Derive v = u + at and s = ut + ½at² from first principles

Question

Derive the first and second equations of motion: $v = u + at$ and $s = ut + \frac{1}{2}at^2$ from first principles, assuming constant acceleration.

(NCERT Class 11, Chapter 3 — Motion in a Straight Line)

Solution — Step by Step

Start with the definition of acceleration

Acceleration is the rate of change of velocity. For constant acceleration $a$ :

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

This is our starting point. Everything else follows from this single definition.

Derive the first equation: v = u + at

Rearrange and integrate:

$dv = a \, dt$

Integrate both sides. At $t = 0$ , velocity is $u$ (initial velocity). At time $t$ , velocity is $v$ .

$\int_u^v dv = \int_0^t a \, dt$

Since $a$ is constant, it comes out of the integral:

$v - u = at$

$\boxed{v = u + at}$

This tells us: final velocity = initial velocity + (acceleration $\times$ time).

Derive the second equation: s = ut + ½at²

Velocity is the rate of change of displacement:

$v = \frac{ds}{dt}$

We already know $v = u + at$ . Substitute:

$\frac{ds}{dt} = u + at$

$ds = (u + at) \, dt$

Integrate both sides. At $t = 0$ , displacement $s = 0$ . At time $t$ , displacement is $s$ .

$\int_0^s ds = \int_0^t (u + at) \, dt$

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

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

Why This Works

Both equations come from the chain: acceleration → velocity → displacement. We integrate once to go from acceleration to velocity (first equation), and integrate again to go from velocity to displacement (second equation). The $\frac{1}{2}$ in the second equation appears naturally from integrating $at$ with respect to $t$ .

The key assumption throughout is constant acceleration. If acceleration varies with time, these simple equations don't hold — you'd need to integrate the specific $a(t)$ function.

Alternative Method — Using the v-t Graph

Draw a velocity-time graph for constant acceleration. It's a straight line from $(0, u)$ to $(t, v)$ .

First equation: The slope of this line is acceleration: $a = (v - u)/t$ , giving $v = u + at$ .

Second equation: Displacement is the area under the v-t graph. This area is a trapezium:

$s = \text{area of rectangle} + \text{area of triangle} = ut + \frac{1}{2}(v - u)t$

Substituting $v - u = at$ :

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

💡 Expert Tip

The graphical method is faster and more intuitive for board exams. For NEET, the calculus method is cleaner and earns full marks. Know both — the examiner can specify which derivation they want.

Common Mistake

⚠️ Common Mistake

Students often forget that these equations are valid only for constant acceleration. If a question says "a particle accelerates uniformly" or "constant acceleration," you can use them. But if acceleration is given as a function of time (like $a = 2t$ ), you must integrate from scratch — plugging into $s = ut + \frac{1}{2}at^2$ with a variable $a$ will give a wrong answer.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the v-t Graph

Common Mistake

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