Equations of motion and numerical applications

A car brakes from $20\ \text{m/s}$ to a stop in $4\ \text{s}$ — how far does it skid before it halts? You could sketch a velocity-time graph and measure the area under it every single time, but physics hands you a shortcut: three ready-made equations of motion that connect initial velocity $u$ , final velocity $v$ , acceleration $a$ , time $t$ and distance $s$ directly, with no graph paper required. They are the practical heart of the CBSE Class 9 chapter Motion, and once you can read a numerical, list its five quantities, and pick the one equation that fits, most problems fall in two or three lines. This article expands the three equations, shows exactly where they come from on a velocity-time graph, works through braking and vertical-motion numericals in full, and flags the sign-convention slip that costs the most marks.

The core insight

For any object moving with uniform acceleration, the three equations of motion — $v = u+at$ , $s = ut+\tfrac12at^2$ , $v^2 = u^2+2as$ — each link exactly four of the five motion quantities, so identifying which three are given and one is asked tells you which equation to use.

The core idea

Uniform acceleration means the velocity changes by an equal amount every second — the velocity-time graph is a straight line. For an object starting at velocity $u$ , accelerating uniformly at $a$ , reaching velocity $v$ after time $t$ , and covering distance $s$ in that time, the three equations of motion hold together the whole story of straight-line motion.

Equation 1 — velocity after time.

v = u + at

where: $u$ is initial velocity, $v$ is final velocity, $a$ is (uniform) acceleration, $t$ is elapsed time. This is just the definition of acceleration, $a = \dfrac{v-u}{t}$ , rearranged — it is also the slope of the velocity-time graph.

Equation 2 — distance covered.

s = ut + \frac{1}{2}at^2

where: $s$ is distance covered in time $t$ . This is the area under the velocity-time graph: a rectangle of area $ut$ (the distance the object would cover at constant speed $u$ ) plus a triangle of area $\tfrac12 t \cdot at = \tfrac12 at^2$ (the extra distance from the speeding-up part).

Equation 3 — velocity without time.

v^2 = u^2 + 2as

This is obtained by eliminating $t$ between equations 1 and 2 algebraically (worked below) — useful whenever a problem never mentions time at all. A practical strategy for any equations-of-motion numerical: first list out the five quantities $u$ , $v$ , $a$ , $t$ , $s$ , writing down the numerical value (with correct sign) for each one the question gives, plus a question mark for the one being asked. Each of the three equations connects exactly four of these five quantities, so once three values are known, the fourth follows from the equation that leaves out the fifth, unrelated quantity. This habit avoids the common error of picking an equation at random and getting stuck partway through a problem. The figure below shows how equations 1 and 2 both read off the same graph.

Figure

A velocity time graph showing a straight sloping line from initial velocity u to final velocity v over time t, with the area under the line split into a rectangle of height u and a triangle on top, illustrating the slope gives acceleration and the total area gives distance

Derivation: eliminating time from the first two equations

Derivation AGetting v squared equals u squared plus 2as

The goal is to combine equation 1, $v = u+at$ , and equation 2, $s = ut+\tfrac12at^2$ , so that time $t$ cancels out completely and only $u$ , $v$ , $a$ and $s$ remain — giving an equation you can use whenever a numerical never states $t$ , such as a braking distance or a free-fall drop given only the height or the impact speed.

Step 1. Start from equation 1 and isolate $t$ by rearranging: $v = u+at \implies at = v-u \implies t = \dfrac{v-u}{a}$ . This expresses time purely in terms of the two velocities and the acceleration.

Step 2. Substitute this expression for $t$ into equation 2, $s = ut + \tfrac12at^2$ , replacing every $t$ : $s = u\left(\dfrac{v-u}{a}\right) + \dfrac12 a\left(\dfrac{v-u}{a}\right)^2$ .

Step 3. Simplify the second term first. Since $\left(\dfrac{v-u}{a}\right)^2 = \dfrac{(v-u)^2}{a^2}$ , the term becomes $\dfrac12 a \cdot \dfrac{(v-u)^2}{a^2} = \dfrac{(v-u)^2}{2a}$ . So $s = \dfrac{u(v-u)}{a} + \dfrac{(v-u)^2}{2a}$ .

