2.4 Motion with Piecewise-Defined Acceleration

A Rocket, Two Rules, One Flight

[Video: A small model rocket sits on a launch pad. The engine ignites. The rocket accelerates upward for several seconds, trailing fire and smoke. Then the flame dies --- the engine has burned out. The rocket keeps climbing, but it is slowing down now, gravity pulling against it. It reaches a peak, hangs for an instant, then falls back toward the ground. A parachute deploys. A timer running along the bottom of the screen marks three distinct phases: "Engine burn," "Coasting upward," "Falling." An acceleration readout in the corner shows a positive value during the burn, then snaps to $-9.8$ m/s$^2$ the moment the engine cuts off.]

Watch the acceleration readout carefully. For the first few seconds, it holds steady at some large positive value --- the engine is pushing. Then, at the moment of engine cutoff, the readout jumps discontinuously to $-9.8$ m/s$^2$. No smooth transition. No gradual fade. One value, then instantly another.

The rocket's flight does not obey a single acceleration rule from start to finish. It obeys one rule during the burn and a different rule after burnout. To describe the full flight, you need both rules --- and you need to know how to stitch them together at the boundary.

This is not a special case. It is how most real motion actually works. A car accelerates from a stoplight, cruises at constant speed, then brakes. A ball is thrown upward, rises under gravity, bounces off the ceiling, and falls back down. A sprinter explodes out of the blocks, hits top speed, and coasts to a stop. In each case, the motion naturally divides into stages, each with its own acceleration.

Sections 2.2 and 2.3 gave you the tools to handle each stage individually. This section is about putting those stages together.

Prediction

Before you read on: At the exact instant the rocket engine cuts off, what happens to the following quantities? For each one, choose: "jumps to a new value instantly," "changes smoothly and continuously," or "I'm not sure."

1. The acceleration
2. The velocity
3. The position

Commit to your answers before continuing.

This is a question that trips up a surprising number of students. Here is a hint: think about what it would mean physically if velocity jumped at engine cutoff. The rocket is at some particular speed the instant before cutoff. If velocity jumped, the rocket would instantaneously change speed --- from, say, 80 m/s to 50 m/s in zero time. That would require infinite acceleration, which is not physical. So velocity cannot jump.

The acceleration, on the other hand, absolutely can jump. One instant the engine is firing and producing a net upward acceleration; the next instant it is off and only gravity acts. The forces change abruptly, so the acceleration changes abruptly. There is nothing unphysical about that.

And position? If position jumped, the rocket would teleport. That doesn't happen.

So: acceleration can be discontinuous at a stage boundary. Velocity and position must be continuous. This is the single most important idea in this section. Everything else follows from it.

Exploration: Building Motion Stage by Stage

[Interactive: Piecewise Motion Builder. The screen is divided into two regions. The top region contains a timeline bar divided into stages. Initially there are two stages: Stage 1 from $t = 0$ to $t = t_1$, and Stage 2 from $t = t_1$ to $t = t_2$. Students can drag the boundary between stages to change $t_1$, and they can drag the right end to change $t_2$. A button lets them add a third stage.

For each stage, a dropdown lets the student choose an acceleration type: "constant" (with a number field), "linear in time" ($a = bt$ with a coefficient field), or "zero." Below the timeline, three synchronized graphs display $a(t)$, $v(t)$, and $x(t)$ in real time.

Key behavior: When the student changes the acceleration in Stage 1, the $v(t)$ and $x(t)$ curves in Stage 1 update. The Stage 2 curves then rebuild automatically, starting from the final velocity and final position of Stage 1. If the student moves the boundary $t_1$, the Stage 1 endpoint values change, and Stage 2 recalculates from the new boundary values.

A highlighted annotation at each stage boundary shows "Boundary conditions: $v(t_1) = \text{...}$, $x(t_1) = \text{...}$" to make the continuity requirement visible.

Guided prompts: - "Set Stage 1 to $a = +10$ m/s$^2$ and Stage 2 to $a = -9.8$ m/s$^2$. Watch the velocity graph. Does it have a corner or a break at the boundary?" - "Now drag the boundary time earlier. What happens to the peak velocity? What happens to the position curve?" - "Add a third stage with $a = 0$. What kind of motion does this represent?" - "Can you create a motion where the object returns to its starting position?"]

Spend a few minutes exploring this. Pay special attention to what happens at the stage boundaries. The velocity graph should have a corner (a sudden change in slope) at each boundary, but no gap. The position graph should be smooth and continuous --- no corners, no gaps. The acceleration graph is the only one that jumps.

