Instantaneous Velocity as a Derivative
Instantaneous velocity is the time derivative of position: (t) = rac{dx}{dt}$.
Example: If (t) = 3t^2 - 2t + 1$ m, then (t) = 6t - 2$ m/s.
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Learn Calculus-Based Kinematics
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Session Length
~17 min
Adaptive Checks
15 questions
Transfer Probes
8
Lesson Notes
Calculus-based kinematics applies differential and integral calculus to describe motion with full generality. Velocity is defined as the time derivative of position, $v(t) = rac{dx}{dt}$, and acceleration as $a(t) = rac{dv}{dt} = rac{d^2x}{dt^2}$. Integration reverses these relationships: given acceleration as a function of time, integrate once to obtain velocity and again to obtain position, applying initial conditions at each step. The chain rule form $a = v rac{dv}{dx}$ enables solving problems where acceleration depends on position rather than time, converting the second-order ODE into a first-order separable equation.
Non-constant acceleration arises naturally in many physical contexts: linear drag ($a = g - bv/m$) produces exponential approach to terminal velocity, quadratic drag ($a = -cv^2$) yields algebraic decay, and spring forces ($a = -omega^2 x$) produce sinusoidal oscillation. In each case the standard constant-acceleration kinematic equations ($v = v_0 + at$, $x = x_0 + v_0 t + frac{1}{2}at^2$) fail, and calculus provides the correct framework.
Graphical interpretation connects these ideas visually: the slope of an $x$-$t$ graph is velocity, the slope of a $v$-$t$ graph is acceleration, and the area under a $v$-$t$ curve over an interval is the displacement during that interval. This topic is essential for AP Physics C: Mechanics and forms the mathematical backbone of Newtonian dynamics.
You'll be able to:
- Compute instantaneous velocity and acceleration by differentiating position functions
- Recover velocity and position from acceleration using integration and initial conditions
- Apply the chain rule form a = v dv/dx to position-dependent acceleration problems
- Distinguish displacement from distance using definite and absolute-value integrals
- Solve first-order separable ODEs from velocity-dependent or position-dependent forces
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Key Concepts
Acceleration as the Second Derivative
Acceleration is (t) = rac{dv}{dt} = rac{d^2x}{dt^2}$.
Example: For (t) = 5t^3$ m, = 15t^2$ m/s and = 30t$ m/s squared.
Integration to Recover Motion
Given acceleration, integrate to get velocity, integrate again to get position. Each step needs an initial condition.
Example: If = 6t$ with (0)=2$, then = 2+3t^2$. With (0)=0$: = 2t+t^3$.
Chain Rule Form
By the chain rule, = v rac{dv}{dx}$. Useful when acceleration depends on position rather than time.
Example: For a spring = -kx/m$, separate: \,dv = (-kx/m),dx$.
Displacement vs Distance
Displacement is the signed integral of velocity. Distance is the integral of the absolute value of velocity.
Example: If velocity changes sign, displacement can be zero while distance is positive.
Non-Constant Acceleration
When acceleration varies, standard kinematic equations fail. Use calculus: set up the ODE and integrate.
Example: Linear drag: = g - bv/m$ integrates to exponential approach to terminal velocity.
Graphical Interpretation
Slope of x-t graph = velocity. Slope of v-t graph = acceleration. Area under v-t graph = displacement.
Example: A curved x-t graph means non-constant velocity.
Separable ODEs in Kinematics
Many problems reduce to separable ODEs. Separate variables and integrate both sides independently.
Example: Quadratic drag = -cv^2$: separate and integrate to get (t) = v_0/(1+v_0 c t)$.
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- Progressive hints (direction, rule, then apply).
- Targeted feedback when a common misconception appears.
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