Kinematic Equations (Uniform Acceleration)

When acceleration is constant, three classical equations describe everything you need: $v = u + at$, $s = ut + \tfrac{1}{2}at^2$, and $v^2 = u^2 + 2as$.

Concept

For motion in 1D with constant acceleration $a$, initial velocity $u$, final velocity $v$ after time $t$, and displacement $s$:

$v = u + at \quad (1)$ $s = ut + \tfrac{1}{2}at^2 \quad (2)$ $v^2 = u^2 + 2as \quad (3)$

A fourth, useful form: average velocity over a uniformly accelerated motion is $(u+v)/2$, so $s = \frac{(u+v)}{2} t.$

Derivation

Calculus Method

Since $a = dv/dt$ is constant: $\int_u^v dv = \int_0^t a\,dt \Rightarrow v - u = at \Rightarrow v = u + at.$

For position, $v = dx/dt$: $\int_0^s dx = \int_0^t (u + at)\,dt = ut + \tfrac{1}{2}at^2.$

For equation (3), use the chain rule form $a = v\,dv/dx$: $\int_u^v v\,dv = \int_0^s a\,dx \Rightarrow \tfrac{1}{2}(v^2 - u^2) = as.$

Graphical Method

On a $v$-$t$ graph with constant acceleration, $v(t)$ is a straight line of slope $a$:

• Eq (1): final $v$ = initial $u$ + slope × time.
• Eq (2): displacement = area under $v$-$t$ = trapezoid = $ut + \tfrac{1}{2} a t^2$.
• Eq (3): eliminate $t$ by combining (1) and (2).

When to Use Which

Worked Example

Q: A train starts from rest and accelerates at $0.5\,\text{m/s}^2$. Find (a) its velocity after 60 s, (b) distance covered, (c) velocity after travelling 1.2 km.

Solution:

(a) $v = u + at = 0 + 0.5(60) = 30$ m/s.

(b) $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(0.5)(60)^2 = 900$ m.

(c) Use (3): $v^2 = 0 + 2(0.5)(1200) = 1200 \Rightarrow v = \sqrt{1200} \approx 34.6$ m/s.

Common Confusions

• These equations only hold for constant acceleration. Don't use them for variable $a$ (use calculus).
• Signs of $a$, $u$, $s$ must be consistent (one positive direction).
• "Distance from rest" implies $u = 0$.

Key Takeaways

• Three equations: $v = u + at$, $s = ut + \tfrac{1}{2}at^2$, $v^2 = u^2 + 2as$.
• Valid only for uniform acceleration.
• Choose equations based on which variable is missing.
• Sign convention must be fixed at the start of the problem.