Speed and Its Rate of Change
A particle’s velocity describes how its position changes and gives the direction of motion along a coordinate line. Its speed describes how fast it moves, without indicating direction. If \(v(t)\) is velocity at time \(t\), then speed is the nonnegative quantity \(|v(t)|\).
Speed can increase while velocity is negative: the particle then moves in the negative coordinate direction and becomes faster. Speed can also decrease while velocity is positive: the particle continues in the positive direction but becomes slower. To decide whether speed is increasing or decreasing, compare the signs of velocity and acceleration.
Acceleration \(a(t)\) is the rate of change of velocity. In particle motion, \(a(t)=v'(t)\). The signs of \(v\) and \(a\) reveal whether velocity’s magnitude is growing or shrinking, provided the velocity is nonzero.
The Sign Test
The factor \(v/|v|\) is \(1\) when velocity is positive and \(-1\) when it is negative. The sign test can therefore be organized as follows, with the zero cases treated separately.
| Velocity \(v\) | Acceleration \(a\) | Conclusion when both are nonzero |
|---|---|---|
| Positive | Positive | Speeding up |
| Positive | Negative | Slowing down |
| Negative | Positive | Slowing down |
| Negative | Negative | Speeding up |
A compact way to remember the rule is to examine the product \(v(t)a(t)\). A positive product means the nonzero signs match, so speed increases. A negative product means the signs differ, so speed decreases. A zero product is not enough to apply that conclusion: it means at least one of the two quantities is zero, and that case needs its own interpretation.
Why the Sign Test Works
When \(v(t)>0\), speed is \(|v(t)|=v(t)\), so its derivative is \(a(t)\). When \(v(t)<0\), speed is \(|v(t)|=-v(t)\), so its derivative is \(-a(t)\). Combining those two cases gives the formula in the theorem.
In the first case, positive acceleration makes speed increase and negative acceleration makes it decrease. In the second case, the signs have the opposite effect: negative acceleration makes speed increase, while positive acceleration makes it decrease. This explains why matching signs indicate speeding up in either direction of motion.
Worked Example: Negative Velocity and Negative Acceleration
Worked Example: Moving Faster in the Negative Direction
A particle has velocity \(v(t)=-3\) meters per second and acceleration \(a(t)=-2\) meters per second squared at a particular time. Is it speeding up or slowing down?
Both values are negative, so their signs match. Equivalently, their product is
Since \(v(t)<0\), the speed is \(-v(t)\) near this time. Its instantaneous rate of change is therefore
The rate is positive, so speed is increasing: the particle is speeding up. The negative velocity still means motion in the negative coordinate direction; it does not mean the speed is negative.
Check: The rule predicts speeding up because the signs match, and the calculated rate of speed change is \(2>0\), confirming that conclusion.
Worked Example: Opposite Signs
Worked Example: Positive Velocity and Negative Acceleration
A particle has velocity \(v(t)=5\) meters per second and acceleration \(a(t)=-1.5\) meters per second squared at a particular time. Determine whether its speed is increasing or decreasing.
The velocity is positive, so speed equals velocity locally. Therefore,
The rate of change of speed is negative, so speed is decreasing. In the sign test, the signs are opposite; their product is
The particle is slowing down at that time. Its velocity is still positive, so it is moving in the positive coordinate direction even as its speed decreases.
Check: A negative derivative of speed means that the speed is decreasing, consistent with the opposite-sign rule.
When Acceleration Is Zero
Suppose \(v(t)\ne0\) but \(a(t)=0\). Then the formula gives an instantaneous rate of change of speed equal to zero. At that instant, speed is neither increasing nor decreasing at a nonzero instantaneous rate. This does not tell us whether speed is increasing or decreasing on nearby intervals: acceleration may change sign around the time in question.
Worked Example: Zero Acceleration at an Instant
Suppose velocity, in meters per second, is \(v(t)=4+(t-1)^2\). Then acceleration is
At \(t=1\), the values are \(v(1)=4+(1-1)^2=4\) and \(a(1)=2(1-1)=0\). Since velocity is positive, speed equals \(v\), and its instantaneous rate of change at \(t=1\) is \(0\).
