Position, Velocity, and Acceleration
A motion model describes an object's position as time changes. Let \(x(t)\) denote its position at time \(t\). The derivative \(x'(t)\) is its velocity: it describes the rate and direction of position change. The derivative of velocity is acceleration. Thus, when \(x\) is twice differentiable,
These relationships connect motion to the shape of the position graph. Acceleration is the second derivative of position, so its sign determines whether the position graph is concave up or concave down on an interval. But acceleration's sign alone does not determine whether the object is speeding up: that also depends on the direction of motion, given by the sign of velocity.
Why Acceleration Determines Concavity
Concavity describes how the slope of a graph changes. For a position graph, the slope at time \(t\) is \(x'(t)=v(t)\). Therefore, if velocity increases as time increases, the slopes of the position graph increase, and the graph is concave up. If velocity decreases, its slopes decrease, and the graph is concave down.
The connection can also be justified directly using the Mean Value Theorem. Suppose \(x\) is twice differentiable on an interval and \(a(t)=x''(t)>0\) throughout it. Choose any two times \(t_1<t_2\) in that interval. Apply the Mean Value Theorem to \(v=x'\) on \([t_1,t_2]\). There is a time \(c\) between \(t_1\) and \(t_2\) such that
Thus \(v(t_2)>v(t_1)\): velocity is increasing, so the slope of the position graph increases and the graph is concave up. If \(a<0\) throughout the interval, the same argument gives \(v(t_2)-v(t_1)<0\), so velocity decreases and the graph is concave down. These conclusions concern intervals where the indicated sign holds, not just a single time.
Concavity Is Not the Same as Speed
Velocity has both a sign and a size. Its sign gives direction along the chosen coordinate axis: \(v>0\) means position is increasing, while \(v<0\) means position is decreasing. Speed is the nonnegative quantity \(|v|\). Acceleration describes the change in velocity, not directly the change in speed.
Where \(v\ne 0\), differentiate the speed \(|v|\). If \(v>0\), then \(|v|=v\), so the derivative of speed is \(a\). If \(v<0\), then \(|v|=-v\), so the derivative of speed is \(-a\). Together, these cases give
Consequently, when \(v\) and \(a\) have the same sign, speed increases; when their signs are opposite, speed decreases. At a time when \(v=0\), this formula does not apply. To describe the motion at or around such a time, examine the signs on either side instead.
| Velocity | Acceleration | Direction and change in speed |
|---|---|---|
| Positive | Positive | Moving in the positive direction; speeding up |
| Positive | Negative | Moving in the positive direction; slowing down |
| Negative | Negative | Moving in the negative direction; speeding up |
| Negative | Positive | Moving in the negative direction; slowing down |
Worked Example: Acceleration, Concavity, and Changing Speed
Worked Example: Acceleration, Concavity, and Changing Speed
Suppose an object's position is \(x(t)=t^3-6t^2+9t\), for \(t\ge 0\). Find its velocity and acceleration, determine the concavity of the position graph, and describe its direction and speed on each relevant time interval.
Differentiate position to obtain velocity:
Differentiate again to obtain acceleration:
Velocity is zero at \(t=1\) and \(t=3\); acceleration is zero at \(t=2\). These times divide the domain into intervals on which the signs are constant. Testing one time in each interval gives:
| Time interval | Sign of \(v\) | Sign of \(a\) | Motion | Concavity of \(x\) |
|---|---|---|---|---|
| \(0<t<1\) | Positive | Negative | Moving positive; slowing down | Concave down |
| \(1<t<2\) | Negative | Negative | Moving negative; speeding up | Concave down |
| \(2<t<3\) | Negative | Positive | Moving negative; slowing down | Concave up |
| \(t>3\) | Positive | Positive | Moving positive; speeding up | Concave up |
For example, at \(t=\tfrac12\), \(v=3(\!-\tfrac12)(\!-\tfrac52)=\tfrac{15}{4}>0\), while \(a=6(\!-\tfrac32)=-9<0\). The object moves in the positive direction but slows down. At \(t=\tfrac52\), \(v=3(\tfrac32)(-\tfrac12)=-\tfrac94<0\) and \(a=3>0\), so it moves in the negative direction while slowing down.
Acceleration changes sign at \(t=2\), from negative to positive, so the position graph changes from concave down to concave up there. In fact, \(x(2)=8-24+18=2\), and \(v(2)=12-24+9=-3\): the object is still moving in the negative direction at the inflection point.
Check: Differentiating \(x(t)=t^3-6t^2+9t\) gives \(3t^2-12t+9\), as listed for \(v\); differentiating that gives \(6t-12\), as listed for \(a\). The signs used in the table follow from the factored velocity and acceleration expressions. In particular, \(v\) changes sign at \(1\) and \(3\), while \(a\) changes sign at \(2\).
