t=\tfrac72\), \(v=\tfrac72\cdot\tfrac32\cdot(-\tfrac12)=-\tfrac{21}{8}<0\), while \(a=\tfrac{147}{4}-42+8=\tfrac{11}{4}>0\), so it slows down. At \(t=\tfrac92\), \(v=\tfrac92\cdot\tfrac52\cdot\tfrac12=\tfrac{45}{8}>0\) and \(a=\tfrac{243}{4}-54+8=\tfrac{59}{4}>0\), so it speeds up. The derivative calculation gives the stated acceleration, and the test values confirm representative classifications in the intervals.
A Common Mistake: Using Velocity Alone to Judge Speed
A positive velocity tells you the direction of motion, not whether the particle is speeding up. For instance, in the first example at \(t=\tfrac12\), velocity is positive but acceleration is negative. The particle moves in the positive direction while slowing down.
A mistaken conclusion would be: “\(v(\tfrac12)>0\), so the particle is speeding up.” That uses direction as if it were speed behavior. The correction is to check both signs: \(v(\tfrac12)=\tfrac54>0\) and \(a(\tfrac12)=-3<0\). Their product is negative, so speed is decreasing.
Another common oversight is to split only at velocity zeros. In the first example, \(v\) has no zero between \(1\) and \(3\), but acceleration changes sign at \(2\). The particle therefore slows down on \((2,3)\) after speeding up on \((1,2)\). A complete analysis checks acceleration changes as well as velocity zeros.
Check Your Understanding
For each question, justify your conclusion using the signs of velocity and acceleration.
- If \(v<0\) and \(a<0\) throughout an interval, which direction does the particle move, and is it speeding up or slowing down?
- If velocity is negative on both sides of a time when \(v=0\), does the particle change direction there? Explain.
- For \(v(t)=(t-2)(t-5)\), find the velocity zeros and acceleration function. Which additional time, if any, must be considered when classifying speed behavior?
- A particle has \(v>0\) and \(a<0\) on an interval. Describe its direction and speed behavior.
- Why can an acceleration zero require an additional split even when there is no velocity zero?
