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Kinematic equations

Four equations describe motion under constant acceleration, and between them they link five quantities: the initial velocity $v_0$, the final velocity $v$, the acceleration $a$, the elapsed time $t$, and the displacement $x - x_0$. The kinematic equations are:

$$v = v_0 + at$$

$$x = x_0 + v_0 t + \tfrac{1}{2}a t^2$$

$$v^2 = v_0^2 + 2a(x - x_0)$$

$$x = x_0 + \frac{v_0 + v}{2}\,t$$

Every one of them holds only while $a$ is constant. Not roughly constant, not constant on average - constant across the whole interval you apply it to. OpenStax University Physics develops all four in section 3.4 under exactly that restriction, and almost every wrong answer people get from them comes from ignoring it.

Each one omits a variable

Count the symbols. Each equation contains four of the five quantities and never all five, and which one is missing is different every time:

That is the whole selection rule. List what the problem gives, note what it asks for, and find the one quantity that appears in neither list. The equation that omits that quantity is the one to use, and it will solve the problem in a single line.

A problem that hands you $v_0$, $a$ and $x - x_0$ and asks for $v$ never mentions time, so $v^2 = v_0^2 + 2a(x-x_0)$ is the tool. Go the other way, find $t$ first from the quadratic, discard the negative root and then compute $v$, and you will get the same number after three times the work and two chances to slip.

Where they come from

Constant acceleration means $a = \Delta v / \Delta t$ over any interval you like, so over the whole interval $a = (v - v_0)/t$. Rearrange that and the first equation is already done. Nothing else is needed.

The fourth comes from a second consequence of constant $a$: velocity changes linearly with time, so the average velocity over the interval is the plain mean of the endpoints, $\bar v = (v_0 + v)/2$. This is false for any other acceleration profile, which is worth remembering, because the equation looks harmless enough to reach for anywhere. Displacement is average velocity times time, so $x - x_0 = \frac{v_0+v}{2}t$.

The remaining two are those two combined. Substitute $v = v_0 + at$ into the fourth:

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

Instead eliminate time, writing $t = (v - v_0)/a$ and putting that into the same equation:

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

which rearranges to $v^2 = v_0^2 + 2a(x - x_0)$. So there is one definition and one fact about averages, and the other two equations are bookkeeping on top of them.

Two stages, worked

A car starts from rest, accelerates at $2.5$ m/s$^2$ for $8.0$ s, then brakes at $4.0$ m/s$^2$ until it stops. How far does it go in total?

Stage one asks nothing about displacement yet, so start with the equation that omits it: $v = 0 + 2.5 \times 8.0 = 20$ m/s. Then take the displacement from the equation that omits final velocity: $x = \tfrac{1}{2} \times 2.5 \times 8.0^2 = 80$ m.

Stage two starts where stage one ended. Now $v_0 = 20$ m/s, $v = 0$, and $a = -4.0$ m/s$^2$. Time is not asked for, so use the equation that drops it:

$$0 = 20^2 + 2(-4.0)d \quad\Longrightarrow\quad d = \frac{400}{8.0} = 50\ \text{m}$$

Total distance $80 + 50 = 130$ m.

Check both stages the other way, with the equation that never touches $a$. Stage one: $\bar v = (0 + 20)/2 = 10$ m/s for $8.0$ s gives $80$ m. Stage two lasts $t = 20/4.0 = 5.0$ s, and $\bar v = 10$ m/s for $5.0$ s gives $50$ m. Same two numbers by a route that uses neither acceleration, which is the check worth doing whenever a stage has both endpoint velocities in hand.

Now the failure. The whole trip lasts $8.0 + 5.0 = 13.0$ s, starts at rest and ends at rest. Treat it as one interval and apply the fourth equation:

$$x = \frac{0 + 0}{2} \times 13.0 = 0\ \text{m}$$

The car travelled 130 m and the equation says zero. The algebra is fine; the hypothesis is not. Acceleration was $+2.5$ m/s$^2$ and then $-4.0$ m/s$^2$, so there is no single $a$ and therefore no single interval. Split at every instant where the acceleration changes, solve each piece on its own, and carry the final velocity of one piece in as the initial velocity of the next. Three stages need three applications. There is no fourth equation clever enough to skip the split, and averaging the two accelerations to $-0.75$ m/s$^2$ does not rescue it either - that gives $x = \tfrac{1}{2}(-0.75)(13.0)^2 = -63$ m, which is wrong in sign as well as size.

Signs on an object that turns around

Free fall is the case people trust and then get wrong, because the object reverses direction partway through. Throw a ball straight up at $15$ m/s, take up as positive, and $a = -9.8$ m/s$^2$ for the entire flight: on the way up, at the top, and on the way down. It never changes. Gravity does not know the ball turned around.

Where is the ball at $t = 2.5$ s?

$$y = 15 \times 2.5 - \tfrac{1}{2} \times 9.8 \times 2.5^2 = 37.5 - 30.625 = 6.88\ \text{m}$$

$$v = 15 - 9.8 \times 2.5 = -9.5\ \text{m/s}$$

Positive height, negative velocity. The ball is 6.88 m up and falling at 9.5 m/s. One application of one equation covered the rise and the fall together, because $a$ really was constant through both.

Check it in two pieces. The top is at $t = 15/9.8 = 1.531$ s and $h = 15^2/(2 \times 9.8) = 11.48$ m. From there the ball falls for $2.5 - 1.531 = 0.969$ s, dropping $\tfrac{1}{2} \times 9.8 \times 0.969^2 = 4.60$ m, leaving it at $11.48 - 4.60 = 6.88$ m. Same number, reached without ever writing a negative velocity.

The wrong version flips the sign of $a$ at the top, on the reasoning that the ball is falling now so gravity ought to be positive, while still measuring $y$ upward. That gives $11.48 + 4.60 = 16.08$ m: the ball climbing through the second half of its flight, 9.2 m too high, and rising faster the longer it has been falling. Choose a positive direction once, before the first line of algebra, and let the signs of $v_0$, $v$, $a$ and $x - x_0$ all follow from that choice. A negative answer for $v$ is the equation telling you which way the object is going, not a slip to be tidied up with an absolute value.

Two dimensions add no new equations. Horizontal and vertical motion share only the clock, so the same four apply independently to $x$ and to $y$, with $a_x = 0$ for a projectile and $a_y = -g$. Working the selection rule twice, once per axis, is the whole of projectile motion. The fastest way to get the rule into your hands is to run through problems where the missing variable changes from one to the next, which is what the one-dimensional kinematics set does, and the simulations let you watch a trajectory integrate while you check your answer against it.

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