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One-Dimensional Kinematics

2.2 Constant Acceleration and Its Exact Solutions

The promise of complete prediction

Drop a ball from your hand. For a brief moment it is motionless, hanging in the air. Then it falls --- slowly at first, then faster and faster. In half a second it has dropped about a meter and a quarter. In a full second, nearly five meters. The motion is not uniform; the ball is speeding up the entire way down.

Now here is the remarkable thing. If you know a single number --- the acceleration due to gravity --- you can predict exactly where that ball will be after 0.3 seconds. After 0.5 seconds. After 1 second. After any time at all (as long as the ball hasn't hit the floor). You can predict its speed at each of those moments, too. One number, and the entire future of the motion is determined.

This is the power of constant acceleration. It is the simplest interesting case of motion --- simpler than anything you will encounter in the real world of varying forces and changing conditions, but rich enough to describe free fall, braking cars, launching rockets (for short durations), and a host of other situations where acceleration doesn't change appreciably.

In this section, you will derive the formulas that make these predictions possible. Not by memorizing them from a table, but by building them yourself from the calculus you already know.

[Video: A ball is released from rest and falls in front of a meter stick. The video plays at half speed, with a digital clock overlay. At $t = 0.1$ s, $0.2$ s, $0.3$ s, and so on, the ball's position is marked with a horizontal line. The distances between consecutive marks grow visibly larger. A voiceover says: "The ball falls farther in each successive tenth of a second. It's not moving at a constant speed --- it's accelerating. But the rate at which it speeds up is constant. That single fact determines everything."]

Prediction: a car from rest

Before we derive anything, make a guess.

Before you read on: A car accelerates from rest at $2$ m/s$^2$. After 5 seconds:

  1. How fast is it going?
  2. How far has it traveled?

Don't compute. Estimate. Write down two numbers before continuing.

[Interactive: Two input fields, one for velocity and one for distance. The student enters their guesses and clicks "Lock in." The guesses are saved and displayed alongside the derived answers later in the section, so the student can see how close their intuition was.]

Most people get the velocity roughly right --- it feels natural that $2$ m/s$^2$ for $5$ seconds should give something around $10$ m/s. The distance is harder. Many students guess $50$ meters (multiplying $10$ m/s by $5$ s). But the car wasn't going $10$ m/s the whole time --- it started from rest and built up speed gradually. The actual distance is less than $50$ m. How much less? The derivation below will give the exact answer.

The guiding question

Why does constant acceleration produce such simple, exact formulas for velocity and position? And what do those formulas actually tell us about the motion?

The answer has two parts: one from calculus, and one from geometry. We will walk through both, side by side, and watch them arrive at the same destination.

Derivation path 1: integration

We start from the definition of acceleration that you met in Chapter 1. Acceleration is the derivative of velocity with respect to time:

$$a(t) = \frac{dv}{dt}$$

If the acceleration is constant --- call it simply $a$ --- then this equation says

$$\frac{dv}{dt} = a$$

This is a statement about rates: velocity changes at a steady rate of $a$ meters per second, every second. To find how velocity accumulates over time, we integrate both sides from $t = 0$ to some later time $t$:

$$\int_0^t \frac{dv}{dt'}\,dt' = \int_0^t a\,dt'$$

The left side is just $v(t) - v(0)$ by the fundamental theorem of calculus. The right side is $a \cdot t$, since $a$ is a constant that pulls out of the integral. Writing $v_0 = v(0)$ for the initial velocity, we get

$$\boxed{v(t) = v_0 + at}$$

This is the first kinematic equation. It says: your velocity at any time equals your starting velocity plus however much the constant acceleration has added (or subtracted) since the start.

Pause and check: Does this match your prediction for the car? With $v_0 = 0$ and $a = 2$ m/s$^2$, after $t = 5$ s we get $v(5) = 0 + 2 \times 5 = 10$ m/s.

Now we repeat the process. Velocity is the derivative of position:

$$v(t) = \frac{dx}{dt}$$

We have just found that $v(t) = v_0 + at$. Substituting:

$$\frac{dx}{dt} = v_0 + at$$

Integrate both sides from $0$ to $t$:

$$\int_0^t \frac{dx}{dt'}\,dt' = \int_0^t (v_0 + at')\,dt'$$

The left side gives $x(t) - x(0)$. On the right, we integrate term by term: $\int_0^t v_0\,dt' = v_0 t$ and $\int_0^t at'\,dt' = \frac{1}{2}at^2$. Writing $x_0 = x(0)$ for the initial position:

$$\boxed{x(t) = x_0 + v_0 t + \tfrac{1}{2}at^2}$$

This is the second kinematic equation. It says: your position at any time equals your starting position, plus the distance you would have covered at your initial velocity, plus an extra term from the acceleration that grows as the square of time.