Worked Example: A Rest Point Without an Inflection Point
Worked Example: A Rest Point Without an Inflection Point
Let \(x(t)=t^2-4t+1\), for \(t\ge 0\). Determine the motion and concavity, paying special attention to the time when the object is momentarily at rest.
Differentiate:
For \(0\le t<2\), \(v<0\) and \(a>0\), so the object moves in the negative direction and slows down. At \(t=2\), \(v=0\), and \(x(2)=4-8+1=-3\). For \(t>2\), \(v>0\) and \(a>0\), so it moves in the positive direction and speeds up. Since \(a=2>0\) on the whole domain, the position graph is concave up throughout.
Thus the object reverses direction at \(t=2\), but the graph does not change concavity there. For a numerical check, \(v(1)=-2\), \(v(2)=0\), and \(v(3)=2\); velocity increases through zero. Also, \(x(1)=1-4+1=-2\), \(x(2)=-3\), and \(x(3)=9-12+1=-2\), consistent with the position decreasing and then increasing.
Check: Differentiating \(t^2-4t+1\) gives \(2t-4\), and differentiating \(2t-4\) gives \(2\). The acceleration remains positive on both sides of \(t=2\), so there is no change in concavity there.
Worked Example: Reading Motion from Sign Data
Worked Example: Reading Motion from Sign Data
A motion model has the following velocity and acceleration signs on three consecutive intervals. Use the data to describe the motion and the concavity of the position graph on each interval.
| Interval | Velocity sign | Acceleration sign |
|---|---|---|
| \(0<t<2\) | Negative | Negative |
| \(2<t<5\) | Negative | Positive |
| \(5<t<7\) | Positive | Positive |
On \(0<t<2\), negative velocity means the object moves in the negative direction. Negative acceleration means velocity is decreasing; because velocity is already negative, its magnitude increases, so the object speeds up. Since \(x''=a<0\), the position graph is concave down.
On \(2<t<5\), the object still moves in the negative direction, but positive acceleration increases velocity toward zero. The signs are opposite, so speed decreases. The position graph is concave up.
On \(5<t<7\), both velocity and acceleration are positive. The object moves in the positive direction and speeds up; the position graph remains concave up.
Check: Multiplying the signs of \(v\) and \(a\) gives positive on the first interval, negative on the second, and positive on the third. These products correctly indicate speeding up, slowing down, and speeding up, respectively. The acceleration signs also match the stated concavity: negative gives concave down, while positive gives concave up. The data describe behavior on the stated intervals; by themselves, they do not specify what happens at the boundary times.
Rest, Reversal, and Inflection
A time with \(v(t)=0\) is a time when the object is momentarily at rest. If velocity changes sign there, the object reverses direction. Neither fact by itself says that the position graph changes concavity. In the second example, velocity changes from negative to positive at \(t=2\), but acceleration stays positive, so the graph stays concave up.
An inflection point is a point where the graph changes concavity. For a twice-differentiable position function, a change in the sign of \(x''=a\) is the relevant evidence for a concavity change. Finding \(a=0\) identifies a candidate time, not a conclusion: the sign of acceleration must be checked on both sides. The first example has an inflection at \(t=2\) because acceleration changes sign there; the second has no inflection at its rest point.
Using Motion Data Carefully
To analyze a motion model, first determine where velocity and acceleration are zero or undefined, if that information is available. These times divide the domain into intervals. On each interval, establish the signs of \(v\) and \(a\), then use the sign pair to describe direction and change in speed. Read concavity directly from acceleration.
Keep the conclusions appropriately local. A sign at one isolated time does not establish a sign throughout an interval. Likewise, sign data for several intervals do not determine behavior at their endpoints unless the model supplies that information. In a physical setting, the coordinate direction is a choice: “positive” and “negative” refer to the chosen axis, while speeding up and slowing down describe changes in the magnitude of velocity.
Compute \(v=x'\) and \(a=v'=x''\) when the position model is given.
Find times where \(v\) or \(a\) is zero or undefined, and check the model's domain.
Use \(v\) for direction, the signs of \(v\) and \(a\) together for speed change, and \(a\) for concavity.
At a rest point or a zero of acceleration, inspect signs on adjacent intervals before claiming a reversal or inflection.
Check Your Understanding
Use the relationships among position, velocity, acceleration, speed, and concavity to answer these questions.
- If \(v>0\) and \(a<0\) throughout an interval, which direction does the object move, and is it speeding up or slowing down?
- What does \(a>0\) throughout an interval imply about the concavity of the position graph and the change in velocity?
- If \(v=0\) at a time, what additional information would show that the object reverses direction there?
- Why is \(a=0\) at a single time insufficient to conclude that the position graph has an inflection point?
- An object has \(v<0\) and \(a<0\) on an interval. Describe its direction, speed change, and position-graph concavity.
