Kinematics in 1-D

Basic Integrals

\[x(t) = x(t_0) + \int_{t_0}^{t} \frac{dx}{dt^\prime} dt^{\prime} \\ v(t) = v(t_0) + \int_{t_0}^{t} \frac{dv}{dt^\prime} dt^{\prime}\]

Notice:

the use of $t^\prime$ instead of $t$ in the integral is deliberate. While both are technically correct, using $t^\prime$ reduces ambiguity as it differentiates the $t$ in the integral’s bound and the entity $t^\prime$ to be integrated over (imagine the latter iterating from $t_0$ to $t$).

Now, consider functions such as $v(x)$ or $a(v)$ instead of $v(t)$ or $a(t)$:

\[\frac{dx}{dt} = v(x) \\ \frac{dx}{v(x)} = dt \\ \int_{x_0}^{x(t)}\frac{1}{v(x)}dx**^{\prime}** = \int_{t_0}^{t} dt^{\prime}\]

After solving this, you will get an expression of $x(t)$.

Example:

Question: Consider Linear Drag: $a = -bv$, where $b$ is a constant. Solve for the expression of $v$ as a function of $t$. (take $t_0 = 0$)
Solution:
\[\frac{dv}{dt} = -bv \\ \frac{1}{v}dv = -b\, dt \\ \int_{v_0}^{v(t)}\frac{1}{v^{\prime}}dv^{\prime} = \int_{t_0=0}^{t} -b\, dt^{\prime} \\ ln\left(\frac{v(t)}{v_0}\right) = -bt \\ v(t) = v_0e^{-bt}\]
And here you see the exponential decaying velocity caused by the drag. As $t \rightarrow \infty$, $v \rightarrow 0$.

Second Order (Linear) Differential Equations

Non-Homogenous Second Order Linear Differential Equations: $ay’‘+by’+cy’=g(t)$ or $a(t)y’‘+b(t)y’+c(t)y’=g(t)$

Homogenous Second Order Linear Differential Equations: $ay’‘+by’+cy’=0$, basically when $g(t)=0$

Homogenous Second Order DE

Common forms include:

$\ddot x +\omega x= 0$, or $\ddot x = -\omega x$

general solution: $x = Asin(\omega t)+ Bcos(\omega t)$

$\ddot x -\frac{1}{\tau} x= 0$, or $\ddot x = \frac{1}{\tau} x$

general solution: $x=Ae^{\frac{t}{\tau}} + Be^{-\frac{t}{\tau}}$