Equations of motion

A car speeding up on a straight road, a stone dropped from a roof, a cyclist braking to a stop — every one of these is uniformly accelerated straight-line motion, and every one of them is governed by the same four equations. Instead of sketching a fresh velocity-time graph and measuring slopes and areas each time, you can go straight to a formula, plug in the known quantities, and read off the answer. That is the entire point of the equations of motion: they package the graph's geometry into algebra you can use in seconds. This article builds the four equations from the velocity-time graph, shows exactly why each one leaves out a different quantity, and works through the two situations ICSE Class 9 tests hardest — straight-line acceleration and motion under gravity. By the end you should be able to look at any numerical, see instantly which three quantities are given and which two are unknown, and pick the one equation that connects them without a wasted step.

The core insight

For a body moving in a straight line with uniform acceleration, the initial velocity $u$ , final velocity $v$ , acceleration $a$ , time $t$ and distance $S$ are linked by four equations — $v=u+at$ , $S=ut+\tfrac12at^2$ , $S=\tfrac12(u+v)t$ , $v^2=u^2+2aS$ — each omitting exactly one of these five quantities.

The core idea

Picture the velocity-time graph of a body that starts at velocity $u$ and, under constant acceleration $a$ , reaches velocity $v$ after time $t$ . Because the acceleration is uniform, this graph is a single straight line rising from $(0,u)$ to $(t,v)$ — and every one of the four equations is just a different reading of that one line.

The slope gives the first equation. The slope of a velocity-time graph is acceleration, so:

a=\frac{v-u}{t} \quad\Rightarrow\quad v=u+at

where: $u$ is initial velocity (m/s), $v$ is final velocity (m/s), $a$ is uniform acceleration (m/s squared), $t$ is time elapsed (s).

The area under the graph gives distance. That area splits into a rectangle of height $u$ and width $t$ , plus a triangle of base $t$ sitting on top with height $v-u=at$ :

S=ut+\tfrac12at^2

where: $S$ is the distance covered (m), and the other symbols are as above. Read the same area instead as a trapezium with parallel sides $u$ and $v$ and width $t$ , and you get the third equation, $S=\tfrac12(u+v)t$ , which needs no $a$ at all.

Eliminating time gives the fourth equation. Substitute $t=(v-u)/a$ from the first equation into the third:

v^2=u^2+2aS

This form is the one to reach for whenever time is not given — for example, when a ball's velocity at a certain height is asked without mentioning how long it took to get there. Each equation leaves out a different one of the five quantities $u,v,a,t,S$ : the first has no $S$ , the second no $v$ , the third no $a$ , the fourth no $t$ . Choosing the right equation is simply a matter of spotting which quantity is missing from the problem and picking the equation that does not need it. Motion under gravity is nothing new physically — it is just $a=g\approx9.8\ \text{m/s}^2$ , directed downward, so if you take "upward" as positive for a thrown object, $a=-g$ while it rises. It helps to notice that all four equations describe exactly the same physical situation — one straight-line velocity-time graph — so they are not four independent facts to memorise separately but four different views of one picture. This is also why any two of them, applied to the same data, must agree: if you compute a distance with the second equation and check it with the third, you should get the same number, and that agreement is a genuine confirmation that your arithmetic (and your choice of signs) was correct. Examiners often reward exactly this kind of cross-check, especially in gravity problems where a sign slip is easy to make and easy to miss without a second calculation to compare against.

Figure

A velocity time graph showing a straight line rising from u at time zero to v at time t, with the area beneath it split into a rectangle and a triangle representing the distance covered

Derivation AWhy v squared equals u squared plus 2aS

The goal is to combine the slope relation and the trapezium-area relation so that time drops out entirely, leaving a formula that connects $u$ , $v$ , $a$ and $S$ alone — the version needed whenever a problem never states how long the motion lasted, such as finding the speed of a falling object at a given height or the stopping distance of a car given only its deceleration.

Step 1. Start from the first equation, $v=u+at$ , and make $t$ the subject:

t=\frac{v-u}{a}

This expression for time in terms of the velocities and the acceleration is the piece we will substitute away.