That $\tfrac{1}{2}at^2$ term is why the ball in the opening video falls farther in each successive interval. The distance doesn't grow linearly --- it grows quadratically. This is the mathematical signature of constant acceleration.

Check the car again: With $x_0 = 0$, $v_0 = 0$, $a = 2$ m/s$^2$, and $t = 5$ s:

$$x(5) = 0 + 0 + \tfrac{1}{2}(2)(25) = 25 \text{ m}$$

The car travels $25$ meters, not $50$. If you guessed $50$, you were imagining the car at full speed the entire time. The acceleration means it spent the early seconds going slowly, so the total distance is only half what the final speed times the total time would suggest.

Derivation path 2: the area under the velocity graph

There is a completely different way to arrive at the same position formula, and it requires no integration at all --- just geometry.

Since $v(t) = v_0 + at$ is a linear function of time, the velocity-versus-time graph is a straight line. It starts at height $v_0$ on the vertical axis and rises (or falls) with slope $a$.

[Video: Animation builds a $v$-vs-$t$ graph. A straight line starts at $v_0$ on the vertical axis and rises to $v_0 + at$ at time $t$. The region between the line and the time axis is shaded. The shaded region is then decomposed into two parts: a rectangle of height $v_0$ and width $t$ (shaded blue), and a triangle sitting on top of the rectangle with base $t$ and height $at$ (shaded orange). Labels appear: "Rectangle area = $v_0 t$" and "Triangle area = $\frac{1}{2}(t)(at) = \frac{1}{2}at^2$." A final label reads: "Total area = displacement = $v_0 t + \frac{1}{2}at^2$."]

You already know from calculus that the area under a velocity curve gives displacement. For a straight-line $v(t)$, this area is a trapezoid, which you can split into a rectangle and a triangle:

  • Rectangle: base $= t$, height $= v_0$. Area $= v_0 t$. This is the displacement you'd get at constant velocity $v_0$.
  • Triangle: base $= t$, height $= at$. Area $= \frac{1}{2}at^2$. This is the extra displacement from the acceleration.

Adding them together and including the initial position:

$$x(t) = x_0 + v_0 t + \tfrac{1}{2}at^2$$

The same formula, derived from a picture instead of an integral. The calculus and the geometry agree, as they must.

Before you read on: Which derivation felt more convincing to you --- the calculus path or the graphical path? There is no right answer. Some people trust algebra; others trust pictures. The point is that both are saying the same thing, and being able to move between them is a sign of real understanding.

The concept: formulas as consequences, not rules

Here is the key idea of this section, and it is worth stating plainly:

The kinematic equations are not rules handed down from authority. They are the inevitable, unavoidable consequence of a single assumption: acceleration is constant.

If you accept that $a$ is constant, then calculus --- or geometry --- forces you to conclude that $v(t)$ is linear and $x(t)$ is quadratic. You have no choice in the matter. The formulas are not something to memorize; they are something that must be true whenever the assumption holds.

This also means you can always re-derive them if you forget. Start from $a = \text{const}$. Integrate once: you get $v(t)$. Integrate again: you get $x(t)$. Every formula is an integral or a derivative of the others. They form a chain:

$$x(t) \xrightarrow{\;\;d/dt\;\;} v(t) \xrightarrow{\;\;d/dt\;\;} a$$

$$a \xrightarrow{\;\;\int dt\;\;} v(t) \xrightarrow{\;\;\int dt\;\;} x(t)$$

The derivative chain from Chapter 1 --- position to velocity to acceleration --- now runs in reverse through integration. This is the first real payoff of the calculus language you built in Chapter 1: the abstract chain of derivatives becomes a concrete tool for solving problems.

The third equation: eliminating time

The two equations we derived both involve time explicitly. But some problems don't give you time, and don't ask for it. A car brakes from $30$ m/s and comes to rest over $45$ meters --- what was the deceleration? You know initial velocity, final velocity, and displacement. Time is irrelevant.

