Kinematics (Calculus-Based)

Definitions via Calculus

$v(t) = \frac{dx}{dt}, \qquad a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

Going from Acceleration to Position

$v(t) = v_0 + \int_0^t a(t')\,dt'$ $x(t) = x_0 + \int_0^t v(t')\,dt'$

Constant Acceleration (Special Case)

Integrating $a = \text{const}$ reproduces the familiar equations:

• $v = v_0 + at$
• $x = x_0 + v_0 t + \frac{1}{2}at^2$
• $v^2 = v_0^2 + 2a(x - x_0)$

Non-Constant Acceleration

When $a(t)$ is a function of time, you must integrate directly. Common cases:

• $a(t) = bt$ → $v(t) = v_0 + \frac{1}{2}bt^2$
• $a(v) = -bv$ (drag-like) → requires separation of variables.

Velocity as a Function of Position

Using the chain rule: $a = v\frac{dv}{dx}$.

This is useful when acceleration depends on position (e.g., springs, gravity).

Projectile Motion

Same decomposition as AP Physics 1, but with calculus: $x(t) = v_{0x}t, \quad y(t) = v_{0y}t - \frac{1}{2}gt^2$ $v_x = v_{0x}, \quad v_y = v_{0y} - gt$

Problem: A particle has acceleration $a(t) = 6t\ \text{m/s}^2$, with $v(0) = 2\ \text{m/s}$ and $x(0) = 0$. Find $v(t)$ and $x(t)$, then find the position at $t = 3\ \text{s}$.

Solution:

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

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

At $t = 3$: $x(3) = 2(3) + 3^3 = 6 + 27 = 33\ \text{m}$.

• Velocity is the derivative of position; acceleration is the derivative of velocity.
• For non-constant acceleration, integrate $a(t)$ to find $v(t)$ and $x(t)$.
• Use $a = v\,dv/dx$ when acceleration depends on position.
• Calculus enables solving problems that algebraic equations cannot.