Velocity and Acceleration Tell Different Parts of the Story
Velocity describes both how quickly an object moves and its direction along a chosen line. Its sign indicates direction: positive velocity means motion in the positive direction, and negative velocity means motion in the negative direction. Speed is the magnitude of velocity, so it measures how quickly the object moves without recording direction.
Acceleration describes how velocity changes with time. A positive acceleration means velocity is increasing; a negative acceleration means velocity is decreasing. Neither statement by itself says whether the object's speed is increasing. To decide that, we must compare the signs of velocity and acceleration.
The reason is that acceleration can change the magnitude of a negative velocity in either direction. For example, a positive acceleration makes a negative velocity less negative, which reduces speed. The sign of acceleration alone is therefore not a reliable indicator of speeding up or slowing down.
Speed as a Function of Velocity
Let \(v(t)\) be the velocity of an object at time \(t\). Its speed is the nonnegative quantity \(S(t)=|v(t)|\). Acceleration is the derivative of velocity, \(a(t)=v'(t)\). These definitions let us calculate the rate at which speed changes.
This formula gives a direct test. If \(S'(t)>0\), speed is increasing; if \(S'(t)<0\), speed is decreasing. In particular, for \(v(t)\ne 0\), the sign of \(S'(t)\) is the sign of the product \(v(t)a(t)\). A positive product means speeding up, and a negative product means slowing down.
Why the Sign Test Works
The formula follows from the two possible expressions for absolute value. When velocity is positive, speed equals velocity. When velocity is negative, speed equals the negative of velocity. Differentiating the appropriate expression gives the speed rate.
In a neighbourhood where velocity remains positive, \(S(t)=|v(t)|=v(t)\). Therefore \(S'(t)=v'(t)=a(t)\). Positive acceleration increases speed; negative acceleration decreases it.
In a neighbourhood where velocity remains negative, \(S(t)=|v(t)|=-v(t)\). Therefore \(S'(t)=-v'(t)=-a(t)\). Positive acceleration decreases speed; negative acceleration increases it.
Together these cases prove the rule whenever velocity is nonzero. For positive velocity, \(v\) and \(a\) have the same sign exactly when \(a>0\), which makes \(S'>0\). For negative velocity, they have the same sign exactly when \(a<0\), which also makes \(S'=-a>0\). Opposite signs give \(S'<0\) in both cases.
Worked Examples with Constant Instantaneous Values
Worked Example: Negative Velocity and Negative Acceleration
At a particular instant, a car has velocity \(v=-6\) metres per second and acceleration \(a=-2\) metres per second squared. Is it speeding up or slowing down?
The velocity is negative, so the car is moving in the negative direction. The acceleration is also negative. Since the signs match, the speed is increasing. We can check using the speed-rate formula:
The rate of change of speed is positive, so the car is speeding up at \(2\ \text{m/s}^2\) at this instant. The units are speed divided by time, as expected for a speed rate.
As a check on the interpretation, a small time later the velocity would be more negative if this acceleration persisted: for a time increase of \(0.1\) seconds, the constant-acceleration estimate gives \(v=-6+(-2)(0.1)=-6.2\ \text{m/s}\). Its magnitude has increased from \(6\) to \(6.2\ \text{m/s}\), consistent with speeding up.
Worked Example: Positive Velocity and Negative Acceleration
A cyclist is moving with velocity \(v=8\) metres per second and has acceleration \(a=-3\) metres per second squared. Determine whether the cyclist is speeding up or slowing down.
The velocity is positive and acceleration is negative, so the signs are opposite. Calculate the speed rate:
The negative result means speed is decreasing at \(3\ \text{m/s}^2\) at this instant. The cyclist is slowing down, even though the acceleration is negative. For a \(0.1\)-second interval over which the acceleration is constant, the velocity changes to \(8+(-3)(0.1)=7.7\ \text{m/s}\). It remains positive, and the speed decreases from \(8\) to \(7.7\ \text{m/s}\), checking the conclusion.
