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Motion with Piecewise Acceleration

Build a piecewise motion model by integrating each acceleration interval and matching the velocity and position at every switching time.

Intermediate 10 min read

What You'll Learn

  • Integrate acceleration separately on each time interval.
  • Carry the velocity at a switching time into the next interval.
  • Match position as well as velocity across acceleration changes.
  • Use a local time variable to simplify formulas after a switch.
  • Apply final-position data to determine an unknown acceleration parameter.
  • Check a piecewise solution by differentiating and testing its boundary values.

Why Acceleration Changes Require Care

In many motion problems, acceleration is not described by one formula for all time. A force may change, an object may enter a new region, or a control system may switch modes. The acceleration can jump at a specified time even though the object does not suddenly change its position or velocity.

The practical method is to integrate on each interval separately. At every switching time, the velocity reached from the interval before the switch becomes the initial velocity for the interval after it. The position must be carried forward in the same way. Integration constants are therefore linked across intervals; they cannot be chosen independently.

Key principle: A change in acceleration can produce a change in the rate of velocity, but it does not by itself produce an instantaneous jump in velocity or position. In a standard piecewise-acceleration model, position and velocity are matched continuously at each switching time.

Piecewise Acceleration and Matching Conditions

Let \(\mathbf{r}(t)\) denote position, \(\mathbf{v}(t)=\mathbf{r}'(t)\) velocity, and \(\mathbf{a}(t)=\mathbf{v}'(t)\) acceleration. Suppose the switching times are \(t_0<t_1<\cdots<t_n\). On each interval, acceleration is specified by a formula \(\mathbf{a}_k(t)\). We assume these formulas are continuous, or at least integrable, on their respective intervals. Their values on opposite sides of a switch need not agree.

Definition: A piecewise-acceleration motion is a motion whose acceleration follows a specified formula on each of a sequence of time intervals. At each finite switching time \(t_k\), position and velocity are matched by requiring the left- and right-hand limits to satisfy \(\mathbf{r}(t_k^-)=\mathbf{r}(t_k^+)\) and \(\mathbf{v}(t_k^-)=\mathbf{v}(t_k^+)\). The acceleration itself may have different one-sided limits.

On a single interval beginning at time \(t_k\), suppose the state at the start is known: position \(\mathbf{r}(t_k)\) and velocity \(\mathbf{v}(t_k)\). For a time \(t\) in that interval, integrating acceleration gives velocity. Integrating velocity then gives position:

$$ \mathbf{v}(t)=\mathbf{v}(t_k)+\int_{t_k}^{t}\mathbf{a}_k(s)\,ds, \qquad \mathbf{r}(t)=\mathbf{r}(t_k)+\int_{t_k}^{t}\mathbf{v}(u)\,du. $$

These equations are also the matching rule in action. Evaluate them at the end of the interval to obtain the velocity and position at the next switch. Use those values as the starting state for the following interval. This avoids treating each interval as an unrelated problem.

A Formula for Position on One Interval

Substituting the velocity integral into the position integral gives a useful expression directly in terms of acceleration. For \(t\geq t_k\) within the interval, it is

$$ \mathbf{r}(t) = \mathbf{r}(t_k)+(t-t_k)\mathbf{v}(t_k) +\int_{t_k}^{t}(t-s)\mathbf{a}_k(s)\,ds. $$

This formula is valid when the acceleration is integrable on the interval. It shows the two contributions to position particularly clearly: the starting velocity contributes \((t-t_k)\mathbf{v}(t_k)\), while acceleration contributes the final integral.

1
Integrate acceleration.
Find the velocity on the current interval, using the velocity already established at its starting time.
2
Integrate velocity.
Find position using the position at the interval’s starting time.
3
Evaluate at the switch.
Record both the position and velocity there, then use them as initial data on the next interval.
4
Check the result.
Differentiate position to recover velocity, differentiate velocity to recover the stated acceleration on each open interval, and compare both states at each switch.

