When acceleration varies with time:

$v(t) = v_0 + \int_{t_0}^t a(t') \, dt'$

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

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

$\boxed{t = 8 \text{ s}}$

(b) Maximum velocity:

$v(t) = \int_0^t (40 - 5t') \, dt' = 40t - \frac{5t^2}{2}$

At t = 8 s: $v(8) = 40(8) - \frac{5(64)}{2} = 320 - 160$

$\boxed{v_{max} = 160 \text{ m/s}}$

(c) Total distance:

$x(t) = \int_0^t \left(40t' - \frac{5t'^2}{2}\right) dt'$

$x(t) = 40 \cdot \frac{t^2}{2} - \frac{5t^3}{6} = 20t^2 - \frac{5t^3}{6}$

At t = 8 s: $x(8) = 20(64) - \frac{5(512)}{6} = 1280 - 426.7$

$\boxed{x = 853 \text{ m}}$

$v\frac{dv}{dx} = -kx$

$v \, dv = -kx \, dx$

Integrating: $\frac{v^2}{2} = -\frac{kx^2}{2} + C$

At x = 0, v = 8: $C = \frac{64}{2} = 32$

$\frac{v^2}{2} = -\frac{4x^2}{2} + 32$

$v^2 = 64 - 4x^2$

$\boxed{v = \sqrt{64 - 4x^2} = 2\sqrt{16 - x^2} \text{ m/s}}$

(b) Maximum displacement:

At maximum displacement, v = 0: $0 = 64 - 4x_{max}^2$

$x_{max}^2 = 16$

$\boxed{x_{max} = 4 \text{ m}}$

(c) Type of motion:

$a = -kx = -(4)x$

This is Simple Harmonic Motion (SHM)!

$\omega^2 = k = 4 \implies \omega = 2 \text{ rad/s}$

The particle oscillates with amplitude A = 4 m and angular frequency ω = 2 rad/s.

General solution: $x(t) = A\sin(\omega t + \phi)$