When acceleration is constant, the messy calculus collapses into three tidy formulas — the kinematic equations. Let $u$ be the initial velocity, $v$ the velocity after time $t$, $a$ the constant acceleration and $s$ the displacement. The three equations are: $v = u + at$, $\;s = ut + \dfrac{1}{2}at^2$, and $v^2 = u^2 + 2as$. They are valid only while $a$ is constant.
Each follows from calculus. Since $a = \dfrac{dv}{dt}$ is constant, integrating gives $v = u + at$ — the first equation. Because $v = \dfrac{dx}{dt} = u + at$, integrating again gives $s = ut + \dfrac{1}{2}at^2$ — the second equation. Eliminating $t$ between the first two yields $v^2 = u^2 + 2as$ — the third equation, the one to reach for whenever time is neither given nor wanted. A useful extra result is the distance in the $n$-th second: $s_n = u + \dfrac{a}{2}(2n - 1)$.
Graphs make motion visible and turn slopes and areas into physics. On a position-time ($x$-$t$) graph, the slope gives velocity: a straight line means uniform velocity, a curve means changing velocity, and a horizontal line means rest. On a velocity-time ($v$-$t$) graph two facts matter at once — the slope gives acceleration, and the area under the graph gives displacement. For uniform acceleration the $v$-$t$ line is straight, and the area beneath it (a trapezium) is exactly $s = ut + \dfrac{1}{2}at^2$. This area method is one of the most powerful shortcuts in kinematics.
Free fall is uniformly accelerated motion under gravity alone. Near the Earth's surface every freely falling body has the same downward acceleration $g \approx 9.8\,\text{m/s}^2$ (often taken as $10\,\text{m/s}^2$), independent of mass. Taking the upward direction as positive, we use $a = -g$. For a body dropped from rest, $u = 0$, so $v = -gt$ and $|s| = \dfrac{1}{2}gt^2$. For a body thrown up, it rises until $v = 0$, reaching height $H = \dfrac{u^2}{2g}$, and the time up equals the time down.
Finally, motion is always relative to a frame, so relative velocity in 1D is just a subtraction of signed velocities: the velocity of $A$ with respect to $B$ is $v_{AB} = v_A - v_B$. If two trains move the same way, the faster overtakes the slower at $v_A - v_B$; if they move oppositely, they approach at $v_A - (-v_B) = v_A + v_B$. Getting the signs right is the whole game.
- First: $v = u + at$. Second: $s = ut + \dfrac{1}{2}at^2$. Third: $v^2 = u^2 + 2as$.
- $x$-$t$ slope $= v$; $v$-$t$ slope $= a$ and area under $v$-$t$ $= s$.
- Free fall: $a = g \approx 9.8\,\text{m/s}^2$ downward, same for all masses.
- Max height of an upthrow: $H = \dfrac{u^2}{2g}$; time up $=$ time down.
- Relative velocity (1D): $v_{AB} = v_A - v_B$ using signed values.