Why the Position Formula Works

The formula follows from integrating velocity, with the acceleration contribution accumulated over the time remaining after each instant \(s\). To see this, substitute \(\mathbf{v}(u)=\mathbf{v}(t_k)+\int_{t_k}^{u}\mathbf{a}_k(s)\,ds\) into the position integral:

$$ \begin{aligned} \mathbf{r}(t) &=\mathbf{r}(t_k)+\int_{t_k}^{t} \left(\mathbf{v}(t_k)+\int_{t_k}^{u}\mathbf{a}_k(s)\,ds\right)du\\ &=\mathbf{r}(t_k)+(t-t_k)\mathbf{v}(t_k) +\int_{t_k}^{t}\int_{t_k}^{u}\mathbf{a}_k(s)\,ds\,du\\ &=\mathbf{r}(t_k)+(t-t_k)\mathbf{v}(t_k) +\int_{t_k}^{t}(t-s)\mathbf{a}_k(s)\,ds. \end{aligned} $$

For the last equality, the integration region consists of pairs with \(t_k\leq s\leq u\leq t\). If the order of integration is reversed, each fixed \(s\) contributes for \(u\) from \(s\) to \(t\), an interval of length \(t-s\). This proves the formula component by component. In particular, acceleration added at time \(s\) affects velocity for the remaining time, and therefore affects position with the factor \(t-s\).

Worked Example: A Two-Interval Vector Motion

Worked Example: A Two-Interval Vector Motion

An object moves in the plane, with time measured in seconds and position in metres. Its acceleration is \(\mathbf{a}(t)=(2,0)\ \text{m/s}^2\) for \(0\leq t<2\), and \(\mathbf{a}(t)=(-1,2)\ \text{m/s}^2\) for \(t\geq2\). Initially, \(\mathbf{r}(0)=(1,-1)\ \text{m}\) and \(\mathbf{v}(0)=(0,3)\ \text{m/s}\). Find position and velocity, and check continuity at \(t=2\).

On the first interval, integrate acceleration and apply the initial velocity:

$$ \mathbf{v}(t)=(0,3)+(2t,0)=(2t,3)\ \text{m/s}. $$

Integrate again and apply the initial position:

$$ \mathbf{r}(t)=(1,-1)+(t^2,3t)=(1+t^2,-1+3t)\ \text{m}. $$

At the switch, these give \(\mathbf{v}(2)=(4,3)\ \text{m/s}\) and \(\mathbf{r}(2)=(5,5)\ \text{m}\). Let \(\tau=t-2\) on the second interval. Starting from the state at the switch, integrate the new acceleration:

$$ \begin{aligned} \mathbf{v}(t)&=(4,3)+(-\tau,2\tau)=(4-\tau,3+2\tau)\ \text{m/s},\\ \mathbf{r}(t)&=(5,5)+(4,3)\tau+\tfrac12(-1,2)\tau^2\\ &=(5+4\tau-\tfrac12\tau^2,\ 5+3\tau+\tau^2)\ \text{m}. \end{aligned} $$

At \(t=2\), \(\tau=0\), so the second formulas give \(\mathbf{v}=(4,3)\ \text{m/s}\) and \(\mathbf{r}=(5,5)\ \text{m}\), exactly matching the first interval. Differentiating the second position formula gives its stated velocity; differentiating that velocity gives \((-1,2)\ \text{m/s}^2\). The corresponding derivatives on the first interval give \((2t,3)\) and \((2,0)\), as required.

Worked Example: A Time-Dependent Acceleration

Worked Example: A Time-Dependent Acceleration

Let \(\mathbf{a}(t)=(2t,0)\ \text{m/s}^2\) for \(0\leq t<2\), and \(\mathbf{a}(t)=(0,-2)\ \text{m/s}^2\) for \(t\geq2\). Given \(\mathbf{r}(0)=(0,2)\ \text{m}\) and \(\mathbf{v}(0)=(1,-1)\ \text{m/s}\), determine the motion on both intervals.

For \(0\leq t<2\), integrate \((2t,0)\) from \(0\) to \(t\):

$$ \mathbf{v}(t)=(1,-1)+(t^2,0)=(1+t^2,-1)\ \text{m/s}. $$

Integrating this velocity gives

$$ \mathbf{r}(t)=(0,2)+(t+\tfrac13t^3,-t) =(t+\tfrac13t^3,2-t)\ \text{m}. $$

At \(t=2\), the state is \(\mathbf{v}(2)=(5,-1)\ \text{m/s}\) and \(\mathbf{r}(2)=(14/3,0)\ \text{m}\). With \(\tau=t-2\), integrate the second acceleration using this state:

$$ \begin{aligned} \mathbf{v}(t)&=(5,-1)+(0,-2\tau)=(5,-1-2\tau)\ \text{m/s},\\ \mathbf{r}(t)&=(14/3,0)+(5,-1)\tau+(0,-\tau^2)\\ &=(14/3+5\tau,-\tau-\tau^2)\ \text{m}. \end{aligned} $$

At the switch, \(\tau=0\), giving \((5,-1)\ \text{m/s}\) and \((14/3,0)\ \text{m}\), so both state variables match. On the first interval, differentiating \(\mathbf{r}\) gives \((1+t^2,-1)\), and differentiating again gives \((2t,0)\). On the second, differentiation gives \((5,-1-2\tau)\) and then \((0,-2)\), confirming both acceleration formulas.

Using a Final Condition to Find an Unknown

Piecewise motion problems may also ask for an unknown acceleration parameter. The parameter is carried through the integrations, and a final position or velocity condition determines its value. The switching rules still apply: first compute the state at the switch, then use that state in the later interval.

Worked Example: Determining an Acceleration from Final Position

An object moves along the \(x\)-axis from rest at \(x(0)=0\). Its acceleration is \(a(t)=k\ \text{m/s}^2\) for \(0\leq t<2\), and \(a(t)=-2\ \text{m/s}^2\) for \(2\leq t\leq4\). Find \(k\) if \(x(4)=4\ \text{m}\).

On the first interval, \(v(0)=0\), so integration gives \(v(t)=kt\ \text{m/s}\). Since \(x(0)=0\), a second integration gives \(x(t)=\tfrac12kt^2\ \text{m}\). Thus at \(t=2\), \(v(2)=2k\ \text{m/s}\) and \(x(2)=2k\ \text{m}\).

For the second interval, put \(\tau=t-2\). Integrating the constant acceleration from the matched velocity gives \(v(t)=2k-2\tau\ \text{m/s}\). Integrating from the matched position gives

$$ x(t)=2k+2k\tau-\tau^2\ \text{m}. $$

At \(t=4\), \(\tau=2\). Therefore \(x(4)=2k+4k-4=6k-4\ \text{m}\). The required final position gives \(6k-4=4\), so \(6k=8\), and \(k=4/3\ \text{m/s}^2\).

Check: substituting this value gives \(x(4)=6(4/3)-4=8-4=4\ \text{m}\), as required. The switch state is \(v(2)=8/3\ \text{m/s}\), \(x(2)=8/3\ \text{m}\). The second-interval formulas at \(\tau=0\) reproduce both values. Differentiating \(x(t)\) gives \(v(t)=2k-2\tau\), and differentiating again gives \(-2\ \text{m/s}^2\); the first-interval derivatives give \(kt\) and \(k\).

A Common Mistake: Restarting from Zero

A frequent error is to integrate the new acceleration after a switch and silently set the new velocity or position equal to zero. That would describe a reset of the motion, not an ordinary change in acceleration. For example, in the first worked example, the velocity at \(t=2\) is \((4,3)\ \text{m/s}\). The incorrect calculation \(\mathbf{v}(t)=(-\tau,2\tau)\) gives \(\mathbf{v}(2)=(0,0)\), contradicting the velocity already reached. The correct expression is \(\mathbf{v}(t)=(4,3)+(-\tau,2\tau)\), which matches at \(\tau=0\).

The same issue occurs with position. The local-time formula for position must include the position at the switch and the velocity there. A safe check is to substitute the switching time into both neighbouring formulas. If either position or velocity differs, the constants have not been matched correctly.

Interpreting the Switch

When acceleration has a jump, the velocity can remain continuous while its derivative changes abruptly. At the switching instant, a two-sided derivative of velocity may fail to exist if the one-sided accelerations differ. The piecewise formulas specify acceleration on the intervals; the value assigned at the single switching instant does not change the integrals or the resulting continuous state.

For several intervals, repeat the same procedure in chronological order. Keep track of the units as well: integrating acceleration over time produces velocity, and integrating velocity over time produces position. These checks are useful, but they do not replace matching the actual values at each switch.

Check Your Understanding

Use the matching conditions and interval-by-interval integrations to answer the following questions.

  1. If acceleration changes at \(t=3\), what data from the first interval are needed to begin the second interval?
  2. Why may acceleration have different one-sided limits at a switch while position and velocity remain continuous?
  3. For constant acceleration \(\mathbf{a}_k\) on an interval starting at \(t_k\), write the velocity formula using the state at \(t_k\).
  4. What does the factor \(t-s\) represent in the acceleration contribution to the position formula?
  5. A proposed second-interval velocity formula gives a different value at the switch from the first-interval formula. What should be checked or corrected?

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