When Does a Particle Change Direction?
A particle moving along a line can be momentarily at rest without reversing its motion. To decide whether it actually changes direction, first find when its velocity is zero, then compare the velocity signs just before and just after each such time.
Recall that positive velocity means motion in the positive coordinate direction, while negative velocity means motion in the negative coordinate direction. At an instant when velocity is zero, the particle is at rest at that instant. That fact alone does not tell us what happens next: it might resume moving in the same direction, or it might move in the opposite direction.
Finding Rest Times and Testing the Signs
If position is given by \(x(t)\), velocity is \(v(t)=x'(t)\). Solve \(v(t)=0\) to find candidate rest times in the time interval under consideration. Then check the sign of \(v\) on each side of each candidate.
For a zero at an interior time \(t=c\), checking the sides means choosing times just below and just above \(c\), within the particle’s time domain. Often it is enough to factor \(v(t)\), mark its zeros on a number line, and test one value in each interval. If velocity is continuous and has no other zeros in one of those intervals, its sign cannot switch within that interval: a continuous function could pass from positive to negative only by taking the value zero somewhere between.
If position \(x(t)\) is given, compute \(v(t)=x'(t)\). If velocity is given directly, use it as written.
Solve \(v(t)=0\), retaining only times in the stated time domain.
Determine whether velocity is positive or negative immediately before and after each interior candidate.
Opposite signs mean a direction change; matching signs mean no direction change at that instant.
This test is local: it concerns motion just before and after a particular time. An endpoint of the stated time interval has only one side within the interval, so the two-sided test cannot classify it as a direction change on that interval alone.
Why Comparing Signs Works
For motion along a line, the sign of velocity records the direction of motion. If \(v(t)>0\) over a stretch of time, position is increasing there, and the particle moves in the positive coordinate direction. If \(v(t)<0\), position is decreasing, and the particle moves in the negative coordinate direction. This is the interpretation of velocity established in the earlier material on direction of motion from velocity signs.
Now consider an interior rest time \(c\) at which velocity is continuous and is nonzero on each side sufficiently close to \(c\). On each side, velocity has one fixed sign. If the signs differ, the particle approaches the instant moving in one coordinate direction and leaves it moving in the other. That is a direction change. If the signs match, it approaches and leaves moving in the same coordinate direction. It is at rest at \(c\), but it does not reverse.
- \(v\) changes sign across \(c\): the particle changes direction at \(c\).
- \(v\) keeps the same sign across \(c\): the particle is momentarily at rest but does not change direction at \(c\).
The continuity condition matters because it lets us use signs on intervals around the candidate time. In the polynomial examples below, velocity is continuous. If the velocity is zero throughout an interval rather than at an isolated instant, the particle is at rest for that interval; it is not merely a momentary stop there. The simple two-sided test applies to an isolated interior rest time.
Worked Examples
Worked Example: A Rest Time Without a Direction Change
A particle’s position, in meters, is modeled by \(x(t)=\frac{1}{3}(t-1)^3\), where \(0\leq t\leq 2\) and \(t\) is measured in seconds. Find its rest time and determine whether its direction changes there.
Differentiate position to find velocity:
Set velocity equal to zero:
Check values on either side of \(1\). At \(t=0.5\) seconds, \(v(0.5)=(0.5-1)^2=0.25\ \text{m/s}\). At \(t=1.5\) seconds, \(v(1.5)=(1.5-1)^2=0.25\ \text{m/s}\). Both values are positive. More generally, a square is positive for every \(t\neq 1\), so velocity is positive on both sides of the rest time.
Therefore the particle is momentarily at rest at \(t=1\) second but does not change direction there. Check the candidate by substitution: \(v(1)=(1-1)^2=0\ \text{m/s}\). The velocity formula also checks against the position model because differentiating \(x(t)=\frac{1}{3}(t-1)^3\) gives \((t-1)^2\), as used above.
Worked Example: A Direction Change at a Rest Time
A particle’s position, in meters, is \(x(t)=t^3-3t^2\), for \(1\leq t\leq 3\), with \(t\) in seconds. Find the interior rest time and decide whether the particle reverses direction.
Differentiate:
Set the factored velocity equal to zero. The algebraic candidates are \(t=0\) and \(t=2\), but only \(t=2\) belongs to the specified interval \(1\leq t\leq 3\). Thus the interior rest time is \(2\) seconds. Substitution confirms it: \(v(2)=3(2)(2-2)=0\ \text{m/s}\).
