From Velocity Signs to Motion Intervals
A particle’s velocity tells us its direction of motion. Positive velocity means motion in the positive direction, and negative velocity means motion in the negative direction. To describe that motion over time, locate the times when velocity is zero and determine its sign between those times.
A zero of velocity is important, but it does not automatically mean that the particle changes direction. The velocity may change sign at that time, or it may have the same sign on both sides. The distinction matters whenever a response asks for intervals of forward or backward motion or for times when direction changes.
At a time when \(v(t)=0\), the particle is momentarily at rest. That instant is not itself a time of forward or backward motion under the strict definitions \(v(t)>0\) and \(v(t)<0\). Whether a zero is also a direction-change time depends on the signs immediately before and after it.
A Sign-Analysis Method
For a velocity function given by a formula, the basic method is to identify its zeros, divide the time domain at those zeros, and determine the sign on each resulting interval. When the function is factored, its factors often make the zeros and sign analysis especially clear.
Use the interval specified in the question. Do not report times outside that interval.
Solve \(v(t)=0\) and retain only solutions in the stated time domain.
Use the zeros as boundaries. On each open interval between consecutive zeros, test one time or otherwise determine the sign.
Positive signs indicate forward motion; negative signs indicate backward motion. Compare the signs on either side of each interior zero to identify direction changes.
State where velocity is zero, and use interval notation or words that make clear whether an endpoint is included in the time domain and whether the particle is actually moving at that instant.
If a zero is inside the domain, the particle is not moving forward or backward at that exact instant. Thus, when describing intervals on which \(v(t)>0\) or \(v(t)<0\), such a zero is excluded. A domain endpoint where the velocity is nonzero can be included. If the question asks for a direction change, report the time of the sign change separately.
Why Checking Signs Between Zeros Works
The sign of a continuous velocity function cannot switch from positive to negative on an interval without the velocity being zero somewhere in between. This is a consequence of the Intermediate Value Theorem: a continuous function taking values of opposite signs must take the value zero at some intermediate time.
This is why one sign test is enough on each interval between consecutive zeros, provided the velocity is continuous there. It also explains why merely finding zeros is not enough: the signs on either side determine whether the particle reverses direction.
Worked Examples
Worked Example: Forward and Backward Intervals from a Factored Velocity
A particle has velocity \(v(t)=(t-1)(t-3)\) for \(0\le t\le4\). Find the intervals on which it moves forward and backward, and identify its direction changes.
First solve \(v(t)=0\):
Both zeros lie in the domain. They divide the domain into \([0,1)\), \((1,3)\), and \((3,4]\) for sign testing.
Test \(t=0\) in the first interval:
So the particle moves forward on \([0,1)\). Test \(t=2\) in the second interval:
So it moves backward on \((1,3)\). Test \(t=4\) in the last interval:
So it moves forward on \((3,4]\). At \(t=1\) and \(t=3\), velocity is zero, as substitution confirms: \(v(1)=(0)(-2)=0\) and \(v(3)=(2)(0)=0\). The signs are positive then negative at \(t=1\), and negative then positive at \(t=3\), so both are direction-change times.
Answer: Forward on \([0,1)\) and \((3,4]\); backward on \((1,3)\). The particle changes direction at \(t=1\) and \(t=3\).
Worked Example: A Zero That Does Not Change Direction
For \(0\le t\le6\), let \(v(t)=(t-2)^2(t-5)\). Find the forward- and backward-motion intervals and determine which zeros are direction changes.
The zeros are obtained from the factors:
Test \(t=1\), which lies before both zeros:
Test \(t=3\), between the zeros:
Test \(t=5.5\), after both zeros:
The velocity is negative on both sides of \(t=2\), so that zero does not mark a direction change. It is zero at the instant \(t=2\), but the particle continues moving backward on either side. At \(t=5\), the sign changes from negative to positive, so the particle reverses direction. Direct substitution confirms \(v(2)=0\) and \(v(5)=0\).