We can build a third equation by eliminating $t$ algebraically. From the first equation:

$$t = \frac{v - v_0}{a}$$

Substitute this into the second equation:

$$x - x_0 = v_0 \left(\frac{v - v_0}{a}\right) + \frac{1}{2}a\left(\frac{v - v_0}{a}\right)^2$$

This looks messy, but it simplifies cleanly. Multiply through and combine terms (the algebra is worth doing once for yourself):

$$x - x_0 = \frac{v_0(v - v_0)}{a} + \frac{(v - v_0)^2}{2a} = \frac{2v_0(v - v_0) + (v - v_0)^2}{2a} = \frac{v^2 - v_0^2}{2a}$$

Rearranging:

$$\boxed{v^2 = v_0^2 + 2a(x - x_0)}$$

This is the third kinematic equation. It connects velocity and displacement without any reference to time. It is not a new piece of physics --- it is a logical consequence of the first two equations. But it is extremely useful for problems where time is unknown or unneeded.

There is also a useful auxiliary relationship. The average velocity under constant acceleration is

$$\bar{v} = \frac{v_0 + v}{2}$$

and the displacement can be written as $\Delta x = \bar{v} \cdot t$. This is sometimes handy, but it is not independent of the other three equations --- it follows from combining the first two.

The complete set

Here are the three independent kinematic equations for constant acceleration, collected in one place:

Equation What it relates Missing quantity
$v = v_0 + at$ Velocity and time Position
$x = x_0 + v_0 t + \frac{1}{2}at^2$ Position and time Final velocity
$v^2 = v_0^2 + 2a(x - x_0)$ Velocity and position Time

Five quantities appear: $x$, $x_0$, $v$, $v_0$, $a$ (and $t$). Each equation involves four of these six quantities and omits one. This means that if you know any three of the five unknowns, you can solve for the other two --- by choosing the equation that contains the quantities you know and the one you want.

Exploration: varying acceleration, varying initial conditions

Now that you have the equations, let's build intuition for what they predict.

[Interactive: Kinematic explorer. A split-screen display. On the left, three sliders control $x_0$, $v_0$, and $a$. On the right, three synchronized panels show: 1. A number line with an animated dot that moves according to the chosen parameters. 2. A position-vs-time graph ($x$ vs $t$). 3. A velocity-vs-time graph ($v$ vs $t$).

The animation runs from $t = 0$ to $t = 6$ s. Default values: $x_0 = 0$, $v_0 = 0$, $a = 2$ m/s$^2$.]

Work through these guided comparisons. In each case, change only the one parameter indicated and observe what happens.

Vary the acceleration (hold $v_0 = 0$, $x_0 = 0$ fixed):

Set $a = 1$ m/s$^2$, then $a = 2$ m/s$^2$, then $a = 4$ m/s$^2$.

  • What happens to the velocity graph? (It stays a straight line through the origin, but the slope changes.)
  • What happens to the position graph? (It stays a parabola opening upward, but the curvature increases.)
  • What changes in the motion? What stays the same?

Now vary the initial velocity (hold $a = 2$ m/s$^2$, $x_0 = 0$ fixed):

Set $v_0 = 0$, then $v_0 = 5$ m/s, then $v_0 = -5$ m/s.

  • With $v_0 = 5$ m/s: the dot starts moving and speeds up. The position graph is a parabola that begins with a positive slope.
  • With $v_0 = -5$ m/s: the dot initially moves to the left, slows down, stops, and reverses direction. Watch the position graph --- it dips below $x_0$ before curving back upward.
  • When does the reversal happen? (When $v = 0$: solve $v_0 + at = 0$ to get $t = 5/2 = 2.5$ s.)

What changes? What stays the same?

Changing $a$ changes the curvature of the position graph and the slope of the velocity graph. Changing $v_0$ shifts the velocity line up or down and changes the initial slope of the position curve. But the shape --- parabolic position, linear velocity --- remains the same in every case. That shape is the signature of constant acceleration.

A historical note

The idea that falling objects accelerate at a constant rate was not obvious. For nearly two thousand years, scholars followed Aristotle's claim that heavier objects fall faster than lighter ones. It was Galileo, in the early 1600s, who challenged this through careful experiment.

