Constant Acceleration — Kinematics, Integration, and Vector Notation

A free preview from The Ultimate Crash Course for STEM Majors by Jonathan David.

Note: The following formulas also apply in the $y$-direction when acceleration is constant. For vertical motion under gravity alone, the sign of the acceleration depends on your choice of coordinates: $a_y=-g$ when upward is positive, or $a_y=g$ when downward is positive.

Constant-Acceleration Formulas

Each equation below assumes constant acceleration in the $x$-direction, with elapsed time $t$ measured from the initial conditions.

Velocity

$$ v_x = v_{x,0} + a_x t $$

Average Velocity

$$ v_{\mathrm{av},x} = v_{x,0} + \frac{1}{2}a_x t = \frac{v_x + v_{x,0}}{2} $$

Displacement

$$ \Delta x = x-x_0 = v_{x,0}t + \frac{1}{2}a_x t^2 $$

Velocity Without Time

$$ v_x^2 = v_{x,0}^2 + 2a_x\Delta x $$

Using initial and final velocity notation:

$$ v_f^2 = v_i^2 + 2a\Delta x $$

Displacement Using Average Velocity

$$ \Delta x = x-x_0 = \left(\frac{v_x+v_{x,0}}{2}\right)t $$

Equivalently:

$$ x_f-x_i = \left(\frac{v_f+v_i}{2}\right)t $$

Note: $v_{x,0}$ denotes the initial velocity in the $x$-direction. Displacement is the signed change in position; it is not always equal to the total distance traveled.

Integration Derivations — Calculus

Note: A derivative describes a rate of change. A derivation is the process of obtaining a result. For constant acceleration $a$, integration gives the velocity and position formulas below.

$$ \frac{dv}{dt}=a \quad\Rightarrow\quad \int_{v_0}^{v(t)}dv=\int_0^t a\,d\tau \quad\Rightarrow\quad v(t)=at+v_0 $$

$$ \frac{ds}{dt}=v(t)=at+v_0 $$ $$ ds=(at+v_0)\,dt $$ $$ s-s_0 =\int_0^t(a\tau+v_0)\,d\tau =\frac{at^2}{2}+v_0t $$

Vector Notation

$$ \mathbf{r} =x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k} \equiv \vec{r} =x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k} $$ $$ \begin{aligned} x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k} &= x(t)\langle1,0,0\rangle +y(t)\langle0,1,0\rangle +z(t)\langle0,0,1\rangle\\ &=\langle x(t),y(t),z(t)\rangle \end{aligned} $$ $$ \therefore\quad \vec{r} =x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k} =\langle x(t),y(t),z(t)\rangle $$

Vector Derivatives

Differentiate each component of the position vector to obtain the velocity vector.

$$ \begin{aligned} \vec{v} &=\frac{d\vec{r}}{dt}\\ &=\frac{dx}{dt}\hat{\imath} +\frac{dy}{dt}\hat{\jmath} +\frac{dz}{dt}\hat{k}\\ &=\langle x'(t),y'(t),z'(t)\rangle\\ &=\langle\dot{x},\dot{y},\dot{z}\rangle \end{aligned} $$