Velocity as derivative of position: $\vec{v}(t) = \frac{d\vec{r}}{dt}$

Acceleration as derivative of velocity: $\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}$

Position from velocity (integration): $\vec{r}(t) = \vec{r}_0 + \int_{t_0}^{t} \vec{v}(t') \, dt'$

Velocity from acceleration (integration): $\vec{v}(t) = \vec{v}_0 + \int_{t_0}^{t} \vec{a}(t') \, dt'$

One-Dimensional Motion

For motion along a straight line (x-axis):

Position function: $x(t)$

Velocity: $v_x(t) = \frac{dx}{dt}$

Acceleration: $a_x(t) = \frac{dv_x}{dt} = \frac{d^2x}{dt^2}$

Example: Polynomial Position Function

If $x(t) = 3t^3 - 2t^2 + 5t + 1$ (in meters), then:

Velocity: $v_x = \frac{dx}{dt} = 9t^2 - 4t + 5 \text{ m/s}$

Acceleration: $a_x = \frac{dv_x}{dt} = 18t - 4 \text{ m/s}^2$

Two-Dimensional Motion

Position vector in 2D: $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$

Velocity vector: $\vec{v}(t) = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} = v_x\hat{i} + v_y\hat{j}$

Acceleration vector: $\vec{a}(t) = \frac{dv_x}{dt}\hat{i} + \frac{dv_y}{dt}\hat{j} = a_x\hat{i} + a_y\hat{j}$

Speed (magnitude of velocity): $v = |\vec{v}| = \sqrt{v_x^2 + v_y^2}$

Projectile Motion

For projectile motion with no air resistance:

Horizontal component:

• $a_x = 0$
• $v_x = v_{0x} = v_0\cos\theta$ (constant)
• $x(t) = x_0 + v_{0x}t$

Vertical component:

• $a_y = -g$
• $v_y(t) = v_{0y} - gt = v_0\sin\theta - gt$
• $y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2$

Variable Acceleration

When acceleration is a function of time, $a(t)$:

$v(t) = v_0 + \int_0^t a(t') \, dt'$

$x(t) = x_0 + \int_0^t v(t') \, dt'$

Example: Time-Dependent Acceleration

If $a(t) = 6t - 4$ m/s², and $v_0 = 2$ m/s, $x_0 = 0$:

$v(t) = 2 + \int_0^t (6t' - 4) \, dt' = 2 + 3t^2 - 4t$

$x(t) = \int_0^t (2 + 3t'^2 - 4t') \, dt' = 2t + t^3 - 2t^2$

$v_t = \frac{g}{b} = \frac{9.8}{0.2}$

$\boxed{v_t = 49 \text{ m/s}}$

(b) Velocity as function of time:

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

Separating variables: $\frac{dv}{g - bv} = dt$

$-\frac{1}{b}\ln(g - bv) = t + C$

At t = 0, v = 0: $C = -\frac{1}{b}\ln(g)$

$\ln\left(\frac{g - bv}{g}\right) = -bt$

$g - bv = ge^{-bt}$

$v = \frac{g}{b}(1 - e^{-bt})$

$\boxed{v(t) = 49(1 - e^{-0.2t}) \text{ m/s}}$

(c) Time to reach 95% of v_t:

$0.95v_t = v_t(1 - e^{-bt})$ $0.95 = 1 - e^{-bt}$ $e^{-bt} = 0.05$ $-bt = \ln(0.05)$

$t = -\frac{\ln(0.05)}{b} = -\frac{-2.996}{0.2}$

$\boxed{t = 15.0 \text{ s}}$