Displacement velocity and acceleration

The Object Model

The object model simplifies analysis by ignoring an object's size, shape, and internal structure. The entire object is treated as a single point located at one position.

This simplification works whenever the object's dimensions are small compared to the distances involved. A car traveling between two cities can be modeled as a point. A spinning basketball being analyzed for rotation cannot — its internal structure matters there.

Throughout kinematics, every object is assumed to follow this model unless stated otherwise. That means one position value, one velocity value, and one acceleration value describe the entire object at any instant.

MisconceptionStudents sometimes think the object model means the object is literally small. It means the object's size is irrelevant to the problem, not that the object has no size.

Exam TipIf a problem says "a car" or "a rocket," treat it as a point unless told otherwise.

Displacement

Displacement is the change in an object's position. It is a vector quantity — it has both magnitude and direction.

Key Equations

Displacement:

$$\Delta x=x-{x}_{0}$$

Variables:

• $\Delta x$ = displacement (m)
• $x$ = final position (m)
• ${x}_{0}$ = initial position (m)

SI unit: metres (m)

ProportionalityDisplacement is directly proportional to the difference between final and initial positions.

Reference sheet status: On the reference sheet (embedded in kinematic equations).

Reading the notation: The symbol $\Delta $ (Greek capital delta) means "change in." So $\Delta x$ means "change in $x$." It always equals the final value minus the initial value.

Displacement depends only on the starting and ending positions. The path taken does not matter. An object that travels 10 m east and then 4 m west has a displacement of 6 m east, even though it covered 14 m of total distance.

A negative displacement means the object moved in the negative direction. Always define a positive direction before solving.

Worked Example: Finding Displacement

Scenario

A drone starts at position ${x}_{0}=12$ m and flies to a final position $x=-5$ m along the $x$-axis.

Displacement:

Equation used

$$\Delta x=x-{x}_{0}$$

Given

$$x=-5\text{ m}$$

$${x}_{0}=12\text{ m}$$

Working

$$\Delta x=-5-12$$

$$\Delta x=-17\text{ m}$$

Answer

$$\Delta x=-17\text{ m}$$

Interpretation

The drone moved 17 m in the negative $x$-direction.

Average Velocity and Average Acceleration

Average velocity and average acceleration describe how position and velocity change over a finite time interval. Both are calculated using initial and final states only.

Key Equations

Average velocity:

$${\vec{v}}_{\text{avg}}=\frac{\Delta \vec{x}}{\Delta t}$$

Variables:

• ${\vec{v}}_{\text{avg}}$ = average velocity (m/s)
• $\Delta \vec{x}$ = displacement (m)
• $\Delta t$ = time interval (s)

SI unit: m/s

Rearrangements:

$$\Delta \vec{x}={\vec{v}}_{\text{avg}}\cdot \Delta t$$

$$\Delta t=\frac{\Delta \vec{x}}{{\vec{v}}_{\text{avg}}}$$

ProportionalityAverage velocity is directly proportional to displacement and inversely proportional to the time interval.

Reference sheet status: On the reference sheet.

Average acceleration:

$${\vec{a}}_{\text{avg}}=\frac{\Delta \vec{v}}{\Delta t}$$

Variables:

• ${\vec{a}}_{\text{avg}}$ = average acceleration (m/s²)
• $\Delta \vec{v}$ = change in velocity (m/s)
• $\Delta t$ = time interval (s)

SI unit: m/s²

Rearrangements:

$$\Delta \vec{v}={\vec{a}}_{\text{avg}}\cdot \Delta t$$

$$\Delta t=\frac{\Delta \vec{v}}{{\vec{a}}_{\text{avg}}}$$

ProportionalityAverage acceleration is directly proportional to the change in velocity and inversely proportional to the time interval. Doubling $\Delta t$ while keeping $\Delta v$ the same halves the average acceleration.

Reference sheet status: On the reference sheet.

Average velocity equals displacement divided by the elapsed time. Because displacement is a vector, average velocity is also a vector. Its direction matches the direction of the net displacement.

Average acceleration equals the change in velocity divided by the elapsed time. An object is accelerating whenever the magnitude or direction of its velocity changes. A car speeding up, a car slowing down, and a car turning at constant speed are all accelerating.

MisconceptionStudents often think acceleration requires a change in speed. An object moving at constant speed along a curved path is accelerating because its direction is changing.

Exam Tip"Acceleration" on the AP exam means any change in the velocity vector — magnitude, direction, or both.

Worked Example: Average Acceleration

Scenario

A cyclist increases velocity from 4.0 m/s east to 10.0 m/s east in 3.0 s. Find the average acceleration.