Worked Example: Use a Velocity Function
Worked Example: Classifying Motion at Several Times
Suppose an object's velocity, in metres per second, is \(v(t)=(t-1)(t-3)\), where \(t\) is measured in seconds. Determine whether it is speeding up or slowing down at \(t=0\), \(t=2\), and \(t=4\).
First expand the velocity function:
Differentiate to obtain the acceleration:
At \(t=0\),
The signs are opposite. Equivalently, \(S'(0)=\operatorname{sign}(3)(-4)=-4\ \text{m/s}^2\). The object is slowing down.
At \(t=2\),
Here the velocity is nonzero but acceleration is zero, so \(S'(2)=\operatorname{sign}(-1)(0)=0\). The instantaneous rate of change of speed is zero. This fact alone does not establish that speed stays constant over an interval.
At \(t=4\),
Both signs are positive, and \(S'(4)=\operatorname{sign}(3)(4)=4\ \text{m/s}^2\). The object is speeding up. In each case the sign of the calculated speed rate agrees with the classification.
What If Velocity or Acceleration Is Zero?
The sign test needs care when either quantity is zero. If \(v\ne0\) and \(a=0\), then the formula gives \(S'=0\): the instantaneous speed rate is zero. This is a statement about the rate at that instant, not a guarantee that the speed is constant over a time interval. Acceleration can be zero at one instant and nonzero at nearby times.
If \(v=0\), the object is momentarily at rest and the sign of velocity cannot be used to assign a direction at that instant. The product \(va\) is zero, but this does not by itself classify the motion as speeding up or slowing down. In fact, the function \(|v(t)|\) need not have a two-sided derivative there. If acceleration is nonzero and velocity changes smoothly through zero, the object may be slowing down as it approaches rest and speeding up after it leaves rest. The exact conclusion depends on the motion near that time.
Common Mistake: Treating Positive Acceleration as Speeding Up
A frequent error is to conclude that positive acceleration always means increasing speed. Consider an object with velocity \(-10\ \text{m/s}\) and acceleration \(+2\ \text{m/s}^2\). The incorrect reasoning is: “Acceleration is positive, so the object is speeding up.” But the velocity is negative, so positive acceleration pushes it toward zero.
The speed is decreasing, so the object is slowing down. For example, if the acceleration remains constant for \(0.1\) seconds, the velocity becomes \(v=-10+(2)(0.1)=-9.8\ \text{m/s}\). Its speed decreases from \(10\) to \(9.8\ \text{m/s}\), confirming the corrected conclusion. Always compare acceleration with the direction indicated by velocity instead of reading acceleration's sign alone.
Why the Distinction Matters
Describing motion accurately requires separating direction from rate of motion. Velocity can increase numerically while speed decreases: a velocity changing from \(-10\) to \(-9\ \text{m/s}\) is increasing as a signed number, but its magnitude is decreasing. Conversely, velocity can decrease numerically while speed increases: a change from \(-9\) to \(-10\ \text{m/s}\) makes velocity more negative and speed larger.
The sign test converts this distinction into a reliable procedure. First identify the sign of velocity, then the sign of acceleration, and finally decide whether the signs match. This classifies the instantaneous change in speed; it does not, without information over an interval, describe the object's entire subsequent motion.
Check Your Understanding
For each question, distinguish what the signs establish at an instant from what would require information over a time interval.
- An object has \(v=5\ \text{m/s}\) and \(a=2\ \text{m/s}^2\). Is its speed increasing or decreasing?
- An object has \(v=-7\ \text{m/s}\) and \(a=3\ \text{m/s}^2\). What is the rate of change of its speed?
- Why does negative acceleration not always mean that an object is slowing down?
- If \(v\ne0\) and \(a=0\) at one instant, what can you conclude about the instantaneous speed rate? What can you not conclude about an interval?
- Why is the same-sign/opposite-sign test not sufficient by itself when \(v=0\)?