Step 2. Start from the third equation, $S=\tfrac12(u+v)t$ , which came from reading the trapezium area of the same velocity-time graph, and substitute the expression for $t$ from Step 1:

S=\tfrac12(u+v)\cdot\frac{v-u}{a}

Step 3. Multiply out $(u+v)(v-u)=v^2-u^2$ using the difference-of-squares identity, so:

S=\frac{v^2-u^2}{2a}

Step 4. Rearrange for $v^2$ by multiplying both sides by $2a$ and adding $u^2$ :

v^2=u^2+2aS

This is the fourth equation of motion, and it is entirely equivalent to the other three — it simply repackages the same graph with time eliminated.

Worked example

Question: A stone is dropped from rest from the top of a tower and hits the ground after 4 s. Taking $g=9.8\ \text{m/s}^2$ , find (a) the velocity with which it hits the ground and (b) the height of the tower. Then find the same height a second way, using the velocity found in (a), to confirm the answer.

Step 1. List what is given. The stone starts from rest, so $u=0\ \text{m/s}$ ; it falls under gravity, so $a=g=9.8\ \text{m/s}^2$ (downward, taken as positive since the motion is entirely downward); and $t=4\ \text{s}$ .

Step 2. Find the final velocity using $v=u+at$ :

v=0+(9.8)(4)=39.2\ \text{m/s}

Step 3. Find the height using $S=ut+\tfrac12at^2$ :

S=(0)(4)+\tfrac12(9.8)(4)^2=0+\tfrac12(9.8)(16)=78.4\ \text{m}

Step 4. Cross-check using $v^2=u^2+2aS$ with the velocity from Step 2: $39.2^2=0^2+2(9.8)S \Rightarrow 1536.64=19.6\,S \Rightarrow S=78.4\ \text{m}$ , which matches Step 3 exactly.

The stone hits the ground at 39.2 m/s, and the tower is 78.4 m tall.

Competitive stretch

If instead you were told only that the stone hit the ground at 39.2 m/s (no time given), you could still find the height directly from $v^2=u^2+2aS$ without ever computing $t$ — a reminder that the fourth equation is the fast route whenever time is missing from the data.

Common traps

Picking an equation before checking which quantity is missing — always list $u,v,a,t,S$ , mark what is given, and choose the equation that skips the one you don't have and don't need.
Sign errors under gravity — fix once whether up or down is positive, and keep every $u$ , $v$ and $a$ consistent with that choice for the whole problem; a body falling from rest should not suddenly get a negative $S$ .
Using $v$ instead of $u$ in $S=ut+\tfrac12at^2$ — this equation needs the initial velocity, not the final one; mixing them up gives a wrong distance even with correct arithmetic elsewhere.
Forgetting that at the highest point of a vertical throw, velocity is momentarily zero but acceleration is still $g$ , acting throughout — gravity never switches off, even for an instant.

Ask the tutor: Stuck on which equation fits a tricky word problem, or want extra practice with sign conventions for objects thrown upward — ask the AI tutor for more worked numericals on this topic.

How do I decide which of the four equations to use in a numerical

List the five quantities $u$ , $v$ , $a$ , $t$ , $S$ and mark which three the question gives you and which one it asks for. That leaves exactly one quantity unused — the one your chosen equation should also leave out. If $t$ is missing, use $v^2=u^2+2aS$ ; if $v$ is missing, use $S=ut+\tfrac12at^2$ ; if $a$ is missing, use $S=\tfrac12(u+v)t$ ; if $S$ is missing, use $v=u+at$ . This mechanical check removes almost all the guesswork.

Why is acceleration due to gravity sometimes positive and sometimes negative in the same chapter

Because the sign of $g$ depends only on the direction you chose as positive for that particular problem, not on gravity itself. If you take downward as positive (typical for a falling or dropped body), $a=+g$ . If you take upward as positive (typical when a body is thrown up), gravity opposes that direction, so $a=-g$ . Pick one convention at the start of a problem and hold it throughout — switching partway through is the single most common source of sign errors.

Do these four equations work if the acceleration keeps changing during the motion

No. All four equations of motion assume the acceleration is constant (uniform) throughout the interval considered. If a body speeds up, then slows down, or its acceleration otherwise varies, you must split the motion into separate stages, each with its own constant acceleration, and apply the equations stage by stage rather than across the whole journey in one shot.

Equations of motion

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads