1.2 Displacement, Velocity, and Acceleration

Displacement is the change in an object's position, a vector you find with $\Delta x = x - x_0$. Average velocity is displacement divided by time, and average acceleration is the change in velocity divided by time.

Multiple-choice questions here ask you to compare distance with displacement, interpret motion descriptions, and calculate averages from given data. Building motion representations and translating between them is a core skill, and these vector definitions set up the graphs and equations in later topics as well as the forces, energy, and momentum that come later in the course.

Vector Definitions for Describing Motion

• Displacement is a vector found with $\Delta x = x - x_0$, while distance is the total path length and is always positive.
• Average velocity is $\vec{v}_{avg} = \frac{\Delta \vec{x}}{\Delta t}$, and average acceleration is $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}$.
• An object is accelerating if the magnitude or direction of its velocity changes, so turning at constant speed still counts as accelerating.
• The object model treats an object as a single point with properties like mass, ignoring size, shape, and internal structure.
• Averaging velocity or acceleration over a very small time interval gives a value close to the instantaneous value.
• Watch your signs: in one dimension, the sign of a vector component shows its direction.

Core Concepts

The object model

When studying motion, you treat an object as a single point, ignoring its size, shape, and internal structure. The point still carries properties like mass and charge. This keeps the analysis focused on the motion of the whole object instead of how each piece moves.

For example, when analyzing the path of a baseball, you do not track how the stitches rotate. You treat the entire ball as one point with mass.

Displacement

Displacement measures how far out of position an object ends up compared to where it started. It only cares about the straight-line difference between the start and end positions, not the path taken to get there.

• Displacement is the change in an object's position.
• Relevant equation: $\Delta x = x - x_0$
  • $\Delta x$ is the displacement
  • $x$ is the final position
  • $x_0$ is the initial position
• Displacement is a vector, so it can be positive, negative, or zero depending on direction.

Distance is different. Distance is the total path length and is always positive. If you walk 3 meters east then 4 meters north, your displacement is 5 meters at about 53° north of east, but the distance you walked is 7 meters.

Average velocity

Average velocity tells you how quickly position changes over a time interval, including direction. Because it has direction, velocity is a vector, which separates it from speed.

• Average velocity is displacement divided by the time interval.
• Relevant equation: $\vec{v}_{avg} = \frac{\Delta \vec{x}}{\Delta t}$
• Units are meters per second (m/s).

A train that travels 30 km east in 45 minutes has an average velocity of 40 km/h east, telling you both how fast and which way it moved.

Average acceleration

Average acceleration describes how quickly velocity changes over a time interval. Like velocity, it is a vector.

• Average acceleration is the change in velocity divided by the time interval.
• Relevant equation: $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}$
• Units are meters per second squared (m/s²).

A rocket that goes from rest to 200 m/s in 10 seconds has an average acceleration of 20 m/s², meaning its velocity grows by 20 m/s each second.

When is an object accelerating?

An object accelerates whenever the magnitude or direction of its velocity changes.

• A change in magnitude means speeding up or slowing down.
• A change in direction means turning, even at constant speed.
• An object with constant velocity (same size and direction) is not accelerating.

A car moving around a circular track at constant speed is still accelerating, because its velocity keeps changing direction as it points tangent to the circle.

Average versus instantaneous values

Average values describe motion over an interval, while instantaneous values describe motion at a single moment. Calculating average velocity or average acceleration over a very small time interval gives a value very close to the instantaneous value at that instant.

For a ball thrown straight up, the instantaneous velocity at its highest point is zero, even though its average velocity over the whole trip is not.

Solving Displacement and Velocity Problems Without Sign Errors

Problem Solving:

• Identify initial and final states first. Most calculations here come down to a final value minus an initial value over a time interval.
• Keep direction attached to your answer. Velocity, acceleration, and displacement are vectors, so include a sign or compass direction.
• Track units carefully: m/s for velocity, m/s² for acceleration.

Common Trap:

• Do not mix up distance and displacement. A round trip can have a large distance but zero displacement.
• Slowing down does not always mean negative acceleration. The sign depends on direction relative to your chosen positive axis, not on whether the object speeds up or slows down.

Practice Problem 1: Displacement Calculation

To solve this problem, we need to find the net displacement vector from the starting point to the final position.

First, let's identify the individual displacements:

• 3 km east: +3 km in x-direction
• 4 km north: +4 km in y-direction
• 1 km west: -1 km in x-direction

Net displacement in x-direction: 3 km - 1 km = 2 km east Net displacement in y-direction: 4 km north

Using the Pythagorean theorem to find the magnitude of the displacement: $|\Delta \vec{r}| = \sqrt{(2 \text{ km})^2 + (4 \text{ km})^2} = \sqrt{4 \text{ km}^2 + 16 \text{ km}^2} = \sqrt{20 \text{ km}^2} = 4.47 \text{ km}$

The direction can be found using trigonometry: $\theta = \tan^{-1}(\frac{4 \text{ km}}{2 \text{ km}}) = \tan^{-1}(2) = 63.4° \text{ north of east}$

Therefore, the displacement is 4.47 km at 63.4° north of east.

The total distance traveled is simply the sum of the distances for each segment: $\text{Total distance} = 3 \text{ km} + 4 \text{ km} + 1 \text{ km} = 8 \text{ km}$

The difference between displacement (4.47 km) and total distance (8 km) illustrates that displacement is the straight-line distance from start to finish, while total distance includes all path segments regardless of direction.

Practice Problem 2: Average Velocity and Acceleration

(a) Displacement:

$\Delta x = x - x_0 = 65 \text{ m} - 5 \text{ m} = 60 \text{ m east}$

(b) Average velocity:

$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{60 \text{ m}}{12 \text{ s}} = 5.0 \text{ m/s east}$

(c) Average acceleration:

$a_{avg} = \frac{\Delta v}{\Delta t} = \frac{10 \text{ m/s} - 4 \text{ m/s}}{3 \text{ s}} = \frac{6 \text{ m/s}}{3 \text{ s}} = 2.0 \text{ m/s}^2 \text{ east}$

Practice Problem 3: Instantaneous Values

Because the average velocity over very small time intervals is close to the instantaneous velocity, these values suggest the runner's instantaneous velocity at $t = 2.0$ s is about 6.0 m/s east. As we shrink the time interval closer and closer to zero, the average velocity gets closer to the true instantaneous value, and the trend here points toward approximately 6.0 m/s east.

Common Misconceptions

• Misconception: Distance and displacement are the same.
  Actually: Distance and displacement are not the same. Distance is the total path length and is always positive, while displacement is a vector that can be positive, negative, or zero.
• Misconception: A negative acceleration always means slowing down.
  Actually: A negative acceleration does not always mean slowing down. It means acceleration points in the negative direction. Whether the object speeds up or slows down depends on the direction of its velocity.
• Misconception: Constant speed means zero acceleration.
  Actually: Constant speed does not mean zero acceleration. If direction changes, the object is accelerating even though its speed stays the same.
• Misconception: Average velocity is the average of two speeds.
  Actually: Average velocity is not the average of two speeds. It is total displacement divided by total time, which can give a different result.
• Misconception: Instantaneous velocity is just a smaller average.
  Actually: Instantaneous velocity is not just a smaller average. The average over a tiny interval approximates it, but the instantaneous value is the exact rate of change at one moment.

Related AP Physics 1 Guides

• 1.1 Scalars and Vectors in One Dimension
• 1.3 Representing Motion
• 1.4 Reference Frames and Relative Motion
• 1.5 Vectors and Motion in Two Dimensions