Answer: Backward on \([0,2)\) and \((2,5)\); forward on \((5,6]\). The only direction change is at \(t=5\). If describing the broader stretch during which its direction is backward, state that it moves backward on both sides of \(t=2\), with a momentary stop at \(t=2\).
Worked Example: Respecting the Domain Endpoints
A particle’s velocity, in meters per second, is \(v(t)=t(t-2)(t-4)\) for \(0\le t\le5\), where \(t\) is measured in seconds. State the forward- and backward-motion intervals and the direction-change times.
The zeros are \(t=0\), \(t=2\), and \(t=4\). The zero at \(t=0\) is a domain endpoint, so there is no earlier part of this motion interval with which to compare signs. Test \(t=1\) in \((0,2)\):
Test \(t=3\) in \((2,4)\):
Test \(t=4.5\) in \((4,5]\):
The signs show that the particle moves forward on \((0,2)\), backward on \((2,4)\), and forward on \((4,5]\). At the domain endpoint \(t=0\), \(v(0)=0( -2)( -4)=0\), so the particle is at rest there, not moving forward at that instant. At the interior zeros, \(v(2)=2(0)(-2)=0\) and \(v(4)=4(2)(0)=0\). The sign changes at both interior zeros, so both are direction changes. The direction-change times are \(2\) seconds and \(4\) seconds.
Answer: Forward on \((0,2)\) and \((4,5]\); backward on \((2,4)\). Direction changes at \(t=2\) seconds and \(t=4\) seconds.
Endpoint and Wording Choices
Intervals must reflect both the stated domain and the meaning of the requested phrase. If the question asks where velocity is positive, the answer is exactly where \(v(t)>0\); zeros are excluded. If it asks for direction-change times, report the interior zeros where the sign switches, not every zero. If it asks for the time interval being considered, preserve its endpoints and do not extend the motion beyond them.
| What the velocity does | What to report |
|---|---|
| Positive on an interval | Forward motion there |
| Negative on an interval | Backward motion there |
| Zero at one instant | The particle is momentarily at rest |
| Changes sign at an interior zero | A direction change occurs there |
| Equals zero but keeps the same sign on both sides | A momentary stop, but no direction change |
Be explicit about units when time is measured in a context. For example, “the particle changes direction at \(t=5\) seconds” states both the time and its unit. Use the form requested by the prompt: an exact time, a decimal to the requested precision, or an interval. Do not round an exact value unless the question calls for an approximation.
Common Mistake: Calling Every Velocity Zero a Reversal
A common incorrect conclusion is to equate “velocity is zero” with “the particle changes direction.” For example, in \(v(t)=(t-2)^2(t-5)\), one might write: “The direction changes at \(t=2\) and \(t=5\), because these are the zeros of velocity.” The zero at \(t=2\) is not a reversal: the test values give \(v(1)=-4\) and \(v(3)=-2\), both negative.
The correction is to compare signs on the two sides of each interior zero. At \(t=2\), the signs are negative and negative, so the particle momentarily stops but does not reverse. At \(t=5\), the signs are negative and positive, so it does reverse. Also avoid including a zero in a strict forward- or backward-motion interval: at that instant, \(v(t)=0\), not \(v(t)>0\) or \(v(t)<0\).
Check Your Understanding
For each question, use the stated time domain and distinguish the sign of velocity from a direction change.
- For \(v(t)=(t-1)(t-4)\) on \(0\le t\le5\), find the forward- and backward-motion intervals and the direction-change times.
- For \(v(t)=(t-3)^2\) on \(0\le t\le5\), when is the particle at rest, and does it change direction there?
- Why is one sign test sufficient on an interval between consecutive zeros when velocity is continuous and has no zero there?
- If the velocity is positive immediately before and immediately after an interior zero, what can you conclude about direction change at that zero?
- In reporting an interval where \(v(t)<0\), should an interior time at which \(v(t)=0\) be included? Explain.