For \(t<1\), \(a(t)=2(t-1)<0\), so speed decreases as time approaches \(1\). For \(t>1\), \(a(t)=2(t-1)>0\), so speed increases after \(1\). Thus the zero acceleration marks an instant with zero rate of speed change, while the behavior on either side is determined by the nearby signs.
Check: Differentiating the given velocity gives \(a(t)=2(t-1)\), which is negative before \(1\), zero at \(1\), and positive after \(1\), matching the stated speed behavior.
When Velocity Is Zero
If \(v(t)=0\), the formula involving \(v/|v|\) does not apply: its denominator would be zero. The sign test therefore cannot classify speed as increasing or decreasing at that instant just by comparing the signs of \(v\) and \(a\). Speed is zero at that time, but its behavior requires care.
If \(v(t_0)=0\) and \(a(t_0)\ne0\), then near \(t_0\) the velocity changes approximately as \(a(t_0)(t-t_0)\). This means that, moving forward from \(t_0\), the particle’s speed grows from zero for times sufficiently close to \(t_0\). However, the two-sided derivative of speed at \(t_0\) need not exist: speed is nonnegative and can have a corner there. A statement about the forward behavior is not the same as applying the nonzero-velocity sign test at \(t_0\).
If both velocity and acceleration are zero, the signs alone do not determine what happens next. The particle may remain at rest for a while, or its speed may begin to grow; further information about the motion is needed. The useful principle is to inspect velocity and acceleration at nearby times whenever the zero values leave the classification unresolved.
Worked Example: Zero Velocity with Nonzero Acceleration
Let \(v(t)=2(t-2)\) meters per second. Then \(a(t)=v'(t)=2\) meters per second squared. At \(t=2\), \(v(2)=2(2-2)=0\) while \(a(2)=2\).
At \(t=2\), the nonzero-velocity sign test does not apply. To check forward behavior, take \(t>2\) close to \(2\). Then \(t-2>0\), so \(v(t)=2(t-2)>0\), and
For \(t>2\), the derivative of this speed is \(2>0\), so the speed increases just after \(t=2\). This conclusion comes from examining nearby times, not from assigning a speeding-up or slowing-down label using \(v(2)=0\) alone.
Check: Differentiating \(v(t)=2(t-2)\) gives \(a(t)=2\). For \(t>2\), the speed is \(2(t-2)\), whose derivative is \(2\), confirming the forward increase.
A Reliable Process
Determine whether velocity and acceleration are positive, negative, or zero at the time or throughout the interval being considered.
Matching nonzero signs mean speed increases; opposite nonzero signs mean it decreases.
If acceleration is zero and velocity is nonzero, the instantaneous speed-change rate is zero. If velocity is zero, the sign test does not apply at that instant.
When a zero value prevents a classification, examine the signs or the speed at nearby times before making a claim about an interval or one-sided behavior.
Common Pitfall: Treating Direction as Speed
A common error is to interpret negative velocity as negative speed, or to assume that positive acceleration always means speeding up. For example, if \(v=-6\) and \(a=2\), a mistaken argument might say, “Acceleration is positive, so the particle is speeding up.” But the signs are opposite, and the speed rate is
The particle is slowing down. Its velocity is negative, but its speed is \(|-6|=6\), not \(-6\). Acceleration indicates how velocity changes; whether that change increases or decreases speed depends on the direction already indicated by velocity.
Also avoid extending the nonzero sign test to a zero value. If \(v=0\), the expression \(v/|v|\) is undefined. If \(a=0\) while \(v\ne0\), the instantaneous speed-change rate is zero, but nearby behavior can still differ on the two sides of that time.
Check Your Understanding
For each situation, classify the speed change if the information allows it. When a zero prevents the sign test, state what additional information would help.
- A particle has \(v=7\) and \(a=3\). Is it speeding up or slowing down?
- A particle has \(v=-4\) and \(a=6\). What is the instantaneous rate of change of its speed, and what does its sign mean?
- If \(v=-2\) and \(a=0\) at an instant, what is the instantaneous rate of change of speed? Does this alone settle the behavior on nearby intervals?
- If \(v=0\) and \(a=5\) at an instant, can the nonzero-velocity sign test be used at that instant? What can be said about sufficiently near forward times under these conditions?
- If \(v=0\) and \(a=0\), do those two values alone determine whether speed will increase just afterward? Explain.