Test one time on each side. At \(t=1.5\) seconds, \(v(1.5)=3(1.5)(1.5-2)=4.5(-0.5)=-2.25\ \text{m/s}\). At \(t=2.5\) seconds, \(v(2.5)=3(2.5)(2.5-2)=7.5(0.5)=3.75\ \text{m/s}\). Velocity is negative before the rest time and positive after it.
The particle therefore changes direction at \(t=2\) seconds: it moves in the negative coordinate direction before that time and in the positive coordinate direction afterward. The derivative check is \(x'(t)=3t^2-6t=3t(t-2)\), matching the velocity used to locate and classify the rest time.
Worked Example: Classifying More Than One Rest Time
Let the position be \(x(t)=\frac{1}{4}(t-1)^4-\frac{2}{3}(t-1)^3\), in meters, on \(0\leq t\leq 4\), where \(t\) is in seconds. Find all rest times and classify each one.
Differentiate each term:
Setting this equal to zero gives \((t-1)^2=0\) or \(t-3=0\). The rest times are \(t=1\) second and \(t=3\) seconds, both interior to the given interval. Direct substitution gives \(v(1)=0^2(-2)=0\ \text{m/s}\) and \(v(3)=2^2(0)=0\ \text{m/s}\).
For \(t=1\), test \(t=0.5\) and \(t=1.5\). The velocities are \(v(0.5)=(-0.5)^2(-2.5)=-0.625\ \text{m/s}\) and \(v(1.5)=(0.5)^2(-1.5)=-0.375\ \text{m/s}\). The signs match, so there is no direction change at \(t=1\) second.
For \(t=3\), test \(t=2.5\) and \(t=3.5\). The velocities are \(v(2.5)=(1.5)^2(-0.5)=-1.125\ \text{m/s}\) and \(v(3.5)=(2.5)^2(0.5)=3.125\ \text{m/s}\). The signs are opposite, so the particle changes direction at \(t=3\) seconds. Differentiating the stated position gives \((t-1)^3-2(t-1)^2\), which factors to the velocity used in both checks.
A Common Mistake: Treating Every Rest Time as a Reversal
A common incorrect conclusion is: “Velocity is zero at \(t=1\), so the particle changes direction at \(t=1\).” For the first example, the calculation behind that claim is \(v(1)=(1-1)^2=0\). That correctly establishes a rest time, but it does not establish a change of direction.
The missing step is checking the signs. In that example, \(v(t)=(t-1)^2>0\) for every \(t\neq 1\). Thus velocity is positive both immediately before and immediately after \(1\). The corrected conclusion is that the particle stops instantaneously and continues in the same direction. To establish a reversal, show opposite signs on the two sides, not merely a zero at the time itself.
Similarly, a sign check should use values from the correct time domain. A root outside the stated interval is not a rest time during the interval being analyzed. And if a candidate lies at an endpoint, do not claim that the two-sided sign test proves a direction change there: one side of the endpoint is outside the given interval.
Why the Distinction Matters
Describing a particle’s motion requires more than listing times when its velocity is zero. The sign comparison tells whether the direction actually changes. That distinction is essential when interpreting a position or velocity model: a rest instant is a fact about velocity at one time, while a reversal is a claim about the motion on both sides of that time.
Acceleration can provide additional information about motion, but a zero of velocity should be classified by checking velocity signs on either side. In particular, do not substitute an acceleration value for this sign comparison. The conclusion sought here is about direction, and direction is indicated by the sign of velocity.
Check Your Understanding
For each question, identify any required sign checks before deciding whether direction changes.
- If \(v(t)=(t-4)^2\) near \(t=4\), is \(t=4\) a rest time? Does the particle change direction there?
- If \(v(t)=(t+1)(t-2)\) on \(0\leq t\leq 3\), which zeros are rest times in the stated interval, and what signs occur on either side of each interior rest time?
- What does a negative velocity immediately before a rest time and a positive velocity immediately after it tell you?
- Why is \(v(c)=0\) alone insufficient to conclude that a particle changes direction at \(t=c\)?
- What limitation applies to using a two-sided sign test when the candidate time is an endpoint of the stated interval?
