Displacement velocity and acceleration

The Object Model and Point Particles

Real objects are complicated — they rotate, flex, and have parts that move differently from one another. Physics simplifies this by using the object model, which ignores the size, shape, and internal structure of an object and treats it as a single point. This point carries all of the object's extensive properties [properties that depend on the amount of matter, such as mass and charge].

Why does this work? When you track a car driving across a city, the length of the car is irrelevant compared to the distance travelled. Shrinking the car to a dot changes nothing about the physics of its journey. The object model is valid whenever the object's dimensions are small compared to the distances involved in the problem.

MisconceptionStudents sometimes think the object model means ignoring mass. It does not. Mass and charge are kept; only size and shape are discarded.

Exam TipIf a question says "treat the object as a particle," it is invoking the object model.

Displacement as Change in Position

Key Equations

Displacement:

$$\vec{d}={x}_{f}-{x}_{i}$$

Variables:

• $\vec{d}$ = displacement (m)
• ${x}_{f}$ = final position (m)
• ${x}_{i}$ = initial position (m)

SI unit: metre (m)

Reference sheet status: Foundational definition — not listed as a standalone equation on the reference sheet.

Every object has a position, which describes where it sits along a chosen coordinate axis relative to a chosen origin. Displacement is the change in an object's position. It equals the final position minus the initial position.

Displacement is a vector quantity. Its sign tells you the direction of the position change along the axis. A positive displacement means the object moved in the positive direction; a negative displacement means the opposite.

Distance, by contrast, is a scalar. It counts the total length of the path travelled and is always positive. An object that walks 3 m east and then 3 m west has zero displacement but 6 m of distance.

Average Velocity

Key Equations

Average velocity:

$${\vec{v}}_{avg}=\frac{\vec{d}}{\Delta t}=\frac{{x}_{f}-{x}_{i}}{\Delta t}$$

Variables:

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

SI unit: metre per second (m/s)

Rearrangements:

$$\vec{d}={\vec{v}}_{avg}\times \Delta t$$

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

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

Reference sheet status: On the reference sheet.

Average velocity is the displacement of an object divided by the time interval over which that displacement occurs. Because displacement is a vector, average velocity is also a vector — it carries a sign that indicates direction.

Average velocity considers only the initial and final states of the object over the interval. It reveals nothing about what happened between those two moments. An object could speed up, slow down, or reverse direction during the interval, yet the average velocity depends only on where it started, where it ended, and how long the trip took.

Average speed, by contrast, equals total distance divided by total time. Average speed is always positive and is often larger than the magnitude of average velocity when the path is not straight.

Examiner InsightAP questions often give a round-trip scenario and ask for both average speed and average velocity. Students who confuse the two lose easy credit.

Exam TipAverage velocity uses displacement; average speed uses distance. State which one you are using.

Worked Example: Average Velocity of a Cyclist

Scenario

A cyclist rides 1200 m east in 80 s, then turns and rides 400 m west in 40 s. Taking east as positive, find her average velocity for the entire trip.

Step-by-step solution:

Finding the total displacement:

$$\vec{d}=(+1200)+(-400)$$

$$\vec{d}=+800\text{ m}$$

Finding the total time:

$$\Delta t=80+40=120\text{ s}$$

Equation used

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

Substitution:

$${\vec{v}}_{avg}=\frac{800}{120}$$

$${\vec{v}}_{avg}=+6.67\text{ m/s}$$

Answer

$${\vec{v}}_{avg}=+6.67\text{ m/s (east)}$$

Interpretation

The cyclist's average velocity is 6.67 m/s in the eastward direction. Her average speed would be larger because the total distance (1600 m) exceeds the displacement (800 m).

Average Acceleration

Key Equations

Average acceleration:

$${\vec{a}}_{avg}=\frac{\Delta \vec{v}}{\Delta t}=\frac{{\vec{v}}_{f}-{\vec{v}}_{i}}{\Delta t}$$

Variables:

• ${\vec{a}}_{avg}$ = average acceleration (m/s²)
• ${\vec{v}}_{f}$ = final velocity (m/s)
• ${\vec{v}}_{i}$ = initial velocity (m/s)
• $\Delta t$ = time interval (s)

SI unit: metre per second squared (m/s²)

Rearrangements:

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

$${\vec{v}}_{f}={\vec{v}}_{i}+{\vec{a}}_{avg}\times \Delta t$$

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

ProportionalityAverage acceleration is directly proportional to the change in velocity and inversely proportional to the time interval. Doubling the change in velocity doubles the acceleration. Doubling the time interval halves the acceleration.

Reference sheet status: On the reference sheet.

Average acceleration is the change in velocity divided by the time interval over which that change occurs. Like velocity, acceleration is a vector — its sign indicates the direction of the velocity change, not the direction of motion.

A common source of confusion: acceleration does not mean "speeding up." An object is accelerating if the magnitude and/or direction of the object's velocity are changing. A car slowing down is accelerating (the velocity magnitude is changing). A satellite moving at constant speed in a circle is accelerating (the velocity direction is changing). Only an object with constant speed moving in a straight line has zero acceleration.

When the acceleration vector points in the same direction as the velocity vector, the object speeds up. When the acceleration vector points opposite to the velocity vector, the object slows down. When the acceleration vector is perpendicular to the velocity vector, the object changes direction without changing speed.

MisconceptionStudents often believe that a negative acceleration means the object is slowing down. A negative acceleration simply means the acceleration points in the negative direction. If the object is also moving in the negative direction, it is actually speeding up.

Exam TipCompare the signs of velocity and acceleration. Same sign → speeding up. Opposite signs → slowing down.

Worked Example: Braking Acceleration

Scenario

A bus travelling at 18 m/s east applies its brakes and comes to rest in 6.0 s. Taking east as positive, find the average acceleration.

Step-by-step solution:

Equation used

$${\vec{a}}_{avg}=\frac{{\vec{v}}_{f}-{\vec{v}}_{i}}{\Delta t}$$

Given

$${\vec{v}}_{i}=+18\text{ m/s}$$

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

$$\Delta t=6.0\text{ s}$$

Substitution:

$${\vec{a}}_{avg}=\frac{0-18}{6.0}$$

$${\vec{a}}_{avg}=\frac{-18}{6.0}$$

Answer

$${\vec{a}}_{avg}=-3.0{\text{ m/s}}^{2}$$

Interpretation

The negative sign means the acceleration is directed west — opposite to the initial velocity. Because velocity and acceleration point in opposite directions, the bus is slowing down.

Instantaneous Values from Small Time Intervals

Averages calculated over an entire trip can hide what happens moment to moment. Instantaneous velocity is the velocity of an object at a single instant in time, and instantaneous acceleration is the acceleration at a single instant.

Calculating average velocity or average acceleration over a very small time interval yields a value that is very close to the instantaneous value. As the time interval $\Delta t$ shrinks toward zero, the average and instantaneous values converge. On a position-time graph, instantaneous velocity equals the slope of the tangent line at that point. On a velocity-time graph, instantaneous acceleration equals the slope of the tangent line at that point.

This idea bridges algebra-based physics and calculus-based physics. For the AP exam, the key practical skill is recognising that a tangent-line slope on a graph gives the instantaneous rate of change.