Topic 1.2 Displacement, Velocity, and Acceleration: define velocity and acceleration as the time derivatives of position and velocity, integrate to recover velocity and position, and apply the constant-acceleration kinematic equations.

What this topic is asking

The College Board (Topic 1.2) wants you to define velocity as the time derivative of position and acceleration as the time derivative of velocity, to integrate to recover velocity and position when the acceleration is a known function of time, to distinguish average from instantaneous quantities, and to apply the constant-acceleration kinematic equations as the special case. This is where AP Physics C: Mechanics diverges from algebra-based physics: the definitions are calculus definitions, and a typical exam problem hands you a position or acceleration function and asks you to differentiate or integrate it.

Velocity and acceleration as derivatives

These are the engine of AP Physics C kinematics. If you are given the position as a function of time, you find velocity by differentiating once and acceleration by differentiating twice. For example, if $x(t) = 4t^2 - 5t$, then $v(t) = 8t - 5$ and $a(t) = 8$ (a constant). The calculus is usually polynomial differentiation, with occasional sines, cosines or exponentials.

Recovering motion by integration

If instead you know the acceleration as a function of time, you build the velocity and position by integrating, using the initial conditions as the constants:

The initial velocity $v_0$ and initial position $x_0$ are the constants of integration; without them the motion is undetermined. This is why exam problems always supply the starting conditions. Graphically, the integral is an area: the area under an acceleration-time graph is the change in velocity, and the area under a velocity-time graph is the displacement.

Average versus instantaneous

An average quantity uses the total change over a whole interval: $v_{avg} = \dfrac{\Delta x}{\Delta t}$ and $a_{avg} = \dfrac{\Delta v}{\Delta t}$. An instantaneous value is the derivative at one moment, the slope of the tangent on the graph. For constant acceleration only, the average velocity equals the simple mean of the endpoints, $v_{avg} = \dfrac{v_0 + v}{2}$. When the acceleration varies, that shortcut fails and you must integrate.

The constant-acceleration equations

When the acceleration is constant, integrating $a$ once and twice produces the familiar kinematic equations:

Each omits one variable: the first has no displacement, the second has no final velocity, the third has no time. List your knowns and the unknown you want, then pick the equation with exactly those. In AP Physics C these are a special case of the calculus, not a starting point: if the acceleration is not constant (a falling object with air resistance, say), you go back to integrating $a(t)$ directly.

Speeding up or slowing down

Because velocity and acceleration are independent vectors, an object can have, for instance, a positive velocity and a negative acceleration at once: it moves forward while slowing. The rule is that an object speeds up when $v$ and $a$ share a sign and slows down when their signs differ. At a turning point the velocity is momentarily zero while the acceleration is nonzero. This sign analysis is exactly what the calculus FRQs reward.

Try this

Q1. A particle has position $x(t) = t^3 - 6t^2 + 9t$ (SI units). Calculate its velocity at $t = 0$ and identify the times it is at rest. [3 points]

• Cue. $v(t) = 3t^2 - 12t + 9$; $v(0) = 9$ m/s. Setting $v = 0$: $3(t-1)(t-3) = 0$, so $t = 1$ s and $t = 3$ s.

Q2. A ball is dropped from rest with $g = 9.8$ m/s squared (constant). Calculate its speed after $2.0$ s. [2 points]

• Cue. $v = v_0 + at = 0 + (9.8)(2.0) = 19.6$ m/s downward.