Average acceleration:

Equation used

$${\vec{a}}_{\text{avg}}=\frac{\Delta \vec{v}}{\Delta t}$$

Given

$$v=10.0\text{ m/s}$$

$${v}_{0}=4.0\text{ m/s}$$

$$\Delta t=3.0\text{ s}$$

Working

$$\Delta v=10.0-4.0=6.0\text{ m/s}$$

$${a}_{\text{avg}}=\frac{6.0}{3.0}$$

Answer

$${a}_{\text{avg}}=2.00{\text{ m/s}}^{2}\text{ east}$$

Interpretation

The cyclist's velocity increased by 2.0 m/s each second, directed east.

Instantaneous Velocity and Acceleration

As a time interval shrinks toward zero, the average velocity over that interval approaches the instantaneous velocity. This is the central idea connecting algebra-based averages to calculus-based instantaneous quantities.

Key Equations

Instantaneous velocity:

$$\vec{v}=\frac{d\vec{r}}{dt} ; {v}_{x}=\frac{dx}{dt}$$

Variables:

• $\vec{v}$ = instantaneous velocity (m/s)
• $\vec{r}$ = position vector (m)
• $t$ = time (s)
• ${v}_{x}$ = $x$-component of velocity (m/s)
• $x$ = position along $x$-axis (m)

SI unit: m/s

Reference sheet status: On the reference sheet.

Instantaneous acceleration:

$$\vec{a}=\frac{d\vec{v}}{dt} ; {a}_{x}=\frac{d{v}_{x}}{dt}$$

Variables:

• $\vec{a}$ = instantaneous acceleration (m/s²)
• $\vec{v}$ = instantaneous velocity (m/s)
• $t$ = time (s)

SI unit: m/s²

Reference sheet status: On the reference sheet.

Calculus form — acceleration as second derivative of position:

$${a}_{x}=\frac{{d}^{2}x}{d{t}^{2}}$$

Reading calculus notation: The symbol $\frac{dx}{dt}$ means the instantaneous rate of change of $x$ with respect to $t$. It is the slope of the $x$-vs-$t$ graph at a single instant. The symbol $\frac{{d}^{2}x}{d{t}^{2}}$ is the second derivative — the rate of change of the rate of change.

Instantaneous velocity is the derivative of position with respect to time. Geometrically, it equals the slope of the position-time graph at a single point. When $x(t)$ is a polynomial in $t$, applying the power rule gives ${v}_{x}(t)$ directly.

Instantaneous acceleration is the derivative of velocity with respect to time. It equals the slope of the velocity-time graph at a single point. Differentiating again, acceleration is the second derivative of position.

The reverse process also works. Integration recovers the original function from its derivative:

$$x(t)=\int {v}_{x}(t) dt ; {v}_{x}(t)=\int {a}_{x}(t) dt$$

Each integration introduces a constant of integration determined by initial conditions. If ${v}_{x}(0)$ is known, it fixes the constant when integrating acceleration. If $x(0)$ is known, it fixes the constant when integrating velocity.

Examiner InsightAP Physics C FRQs often give a position function like $x(t)=A{t}^{3}+Bt$ and ask for velocity and acceleration at a specific time. Differentiate first, then substitute.

Exam TipAlways show the differentiation step explicitly — do not jump to the numerical answer.

Derivation: Velocity and Acceleration from a Position Function

Starting point: Definition of instantaneous velocity.

$${v}_{x}=\frac{dx}{dt}$$

Step 1 — State the given position function

$$x(t)=5{t}^{3}-2t+1$$

Step 2 — Differentiate $x(t)$ with respect to $t$ using the power rule

Each term $A{t}^{n}$ differentiates to $nA{t}^{n-1}$.

$${v}_{x}(t)=\frac{d}{dt}(5{t}^{3}-2t+1)=15{t}^{2}-2$$

Step 3 — Differentiate ${v}_{x}(t)$ to find acceleration

$${a}_{x}(t)=\frac{d{v}_{x}}{dt}=\frac{d}{dt}(15{t}^{2}-2)=30t$$

Result

$${v}_{x}(t)=15{t}^{2}-2 ; {a}_{x}(t)=30t$$

At $t=0$, the velocity is $-2$ m/s and the acceleration is zero. Both velocity and acceleration increase with time.

Worked Example: Integration to Find Velocity

Scenario

A particle has acceleration ${a}_{x}(t)=6t$ m/s² and initial velocity ${v}_{x}(0)=3$ m/s. Find ${v}_{x}(t)$.

Step-by-step solution

Equation used

$${v}_{x}(t)=\int {a}_{x}(t) dt$$

Step 1 — Set up the integral:

$${v}_{x}(t)=\int 6t dt$$

The integration variable is $t$. The integrand $6t$ represents how acceleration varies with time.

Step 2 — Evaluate the integral using the power rule:

$${v}_{x}(t)=3{t}^{2}+C$$

Step 3 — Apply the initial condition ${v}_{x}(0)=3$:

$$3=3(0{)}^{2}+C$$

$$C=3$$

Answer

$${v}_{x}(t)=3{t}^{2}+3\text{ m/s}$$

Interpretation

The velocity starts at 3 m/s and increases as ${t}^{2}$ because the acceleration itself grows linearly with time.