Kinematic equations

A car brakes, a lift rises, a stone falls — as long as the acceleration stays constant, the entire motion is pinned down by just three algebraic equations. Instead of tracking position instant by instant, the kinematic equations let you plug in whatever you know — initial velocity, final velocity, acceleration, time, displacement — and solve directly for whatever you don't. This is the working core of Unit 2, Kinematics, under Motion in a Straight Line: three equations, derivable both from a $v$ – $t$ graph and from calculus, applied constantly to free fall and vertical throws. Get comfortable picking the right equation for a given unknown, and most one-dimensional motion numericals reduce to careful substitution and arithmetic. Below: where the three equations come from, a full derivation of the displacement equation, a worked vertical-throw problem, and the traps that cost the most marks.

The core insight

For constant acceleration $a$ , the three kinematic equations $v = u+at$ , $s = ut+\tfrac12at^2$ , $v^2 = u^2+2as$ connect initial velocity $u$ , final velocity $v$ , time $t$ , and displacement $s$ , so any two known quantities fix the rest — including motion under gravity, where $a = \pm g$ .

The core idea

For a body moving in a straight line with constant (uniform) acceleration $a$ , starting at velocity $u$ , reaching velocity $v$ after time $t$ , and covering displacement $s$ , three equations hold simultaneously.

v = u + at \tag{1}

where: $u$ is initial velocity, $v$ is final velocity, $a$ is the (constant) acceleration, $t$ is elapsed time.

s = ut + \tfrac{1}{2}at^2 \tag{2}

where: $s$ is displacement over time $t$ ; this follows from (1) once you know displacement is the area under a $v$ – $t$ graph.

v^2 = u^2 + 2as \tag{3}

where: this is equations (1) and (2) combined with $t$ eliminated — useful whenever time is not given or not needed. A fourth, less-quoted relation drops out of the same two equations: $s = \left(\dfrac{u+v}{2}\right)t$ , the displacement as average velocity times time. It holds only because velocity grows linearly with time under constant $a$ , so the average of $u$ and $v$ really is the mean velocity over the interval. These equations govern motion under gravity on a vertical line by simply setting $a = g = 9.8\ \text{m/s}^2$ , with a sign attached once you fix a positive direction — commonly upward positive, making $a = -g$ for any body in free flight, since gravity decelerates upward motion and accelerates downward motion equally. Get the sign convention wrong and every answer that follows inherits the error. Notice, too, that all three equations reduce to the same underlying idea stated three different ways: velocity changes steadily with time (equation 1), displacement accumulates as the area swept under that changing velocity (equation 2), and eliminating time altogether links displacement directly to the change in velocity squared (equation 3). None of them introduces new physics beyond "acceleration is constant" — they are three algebraic faces of one assumption, chosen so that whichever variable a problem leaves out, one of the three equations avoids needing it.

Figure

A velocity time graph showing a straight sloped line from initial velocity u at time zero to final velocity v at time t, with the acceleration as the slope and the shaded trapezium area under the line representing displacement s

Derivation ADeriving displacement from the velocity-time graph

Show that the area under a straight-line velocity-time graph for uniformly accelerated motion, a trapezium of parallel sides $u$ and $v$ and width $t$ , reduces algebraically to the standard displacement equation $s = ut + \tfrac12at^2$ , once the final velocity is written in terms of the acceleration and elapsed time.

Step 1. Plot velocity against time for a body starting at $u$ and accelerating uniformly to $v$ over a time $t$ . Because acceleration is constant, this graph is a straight line, and displacement equals the area swept out under it (velocity times time, summed continuously).

Step 2. The region under the line is a trapezium with parallel sides of length $u$ (at $t=0$ ) and $v$ (at time $t$ ), and width $t$ along the time axis. The area of a trapezium is half the sum of the parallel sides times the width, so $s = \tfrac12(u+v)t$ .

Step 3. Bring in equation (1), $v = u + at$ , which comes directly from the slope of this same line: slope $= (v-u)/t = a$ , rearranged. Substitute this expression for $v$ into the trapezium area from Step 2: $s = \tfrac12\big(u + (u+at)\big)t = \tfrac12(2u+at)t$ .