Galileo couldn't directly time a free-falling ball --- objects fall too fast for the crude clocks of his era. Instead, he slowed things down. He rolled polished bronze balls down inclined ramps of various angles and timed them using a water clock: a bucket with a small hole, where the mass of water that drained during the ball's descent measured the elapsed time.

His key finding: the distance traveled is proportional to the square of the time. Double the time, and the ball rolls four times as far. Triple the time, nine times as far. This is exactly the $\frac{1}{2}at^2$ relationship you derived above. Galileo established the fact experimentally; the kinematic equations encode his discovery in the language of calculus that would not be invented until decades later by Newton and Leibniz.

The kinematic equations are not a modern convenience. They are the mathematical distillation of one of the first great experiments in physics.

Practice

Layer 1: Concrete

Problem 1. A ball is dropped from rest from the top of a building 45 m tall. Take $g = 10$ m/s$^2$ downward.

(a) How long does it take to reach the ground?

(b) How fast is it moving when it hits the ground?

Check your answer Set up coordinates with $x$ measured downward from the release point, so $a = g = 10$ m/s$^2$, $v_0 = 0$, and $x_0 = 0$. (a) Use $x = x_0 + v_0 t + \frac{1}{2}at^2$: $$45 = 0 + 0 + \frac{1}{2}(10)t^2 = 5t^2$$ $$t^2 = 9, \quad t = 3 \text{ s}$$ (b) Use $v = v_0 + at$: $$v = 0 + 10 \times 3 = 30 \text{ m/s}$$ You could also use the third equation as a check: $v^2 = 0 + 2(10)(45) = 900$, so $v = 30$ m/s.

Problem 2. A car traveling at $25$ m/s applies its brakes and decelerates at $5$ m/s$^2$.

(a) How long does it take to stop?

(b) How far does it travel while braking?

Check your answer Take the direction of motion as positive, so $v_0 = 25$ m/s and $a = -5$ m/s$^2$ (opposing the motion). (a) At the moment the car stops, $v = 0$: $$0 = 25 + (-5)t \implies t = 5 \text{ s}$$ (b) Use $x = x_0 + v_0 t + \frac{1}{2}at^2$ with $x_0 = 0$: $$x = 25(5) + \frac{1}{2}(-5)(25) = 125 - 62.5 = 62.5 \text{ m}$$ Or use the time-free equation: $0 = 25^2 + 2(-5)(x)$, so $x = 625/10 = 62.5$ m.

Layer 2: Pattern --- choosing the right equation

Problem 3. A ball is thrown straight up with an initial speed of $20$ m/s. How high does it rise? (Take $g = 10$ m/s$^2$.)

Before you compute: Which kinematic equation should you choose? Notice that the problem gives you $v_0$, tells you $v = 0$ at the top, and asks for displacement. Time is not given and not asked for. Which equation avoids time entirely?

Check your answer Use $v^2 = v_0^2 + 2a(x - x_0)$. Taking upward as positive: $v_0 = 20$ m/s, $v = 0$ at the highest point, and $a = -10$ m/s$^2$. $$0 = 20^2 + 2(-10)(x - 0)$$ $$0 = 400 - 20x$$ $$x = 20 \text{ m}$$ Choosing the right equation saved you from finding $t$ as an intermediate step. The general strategy: look at what you know and what you want, identify the variable you *don't* need, and pick the equation that leaves it out.

Layer 3: Structure --- why exactly three equations?

Before you read on: Why are there exactly three independent kinematic equations, and not four or five? Could there be more?

Check your answer There are five kinematic quantities that describe constant-acceleration motion: $x$, $v$, $t$, plus the given constants $x_0$, $v_0$, and $a$. Grouping the unknowns as $\{x, v, t\}$ (the three things that change), each equation connects all three to the known constants but can be written so that one of the three unknowns is absent. More fundamentally, there are only *two* independent relationships --- the results of two integrations: 1. $v = v_0 + at$ (integrating $a = dv/dt$) 2. $x = x_0 + v_0 t + \frac{1}{2}at^2$ (integrating $v = dx/dt$) The third equation ($v^2 = v_0^2 + 2a\Delta x$) is derived by eliminating $t$ from these two. It is *dependent* on them --- algebraically convenient, but not a new piece of information. So the answer is: there are two independent equations (from two integrations), and a third that is a derived shortcut. You cannot produce a fourth independent equation because there are no more integrations to perform. The constant-acceleration assumption, combined with calculus, yields exactly what we have and nothing more.

