Equations of Uniformly Accelerated Motion by Calculus Method

Consider one-dimensional motion with constant acceleration $a$. Let $u$ be the velocity at time $t=0$ and $v$ the velocity at a later time $t$.

1. Velocity-time relation. Acceleration is the first derivative of velocity: $a=dv/dt$, i.e. $dv=a\,dt$. Integrating from $t=0$ (where $v=u$) to time $t$ (where the velocity is $v$), and treating $a$ as constant so it comes outside the integral,

   $\int_u^v dv = a\int_0^t dt \quad\Longrightarrow\quad v-u=at$

   $\boxed{v = u+at} \qquad(2.7)$

   (If $a$ depended on time, it could not be pulled outside the integral this way.)
2. Displacement-time relation. Velocity is the first derivative of displacement: $v=ds/dt$, i.e. $ds=v\,dt=(u+at)\,dt$. Assuming the particle starts at the origin ($s=0$) at $t=0$ and reaches displacement $s$ at time $t$,

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

   $\boxed{s = ut+\tfrac12 at^2} \qquad(2.8)$

3. Velocity-displacement relation. Writing $a=\dfrac{dv}{dt}=\dfrac{dv}{ds}\dfrac{ds}{dt}=v\dfrac{dv}{ds}$ (since $ds/dt=v$), we get $a\,ds = v\,dv = \tfrac12 d(v^2)$, i.e. $ds=\dfrac{1}{2a}d(v^2)$. Integrating as $v^2$ runs from $u^2$ to $v^2$ while $s$ runs from $0$ to $s$,

   $s=\frac{1}{2a}(v^2-u^2) \quad\Longrightarrow\quad$

   $\boxed{v^2 = u^2+2as} \qquad(2.9)$

   (iv) Displacement in terms of $u$, $v$ and $t$. From (2.7), $at=v-u$; substituting into (2.8), $s=ut+\tfrac12(v-u)t$, which simplifies to

   $\boxed{s = \left(\frac{u+v}{2}\right)t} \qquad(2.10)$

   Equations (2.7)-(2.10) are the four kinematic equations of motion, valid only for straight-line motion with constant acceleration (they do not apply to circular or oscillatory motion, which need their own treatment). Case (1): a body falling from height $h$ (free fall). Choose the downward direction as the positive $y$-axis. The acceleration due to gravity, $g$, is constant near Earth's surface and points in this chosen positive direction, so $a_y=g$ (and $a_x=a_z=0$). If the object is simply released from rest ($u=0$) at $y=0$ (the drop point), the equations above specialise to

   $v=gt \quad(2.14), \qquad y=\tfrac12gt^2 \quad(2.15), \qquad v^2=2gy \quad(2.16)$

   The time $T$ to fall a height $h$ (i.e. $y=h$ when $t=T$) follows from (2.15):

   $h=\tfrac12gT^2 \quad\Longrightarrow\quad T=\sqrt{\dfrac{2h}{g}} \quad(2.17)-(2.18)$

   — a taller drop takes longer to complete. The speed on reaching the ground ($y=h$) follows from (2.16):

   $v_{ground}=\sqrt{2gh} \quad(2.19)$

   — a body dropped from a greater height reaches the ground faster. (Crucially, none of these equations contain the mass of the falling object — in vacuum, a feather and an iron ball dropped from the same height take exactly the same time to fall and land with exactly the same speed; Galileo's famous conclusion.) If instead the object is thrown downward with some initial speed $u$ (still along the same positive-downward axis), the more general forms apply: $v=u+gt$ (2.11), $y=ut+\tfrac12gt^2$ (2.12), $v^2=u^2+2gy$ (2.13). Case (2): a body thrown vertically upward. Now choose the upward direction as positive $y$; gravity then acts in the negative $y$-direction, so $a=-g$. For an object launched from the ground with initial speed $u$ (neglecting air resistance): …