Step 4. Expand: $s = \tfrac12(2u)t + \tfrac12(at)t = ut + \tfrac12at^2$ , which is exactly equation (2). Eliminating $t$ instead between (1) and (2) — substitute $t=(v-u)/a$ into $s=\tfrac12(u+v)t$ — yields $s=\tfrac12(u+v)\cdot\tfrac{v-u}{a}=\tfrac{v^2-u^2}{2a}$ , rearranging to equation (3), $v^2=u^2+2as$ . All three equations therefore come from a single graph.

Worked example

A ball is thrown vertically upward from the ground with initial speed $24.5\ \text{m/s}$ . Taking $g = 9.8\ \text{m/s}^2$ and upward as positive, so $a = -9.8\ \text{m/s}^2$ , find (a) the maximum height reached and (b) the total time to return to the ground.

Step 1. At maximum height the velocity is momentarily zero, so $v=0$ , $u=24.5\ \text{m/s}$ , $a=-9.8\ \text{m/s}^2$ , and $s$ is the unknown maximum height. Equation (3) needs no time value, so use it directly.

Step 2. $0 = (24.5)^2 + 2(-9.8)s$ , so $s = \dfrac{(24.5)^2}{2(9.8)} = \dfrac{600.25}{19.6} = 30.625\ \text{m}$ . The ball rises to a maximum height of about 30.6 m.

Step 3. For the time to reach that height, use equation (1): $0 = 24.5 + (-9.8)t \Rightarrow t = 2.5\ \text{s}$ . By the symmetry of motion under gravity, with no air resistance, the fall back down takes exactly as long as the rise, so the total time of flight is $2 \times 2.5 = \textbf{5 s}$ .

Competitive stretch

Find the ball's velocity when it has fallen back to a height of $15\ \text{m}$ above the ground, on the way down. Take $s = 15\ \text{m}$ measured from the ground, so the displacement from the throw point is still $+15\ \text{m}$ in the upward-positive convention. Equation (3) gives $v^2 = (24.5)^2 - 2(9.8)(15) = 600.25 - 294 = 306.25$ , so $v = \pm 17.5\ \text{m/s}$ ; the negative root, $v=-17.5\ \text{m/s}$ , is the physical answer on the way down, meaning the ball moves at $17.5\ \text{m/s}$ downward.

Common traps

Applying these equations when acceleration is not constant — a changing $a(t)$ needs calculus, not the standard three equations.
Sign slips in vertical-motion problems — forgetting to make $g$ negative when upward is chosen positive (or vice versa), which flips the size or sign of the answer.
Confusing displacement $s$ , which can be negative or zero even after real motion, with total distance travelled — especially when a body reverses direction, like a ball thrown up and coming back down.
Reaching for two equations and solving simultaneously when a single equation, chosen to avoid the missing unknown, would do the job directly.
Forgetting to check units before substituting — mixing $\text{km/h}$ with $\text{m/s}^2$ and seconds in the same equation is a common source of answers that are wrong by a clean factor like 3.6.

Ask the tutor: if picking the right equation still feels like a guess, ask the AI tutor to work through a few mixed numericals using the "which variable is missing" method.

Why do the kinematic equations only work for constant acceleration

Every one of the three equations is derived from a straight-line velocity-time graph, or equivalently from integrating a constant $a$ . If acceleration itself is changing, the graph is curved, the trapezium-area argument breaks down, and $a$ can no longer be pulled out as a single number in the algebra. For variable acceleration you would instead need calculus, integrating $a(t)$ directly, which is beyond this equation set.

How do I choose the right equation quickly in an exam

List the five quantities $u, v, a, t, s$ and mark which one is missing from the problem. Equation (1) skips $s$ , equation (2) skips $v$ , and equation (3) skips $t$ . Pick whichever equation excludes the unknown you were never given, and you avoid solving two equations together. This single habit saves the most time in numericals.

Does displacement being negative mean something went wrong

No — a negative $s$ simply means the body ended up on the opposite side of your chosen positive direction from where it started. In the worked example, taking upward as positive, a ball that ends up below the throw point would show negative $s$ . What matters is staying consistent with one sign convention throughout a single problem, not avoiding negative numbers.

Kinematic equations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads