Kinematics with calculus Knowledge-Based Study

Study notes

1. In one-dimensional kinematics, calculus provides the exact mathematical framework for describing motion when velocity or acceleration changes continuously over time.

   Instantaneous Velocity

   Instantaneous velocity is defined as the first derivative of position with respect to time:

   $$v(t) = \frac{dx}{dt}$$

   Instantaneous Acceleration

   Instantaneous acceleration is defined as the time rate of change of velocity, which corresponds to the second derivative of position with respect to time:

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

   Geometrically, velocity represents the slope of the tangent line on a position-time graph, while acceleration represents the slope of the tangent line on a velocity-time graph.

   When acceleration or velocity is known as a function of time, integration enables the determination of velocity and position by accumulating infinitesimal changes.

   Finding Velocity from Acceleration

   Integrating acceleration between an initial time $t = 0$ and time $t$ yields the change in velocity:

   $$v(t) - v_0 = \int_{0}^{t} a(t') \, dt'$$

   Finding Position from Velocity

   Integrating velocity over the same interval gives the displacement:

   $$x(t) - x_0 = \int_{0}^{t} v(t') \, dt'$$

   The constants of integration, $x_0$ and $v_0$, correspond to the initial position and initial velocity of the object, which must be specified to obtain unique equations of motion.

   When acceleration is given as a function of position, $a(x)$, rather than time, the time differential must be eliminated using the chain rule:

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

   Separating variables and integrating both sides between initial and final states gives:

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

   This kinematic relationship connects position-dependent forces directly to velocity without requiring time as an intermediate variable.

   Consider a particle moving along a straight line with time-dependent acceleration $a(t) = 6t - 4$ (in SI units), with initial conditions $v(0) = 3\text{ m/s}$ and $x(0) = 2\text{ m}$.

   Step 1: Velocity Function

   Integrate the acceleration with respect to time:

   $$v(t) = v_0 + \int_{0}^{t} (6t' - 4) \, dt' = 3 + \left[ 3t^2 - 4t \right] = 3t^2 - 4t + 3$$

   Step 2: Position Function

   Integrate the velocity function to find position:

   $$x(t) = x_0 + \int_{0}^{t} (3t'^2 - 4t' + 3) \, dt' = 2 + \left[ t^3 - 2t^2 + 3t \right] = t^3 - 2t^2 + 3t + 2$$

   This demonstrates how integration of arbitrary continuous functions systematically produces the kinematic equations of motion.