$x(t) = x_0 + \int_{t_0}^t v(t') \, dt'$

Example: Linear Time Dependence

If $a(t) = At + B$ where $A$ and $B$ are constants:

$v(t) = v_0 + \int_0^t (At' + B) \, dt' = v_0 + \frac{1}{2}At^2 + Bt$

$x(t) = x_0 + \int_0^t \left(v_0 + \frac{1}{2}At'^2 + Bt'\right) dt' = x_0 + v_0t + \frac{1}{6}At^3 + \frac{1}{2}Bt^2$

Acceleration as a Function of Position: a(x)

When acceleration depends on position, use the chain rule:

$a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$

Therefore: $v \, dv = a(x) \, dx$

Integrating both sides: $\int_{v_0}^v v' \, dv' = \int_{x_0}^x a(x') \, dx'$

$\frac{1}{2}(v^2 - v_0^2) = \int_{x_0}^x a(x') \, dx'$

Example: Spring Force

For a spring with $a = -\frac{k}{m}x$:

$v \, dv = -\frac{k}{m}x \, dx$

$\int_{v_0}^v v' \, dv' = -\frac{k}{m}\int_{x_0}^x x' \, dx'$

$\frac{1}{2}(v^2 - v_0^2) = -\frac{k}{2m}(x^2 - x_0^2)$

$v^2 = v_0^2 - \frac{k}{m}(x^2 - x_0^2)$

Acceleration as a Function of Velocity: a(v)

When acceleration depends on velocity:

$a = \frac{dv}{dt} = a(v)$

Separate variables: $\frac{dv}{a(v)} = dt$

Integrate: $\int_{v_0}^v \frac{dv'}{a(v')} = \int_0^t dt' = t$

Example: Linear Drag Force

For drag force $a = -bv$ (where $b > 0$):

$\frac{dv}{dt} = -bv$

$\frac{dv}{v} = -b \, dt$

$\int_{v_0}^v \frac{dv'}{v'} = -b\int_0^t dt'$

$\ln\frac{v}{v_0} = -bt$

$v(t) = v_0e^{-bt}$

To find position, integrate velocity:

$x(t) = \int_0^t v_0e^{-bt'} \, dt' = \frac{v_0}{b}(1 - e^{-bt})$

Quadratic Drag Force

For drag proportional to $v^2$: $a = -bv^2$

$\frac{dv}{dt} = -bv^2$

$\int_{v_0}^v \frac{dv'}{v'^2} = -b\int_0^t dt'$

$-\frac{1}{v} + \frac{1}{v_0} = -bt$

$v(t) = \frac{v_0}{1 + bv_0t}$

Falling with Air Resistance

Terminal velocity occurs when drag force equals gravitational force:

For linear drag: $a = g - bv$

At terminal velocity: $v_t = \frac{g}{b}$

General solution: $v(t) = v_t(1 - e^{-bt}) + v_0e^{-bt}$

For quadratic drag: $a = g - bv^2$

Terminal velocity: $v_t = \sqrt{\frac{g}{b}}$