Step 4. Put both terms over the common denominator $2a$ : $s = \dfrac{2u(v-u) + (v-u)^2}{2a}$ . Factor out $(v-u)$ from the numerator: $s = \dfrac{(v-u)\left[2u + (v-u)\right]}{2a} = \dfrac{(v-u)(v+u)}{2a}$ , using $2u + v - u = u+v$ .

Step 5. Recognise $(v-u)(v+u) = v^2 - u^2$ (difference of squares), so $s = \dfrac{v^2-u^2}{2a}$ . Multiplying both sides by $2a$ gives the third equation: $2as = v^2 - u^2$ , or rearranged, $v^2 = u^2 + 2as$ — exactly the result we set out to reach, with no explicit $t$ anywhere.

Worked example

Question: A car moving at $20\ \text{m/s}$ applies brakes and decelerates uniformly, coming to rest after $4\ \text{s}$ . Find (a) the retardation, and (b) the distance covered before stopping.

Step 1. List the knowns with a sign convention: take the direction of motion as positive, so $u = 20\ \text{m/s}$ , $v = 0\ \text{m/s}$ (comes to rest), $t = 4\ \text{s}$ , and $a$ , $s$ are unknown.

Step 2. Find $a$ using $v = u+at$ , since it contains $u$ , $v$ , $a$ , $t$ — exactly the four quantities we have or want: $0 = 20 + a(4) \implies a = \dfrac{-20}{4} = -5\ \text{m/s}^2$ . The negative sign confirms it is a retardation of $5\ \text{m/s}^2$ .

Step 3. Find $s$ using $v^2 = u^2+2as$ , since it avoids needing $t$ again as an independent check: $0^2 = 20^2 + 2(-5)s \implies 0 = 400 - 10s \implies s = 40\ \text{m}$ .

Step 4. Cross-check with $s = ut+\tfrac12at^2$ : $s = (20)(4) + \tfrac12(-5)(4)^2 = 80 - 40 = 40\ \text{m}$ — matches.

The car's retardation is $5\ \text{m/s}^2$ , and it skids $40\ \text{m}$ before stopping.

Competitive stretch

If the same car had been moving at double the speed, $40\ \text{m/s}$ , with the same retardation of $5\ \text{m/s}^2$ , the braking distance would be $s = \dfrac{v^2-u^2}{2a} = \dfrac{0 - 1600}{-10} = 160\ \text{m}$ — four times farther, not twice, because $s \propto u^2$ in $v^2 = u^2+2as$ . This $u^2$ dependence is exactly why doubling a vehicle's speed roughly quadruples its stopping distance on the road.

Common traps

Plugging numbers into the wrong equation — first note which three of $u, v, a, t, s$ are given and which one is asked, then use the equation containing exactly those four.
Forgetting to convert units (km/h to m/s, minutes to seconds) before substituting — the equations are only valid in consistent SI units.
Losing track of sign convention in retardation or free-fall problems — fix a positive direction once and apply it to $u$ , $v$ and $a$ throughout.
Treating these equations as valid for changing (non-uniform) acceleration — they hold only when acceleration is constant.

Ask the tutor: Stuck on which equation of motion fits a particular numerical, or want your working checked step by step — ask the AI tutor to walk through it with you.

Why do we call it retardation instead of just negative acceleration?

They mean the same physics — acceleration acting opposite to the direction of motion, so speed decreases. Retardation is simply the everyday word for a negative acceleration once you have fixed the direction of motion as positive. In equations you still just plug in $a$ as a negative number; there is no separate formula for retardation, only a sign to track carefully.

Can I use these three equations for a ball thrown at an angle, like a projectile?

Not directly along the slanted path, but yes if you split the motion into a horizontal part (constant velocity, no acceleration) and a vertical part (uniform acceleration $g$ downward), then apply the equations separately to each part. Projectile motion at an angle is beyond this chapter's scope, but the same three equations remain the tool once motion is broken into straight-line pieces.

Which equation should I start with if a question does not obviously give four quantities?

Reread the question and write down every quantity it states or implies — "starts from rest" means $u=0$ , "comes to rest" means $v=0$ , and "dropped" usually means $u=0$ with $a=g$ downward. Once you have listed all known values against $u, v, a, t, s$ , exactly one equation will contain your three knowns plus the one unknown you need; use that one first and, if time permits, verify with a second equation.