Concept Reveal: The Piecewise Strategy

What you just explored has a clean, repeatable structure. Here it is, stated as a general method.

The setup

Suppose an object's acceleration is defined by different rules in different time intervals:

$$a(t) = \begin{cases} a_1(t) & \text{for } 0 \leq t < t_1 \ a_2(t) & \text{for } t_1 \leq t < t_2 \ a_3(t) & \text{for } t_2 \leq t \leq t_3 \end{cases}$$

Each $a_i(t)$ might be a constant, a function of time, or anything else. The tools from Sections 2.2 and 2.3 handle each piece individually.

The procedure

Stage 1 ($0 \leq t < t_1$): Start from the given initial conditions $v(0) = v_0$ and $x(0) = x_0$. Use the acceleration $a_1(t)$ to find velocity and position in this interval:

$$v_1(t) = v_0 + \int_0^t a_1(t')\,dt'$$

$$x_1(t) = x_0 + \int_0^t v_1(t')\,dt'$$

At the end of Stage 1, evaluate the final velocity and position:

$$v(t_1) = v_1(t_1), \qquad x(t_1) = x_1(t_1)$$

Stage 2 ($t_1 \leq t < t_2$): Now use $a_2(t)$, but start from the values that Stage 1 ended with:

$$v_2(t) = v(t_1) + \int_{t_1}^t a_2(t')\,dt'$$

$$x_2(t) = x(t_1) + \int_{t_1}^t v_2(t')\,dt'$$

Stage 3 and beyond: repeat. Each stage inherits its initial conditions from the previous stage's final values.

The boundary conditions

The rules that glue the stages together are called continuity conditions (or matching conditions):

• Position is continuous: $x_1(t_1) = x_2(t_1)$. The object does not teleport.
• Velocity is continuous: $v_1(t_1) = v_2(t_1)$. The object does not instantaneously change speed.
• Acceleration may be discontinuous. Forces can switch on or off abruptly, so $a$ can jump at a boundary.

These are not optional conventions. They are physical requirements. An object has a definite location and a definite velocity at every instant, and those quantities change continuously in time because forces are finite. (In more advanced physics, you will encounter situations where even these assumptions break down --- collisions, for instance --- but for now, continuity of $x$ and $v$ is the law.)

The piecewise strategy in one sentence: Solve each stage using the tools you already have, and let the end of each stage set the beginning of the next.

A Complete Example: The Rocket Flight

Let's work through the rocket problem from the opening video, start to finish.

Setup: A rocket launches vertically from rest. During the first 10 seconds, the engine produces a constant upward acceleration of $a_1 = 20$ m/s$^2$ (this is the net acceleration, already accounting for gravity during the burn). At $t = 10$ s, the engine cuts off. From that point on, the only acceleration is gravity: $a_2 = -9.8$ m/s$^2$.

Question: How high does the rocket go? When does it reach its peak?

Stage 1: Engine burn ($0 \leq t \leq 10$ s)

Initial conditions: $v(0) = 0$, $x(0) = 0$ (taking the launch pad as the origin, upward as positive).

With constant acceleration $a_1 = 20$ m/s$^2$:

$$v_1(t) = 0 + 20t = 20t$$

$$x_1(t) = 0 + \frac{1}{2}(20)t^2 = 10t^2$$

At $t_1 = 10$ s:

$$v(10) = 20(10) = 200 \text{ m/s}$$

$$x(10) = 10(10)^2 = 1000 \text{ m}$$

So at the moment of engine cutoff, the rocket is 1000 m above the ground and traveling upward at 200 m/s.

Stage 2: Coasting under gravity ($t \geq 10$ s)

Initial conditions for this stage: $v(10) = 200$ m/s, $x(10) = 1000$ m.

With constant acceleration $a_2 = -9.8$ m/s$^2$:

$$v_2(t) = 200 + (-9.8)(t - 10) = 200 - 9.8(t - 10)$$

$$x_2(t) = 1000 + 200(t - 10) + \frac{1}{2}(-9.8)(t - 10)^2$$

Notice the form: $t - 10$ appears everywhere, not bare $t$. This is because Stage 2 starts at $t = 10$ s, not at $t = 0$. If you use bare $t$ in the constant-acceleration formulas, you will get the wrong answer. This is one of the most common errors in piecewise problems.

Before you read on: Compute the time at which the rocket reaches its maximum height. Then compute that maximum height. Try it yourself before checking below.

Finding the peak: At maximum height, velocity is zero:

$$v_2(t_{\text{peak}}) = 0 \implies 200 - 9.8(t_{\text{peak}} - 10) = 0$$

$$t_{\text{peak}} - 10 = \frac{200}{9.8} \approx 20.4 \text{ s}$$

$$t_{\text{peak}} \approx 30.4 \text{ s}$$

Finding the maximum height:

$$x_2(30.4) = 1000 + 200(20.4) + \frac{1}{2}(-9.8)(20.4)^2$$

$$= 1000 + 4080 - 2039 \approx 3041 \text{ m}$$

The rocket climbs for about 30 seconds total and reaches a height of roughly 3 km. Notice how the continuity conditions --- $v(10) = 200$ m/s and $x(10) = 1000$ m --- carried the information from Stage 1 into Stage 2. Without them, Stage 2 would have no way to know how the engine burn went.

[Video: Animation of the rocket flight with synchronized graphs of $a(t)$, $v(t)$, and $x(t)$. The acceleration graph shows a step function: $+20$ for $0 \leq t \leq 10$, then $-9.8$ for $t > 10$. The velocity graph is a line rising from 0 to 200 m/s, then declining linearly. The position graph is a parabola curving upward during Stage 1, then a steeper parabola that levels off and reaches a peak. Vertical dashed lines at $t = 10$ s mark the stage boundary on all three graphs. Labels call out the continuity: "velocity is continuous here" and "position is continuous here," while the acceleration graph shows a visible jump.]

A Subtlety Worth Pausing On

Look at the velocity graph from the rocket example. At $t = 10$ s, the graph has a corner --- the slope changes abruptly from $+20$ to $-9.8$. The velocity itself is continuous (no gap), but its derivative (the acceleration) is not.

Now look at the position graph at $t = 10$ s. The position is continuous and smooth --- no corner, no kink. This is because position is the integral of velocity, and integrating a continuous function always produces a smooth result.

So the hierarchy looks like this:

Quantity	Behavior at a stage boundary
$a(t)$ (acceleration)	Can jump discontinuously
$v(t)$ (velocity)	Continuous, but may have a corner (slope change)
$x(t)$ (position)	Continuous and smooth (no corner)

Each level of integration adds one degree of smoothness. This is not a coincidence --- it is a general fact about integrals, and it will come up again when you study forces and impulse.

Connection to Previous Sections

In Section 2.2, you learned to solve problems with constant acceleration --- one rule, start to finish. In Section 2.3, you extended that to nonuniform acceleration: the acceleration could be any function of time, but it was still one function for the entire motion.

This section stitches those tools together. Each stage of a piecewise problem is just a Section 2.2 or Section 2.3 problem on its own. What is new here is not the calculus within each stage --- it is the organizational work of setting up the stages, enforcing the continuity conditions, and tracking the information that passes from one stage to the next.

In other words, the hard part of piecewise motion is not the physics or the math. It is the bookkeeping.

Spaced Recall

Before we move to practice, take a moment to recall some earlier ideas.

From Section 2.2: Write down the three constant-acceleration kinematic equations from memory. What assumption must be true for those equations to apply?

From Section 1.3: What is the difference between a reference frame and a coordinate system? In the rocket problem above, what was our choice of coordinate system?

If either of these feels fuzzy, revisit the relevant section. These ideas are cumulative --- you will need them repeatedly.

Metacognition Check

Pause and reflect: What do you find hardest about piecewise problems so far?

If you said "the calculus" or "solving the integrals," that is worth noting --- but those are Section 2.2 and 2.3 skills, and you can review them.

If you said "keeping track of which stage I'm in" or "remembering what carries over at the boundary," then you have identified the real challenge. The hardest part of piecewise problems is not the calculus --- it is organizing the stages and tracking what carries over. If you feel lost in a piecewise problem, resist the urge to do more algebra. Instead, draw a clearer diagram of the stages: label the time intervals, write down the acceleration rule for each one, and explicitly list the boundary values. The algebra will follow once the organization is clear.

Practice

Layer 1: Concrete --- A Two-Stage Problem

A car starts from rest and accelerates at $a_1 = 3$ m/s$^2$ for 8 seconds. Then the driver hits the brakes, and the car decelerates at $a_2 = -5$ m/s$^2$ until it stops.

(a) What is the car's velocity at the end of Stage 1?

(b) How far does the car travel during Stage 1?

(c) How long does Stage 2 last?

(d) How far does the car travel during Stage 2?

(e) What is the total distance traveled?

Check your answer

**(a)** At the end of Stage 1: $$v(8) = 0 + 3(8) = 24 \text{ m/s}$$ **(b)** Distance during Stage 1: $$x_1(8) = 0 + \frac{1}{2}(3)(8)^2 = 96 \text{ m}$$ **(c)** Stage 2 starts with $v = 24$ m/s and $a = -5$ m/s$^2$. The car stops when $v = 0$: $$0 = 24 + (-5)\Delta t_2 \implies \Delta t_2 = \frac{24}{5} = 4.8 \text{ s}$$ **(d)** Distance during Stage 2: $$\Delta x_2 = 24(4.8) + \frac{1}{2}(-5)(4.8)^2 = 115.2 - 57.6 = 57.6 \text{ m}$$ **(e)** Total distance: $96 + 57.6 = 153.6$ m. Notice how the velocity at the end of Stage 1 ($24$ m/s) became the initial velocity for Stage 2. That handoff is the continuity condition at work.

Layer 2: Pattern --- A Three-Stage Problem with Nonuniform Acceleration

A subway train starts from rest at a station. Its motion has three stages:

• Stage 1 ($0 \leq t \leq 4$ s): The acceleration increases linearly from zero, $a(t) = 1.5t$ m/s$^2$.
• Stage 2 ($4 \leq t \leq 20$ s): The train cruises at constant velocity.
• Stage 3 ($20 \leq t \leq 26$ s): The train brakes with constant deceleration until it stops.

(a) Find the velocity at the end of Stage 1 by integrating $a(t) = 1.5t$.

(b) What is the constant velocity during Stage 2?

(c) Find the position at the end of Stage 1, then at the end of Stage 2.

(d) Find the constant deceleration in Stage 3 that brings the train to rest in exactly 6 seconds.

(e) Sketch qualitative graphs of $a(t)$, $v(t)$, and $x(t)$ for the entire trip.

Check your answer

**(a)** Integrating $a(t) = 1.5t$ with $v(0) = 0$: $$v_1(t) = \int_0^t 1.5t'\,dt' = \frac{1.5}{2}t^2 = 0.75t^2$$ At $t = 4$ s: $v(4) = 0.75(16) = 12$ m/s. **(b)** By the continuity condition, Stage 2 begins at $v = 12$ m/s. Since Stage 2 has constant velocity (meaning $a = 0$), the velocity throughout Stage 2 is $12$ m/s. **(c)** Position at the end of Stage 1: $$x_1(t) = \int_0^t 0.75t'^2\,dt' = 0.25t^3$$ $$x(4) = 0.25(64) = 16 \text{ m}$$ Position at the end of Stage 2: The train travels at 12 m/s for $(20 - 4) = 16$ s: $$x(20) = 16 + 12(16) = 208 \text{ m}$$ **(d)** Stage 3 starts at $v = 12$ m/s and must reach $v = 0$ in 6 s: $$0 = 12 + a_3(6) \implies a_3 = -2 \text{ m/s}^2$$ **(e)** The qualitative features to capture in your sketch: - $a(t)$: a rising line (Stage 1), then zero (Stage 2), then a negative constant (Stage 3). - $v(t)$: a parabolic curve rising from zero (Stage 1), then a flat line at 12 m/s (Stage 2), then a straight line declining to zero (Stage 3). Continuous everywhere, with corners at $t = 4$ and $t = 20$. - $x(t)$: a cubic curve bending upward (Stage 1), then a straight line with slope 12 (Stage 2), then a parabola that levels off (Stage 3). Smooth and continuous everywhere.

Layer 3: Structure --- Why Must Velocity Be Continuous?

Explain, in physical terms, why position and velocity must be continuous at stage boundaries, but acceleration does not have to be.

Your explanation should address: (a) What would it mean physically if position were discontinuous? (b) What would it mean physically if velocity were discontinuous? (c) Why is discontinuous acceleration physically reasonable?

Check your answer

**(a)** Discontinuous position would mean the object teleports --- it is at one location at time $t_1^-$ and at a different location at time $t_1^+$, with no time in between to travel the distance. This is not physically possible for any real object. **(b)** Discontinuous velocity would mean the object's speed changes by a finite amount in zero time. By Newton's second law ($F = ma$), a finite change in velocity in zero time would require infinite acceleration, which would require infinite force. No finite force can produce a velocity jump. (In idealized collision models, we sometimes *allow* velocity jumps as an approximation, but that is a deliberate modeling choice, not physical reality.) **(c)** Acceleration is proportional to the net force on the object. Forces can change abruptly --- an engine turns off, a rope snaps, you step off a ledge. When the forces change abruptly, the acceleration changes abruptly. There is nothing unphysical about a sudden change in force, so there is nothing unphysical about a sudden change in acceleration. The underlying mathematical reason is that velocity is the integral of acceleration, and integrals of bounded functions are always continuous. Position is the integral of velocity, so it is continuous too. But acceleration is the *derivative* of velocity, and derivatives of continuous functions need not be continuous (think of $|t|$ at $t = 0$).

Layer 4: Creation --- Design a Three-Stage Flight

Design a three-stage vertical motion (with specific numerical values for acceleration and stage durations) in which an object:

1. Launches from the ground ($x = 0$) at rest ($v = 0$),
2. Reaches a maximum height during the motion,
3. Returns to the ground ($x = 0$) at the end of the third stage.

Your design should specify the acceleration and duration of each stage. Then verify that the object actually returns to $x = 0$ by computing $x(t)$ at the end of each stage.

Hints and a sample solution

**Hints:** The object must end at $x = 0$, which means the total displacement over all three stages must be zero. The object must reach a maximum height, which means velocity must be positive for a while (climbing), reach zero (the peak), and then become negative (falling). One natural design: a powered ascent, a coasting phase under gravity, and a free fall or braked descent. **One possible solution:** - **Stage 1** ($0 \leq t \leq 5$ s): $a = +8$ m/s$^2$. $v(5) = 40$ m/s, $x(5) = 100$ m. - **Stage 2** ($5 \leq t \leq 15$ s): $a = -9.8$ m/s$^2$. $v(15) = 40 - 9.8(10) = -58$ m/s, $x(15) = 100 + 40(10) - \frac{1}{2}(9.8)(100) = 100 + 400 - 490 = 10$ m. The velocity passes through zero at $t = 5 + 40/9.8 \approx 9.08$ s, so the peak height occurs during this stage. - **Stage 3** ($15 \leq t \leq t_f$): $a = -9.8$ m/s$^2$ (continued free fall). We need $x(t_f) = 0$: $10 + (-58)(t_f - 15) + \frac{1}{2}(-9.8)(t_f - 15)^2 = 0$. Solving: $4.9\tau^2 + 58\tau - 10 = 0$ where $\tau = t_f - 15$. The positive root gives $\tau \approx 0.17$ s, so $t_f \approx 15.17$ s. Your design will almost certainly look different from this one, and that is fine. The important thing is that the continuity conditions are satisfied at every boundary and the final position is zero.

A Practical Checklist

When you encounter a piecewise motion problem, here is a reliable procedure:

1. Identify the stages. Read the problem and determine how many stages there are and what triggers the transition between them (a time, a speed, a position).

2. For each stage, write down the acceleration rule. Is it constant? A function of time? Zero?

3. Start with Stage 1. Use the given initial conditions and the acceleration rule to find $v(t)$ and $x(t)$ for that stage.

4. Evaluate the boundary values. At the end of Stage 1, compute the final velocity and final position. These become the initial conditions for Stage 2.

5. Repeat for each subsequent stage. Always use $t - t_{\text{start}}$ (or integrate from $t_{\text{start}}$ to $t$) to account for the stage not starting at $t = 0$.

6. Answer the question. Once you have $v(t)$ and $x(t)$ for all stages, you can find whatever the problem asks for --- peak height, total time, stopping distance, and so on.

If you get lost, go back to step 1. A clear diagram of the stages, with the acceleration, boundary times, and boundary values labeled, will do more for you than any amount of algebraic manipulation.

Reflection

What is the general procedure for attacking a piecewise motion problem?

Try to state it in your own words before looking at the checklist above. If your procedure matches the checklist, you have internalized the method. If it doesn't, compare the two versions and see where they differ.

Also consider: What is the one thing that connects all the stages together? It is not the acceleration --- that changes from stage to stage. It is the continuity conditions. The velocity and position at the end of one stage become the initial conditions for the next. That handoff is the thread running through the entire problem.

Looking Ahead

You now have the tools to handle any one-dimensional motion described by a known acceleration, whether that acceleration is constant, time-dependent, or piecewise. But there is a class of questions we have been answering along the way without fully examining them: When does the object reach its highest point? When does it change direction? Where is it moving fastest?

These are questions about turning points and extrema --- special moments in the motion where something interesting happens. In the next section, we step back from computation and focus on qualitative analysis: extracting the key features of a motion from its equations and graphs without necessarily solving for every detail. You already used one of these tools when you set $v = 0$ to find the rocket's peak altitude. Section 2.5 develops that idea systematically --- and shows you how much of the motion's story you can read without grinding through algebra.