Layer 4: Debug

A student is solving this problem: "A ball is thrown downward at $5$ m/s from a height of $20$ m. What is its speed when it has risen to $25$ m?"

The student writes:

$$v^2 = v_0^2 + 2a(x - x_0) = 5^2 + 2(-10)(25 - 20) = 25 - 100 = -75$$

The student gets $v^2 = -75$ and says, "I must have made a sign error."

Did the student make a sign error? What does the negative value under the square root mean physically?

Check your answer The student did not make a sign error. The arithmetic is correct --- and the negative value of $v^2$ is the answer, not a mistake. The ball is thrown *downward*. It cannot rise to $25$ m, because it never goes up at all. The problem is asking about a situation that never happens. A ball thrown downward from $20$ m will go to $19$ m, then $18$ m, then hit the ground. It will never reach $25$ m. The equation $v^2 = v_0^2 + 2a\Delta x$ is telling you exactly this: there is no real velocity that corresponds to the ball being at $25$ m with these initial conditions. A negative value of $v^2$ means the object *never reaches that position*. The mathematics is not broken --- it is faithfully reporting that the physical scenario is impossible. This is a powerful diagnostic. If you get $v^2 < 0$, don't hunt for sign errors. Ask yourself: *is it physically possible for the object to reach that position?* The equation is answering that question for you.

Variation: what the assumption buys you

Everything in this section rests on a single assumption: acceleration is constant. Let's be explicit about what that assumption gives you and what would break if it were violated.

Constant acceleration means:

  • $v(t)$ is a straight line. You can read the slope (acceleration) and intercept (initial velocity) directly from a $v$-vs-$t$ graph.
  • $x(t)$ is a parabola. The motion has a clean, predictable shape.
  • Three simple algebraic equations relate the five kinematic variables. Any problem can be solved with algebra alone --- no calculus required after the initial derivation.

If acceleration is not constant --- if it depends on time, or on velocity, or on position --- then:

  • $v(t)$ is no longer a straight line. The velocity graph curves.
  • $x(t)$ is no longer a parabola. The position graph has a more complex shape.
  • The three kinematic equations no longer apply. You must return to the integrals: $v(t) = v_0 + \int_0^t a(t')\,dt'$ and $x(t) = x_0 + \int_0^t v(t')\,dt'$.

The kinematic equations are not universal truths. They are consequences of a specific assumption. Use them only when that assumption is justified, and be prepared to abandon them when it is not.

[Interactive: Constant vs. non-constant comparison. Two side-by-side panels. The left panel shows motion under constant acceleration $a = 2$ m/s$^2$. The right panel shows motion under an acceleration that increases linearly: $a(t) = 0.4t$ m/s$^2$, chosen so that both panels produce the same final velocity at $t = 5$ s. Both panels display $a$-vs-$t$, $v$-vs-$t$, and $x$-vs-$t$ graphs. Students observe: "The velocity graphs look different --- one is a straight line, the other is a curve. The position graphs look different --- one is a parabola, the other is a steeper curve. The kinematic equations describe the left panel perfectly and the right panel not at all."]

Reflection

Think about the central assumption of this section: acceleration is constant.

What real-world situation would violate this assumption most dramatically?

Consider: a car in traffic, a leaf falling from a tree, a ball rolling on a rough surface, a rocket burning fuel. In which of these cases would you trust the kinematic equations, and in which would you hesitate? What is it about each situation that makes the assumption reasonable or unreasonable?

The formulas you derived are powerful --- but only within their domain of validity. The most important thing a physicist can say about a formula is not "here is how to use it" but "here is when it breaks."

Looking Ahead

You have just seen the best-case scenario for kinematics: acceleration is constant, and two straightforward integrations produce exact, closed-form solutions. The formulas are simple, the predictions are precise, and every problem reduces to algebra.

But constant acceleration is a special case --- an unusually generous one. In the next section, we confront what happens when acceleration varies with time. A car whose acceleration fades as it approaches top speed. A rocket that accelerates faster as it burns fuel and gets lighter. In these situations, the kinematic equations of this section are useless. You will need to return to the integral relationships and work with them directly.

The good news: the calculus framework you built here --- integrating $a$ to get $v$, integrating $v$ to get $x$ --- still works perfectly. Constant acceleration was just the case where the integrals happened to be easy. The next section is about what happens when they aren't.