AP Physics C: Mechanics — Unit 1: Kinematics Breakdown

Fundamental Concepts: Position, Velocity, and Acceleration

One-dimensional kinematics describes motion along a straight line (e.g., the x-axis or y-axis) without considering the forces causing the motion. In AP Physics C, unlike Physics 1, we define these quantities using differential calculus.

Position and Displacement

Position ($x$): The location of an object relative to a defined origin at a specific time $t$.
Displacement ($\Delta x$): The change in position. It is a vector quantity.
\Delta x = xf - xi
Distance: The total path length traveled. It is a scalar and is never negative.

Velocity vs. Speed

Average Velocity: The total displacement divided by the time interval.
v_{avg} = \frac{\Delta x}{\Delta t}
Instantaneous Velocity ($v$): The velocity at a specific moment in time. Following calculus definitions, it is the derivative of position with respect to time.
v(t) = \frac{dx}{dt}
Speed: The magnitude of the instantaneous velocity vector ($|v|$). Speed is a scalar.

Acceleration

Average Acceleration: The change in velocity over a time interval.
a_{avg} = \frac{\Delta v}{\Delta t}
Instantaneous Acceleration ($a$): The rate of change of velocity. It is the derivative of velocity, or the second derivative of position.
a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}

Graphs showing the relationship between Position, Velocity, and Acceleration

Motion with Constant Acceleration

When acceleration is constant (i.e., $a(t) = \text{constant}$), we can derive the "Big Three" kinematic equations. These are frequently used for free-fall problems and uniform interactions.

Warning: These equations ONLY apply when acceleration is constant.

The Kinematic Equations

Equation	Variables Related	Missing Variable
vx = v{x0} + a_xt	Velocity, Time, Accel	Position ($x$)
x = x0 + v{x0}t + \frac{1}{2}a_xt^2	Position, Time, Accel	Final Velocity ($v$)
vx^2 = v{x0}^2 + 2ax(x - x0)	Velocity, Position, Accel	Time ($t$)

Notation Refresher:

$x_0$: Initial position at $t=0$
$v_{x0}$: Initial velocity at $t=0$
$a_x$: Constant acceleration

Free Fall

Free fall is the most common example of constant acceleration. In the absence of air resistance, all objects near Earth's surface accelerate downward at a constant rate $g$.

Value: $g \approx 9.8 \; m/s^2$ (often approximated as $10 \; m/s^2$ on multiple choice depending on exam instructions).
Direction: Gravity acts downward. If you define "up" as positive, then $a_y = -g = -9.8 \; m/s^2$.

Diagram of a ball thrown upward showing velocity and acceleration vectors

Example: The Vertical Toss

A ball is thrown straight up with an initial speed of $20 \; m/s$. How high does it go?

Solution:

Define coordinates: Let ground $= 0$, Up $= +y$.
Identify variables: $v{y0} = 20$, $ay = -9.8$, $v_{final} = 0$ (at the peak).
Select Equation: $vy^2 = v{y0}^2 + 2a_y(\Delta y)$
Substitute: $0 = (20)^2 + 2(-9.8)(\Delta y)$
Solve: $-400 = -19.6(\Delta y) \Rightarrow \Delta y \approx 20.4 \; m$

Motion with Non-Constant Acceleration (Calculus)

This section distinguishes AP Physics C from lower-level physics. If acceleration changes with time (e.g., air resistance, varying engine thrust), you cannot use the kinematic equations above. You must use integration.

From Acceleration to Velocity

Since $a = \frac{dv}{dt}$, we can integrate acceleration to find velocity:
v(t) = \int a(t) \, dt + C
Or, using definite limits:
vf - vi = \int{ti}^{t_f} a(t) \, dt
Note: The area under the Acceleration vs. Time graph represents the change in velocity ($\Delta v$).

From Velocity to Position

Since $v = \frac{dx}{dt}$, we can integrate velocity to find position:
x(t) = \int v(t) \, dt + C
Or, using definite limits:
xf - xi = \int{ti}^{t_f} v(t) \, dt
Note: The area under the Velocity vs. Time graph represents the change in position (displacement).

Graph showing area under a curve for calculus integration

Worked Example: Non-Constant Acceleration

An object moves with acceleration given by $a(t) = 6t \; m/s^2$. At $t=0$, the object is at rest at the origin.

Find the velocity function $v(t)$.
v(t) = \int 6t \, dt = 3t^2 + C1 Apply initial condition: at $t=0, v=0$. Therefore $C1 = 0$.
v(t) = 3t^2
Find the position function $x(t)$.
x(t) = \int 3t^2 \, dt = t^3 + C2 Apply initial condition: at $t=0, x=0$. Therefore $C2 = 0$.
x(t) = t^3

Graphical Analysis Summary

Understanding the relationship between graphs is crucial.

Graph Type	Slope Represents	Area Under Curve Represents
Position vs. Time ($x-t$)	Velocity ($v$)	N/A (physically meaningless)
Velocity vs. Time ($v-t$)	Acceleration ($a$)	Displacement ($\Delta x$)
Acceleration vs. Time ($a-t$)	Jerk (change in accel)	Change in velocity ($\Delta v$)

Common Mistakes & Pitfalls

Using Kinematic Equations when $a \neq constant$: This is the most common error. If $a$ depends on time ($t$) or position ($x$), you MUST use calculus (integration), not the algebraic equations.
Confusing Displacement with Distance: If an object moves forward and then backward to its starting point, displacement is 0, but distance is positive. $\int |v(t)| dt$ is distance, while $\int v(t) dt$ is displacement.
Forgetting the $+C$: When performing indefinite integration, always add the constant of integration ($C$) and solve for it using the Initial Conditions given in the problem statement.
Sign Errors in Free Fall: Define your coordinate system immediately. If Up is positive, then $g$ is negative. If Down is positive, $g$ is positive. Be consistent.