1.2 Displacement, Velocity, and Acceleration

Displacement, velocity, and acceleration describe how an object's position changes over time. In AP Physics C: Mechanics, the key calculus connection is that velocity is the derivative of position, acceleration is the derivative of velocity, and displacement can be found by integrating velocity.

Start by treating the object as a point particle when its size and shape do not matter. Then distinguish average quantities over a time interval from instantaneous quantities at one moment.

Change in Object Position

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Object Model Simplification

In physics, we often need to analyze complex objects with many moving parts. To make this analysis manageable, we use a simplified model.

The point particle model treats complex objects as single points with properties like mass and charge, but without physical dimensions.

• This simplification ignores the object's size, shape, and internal structure
• The model places all the object's mass concentrated at a single point in space
• For rotational motion, we may need to consider the object's dimensions, but for linear motion, the point model is usually sufficient
• This simplification works well when an object's size is much smaller than the distance it travels

Displacement Definition

Displacement measures how far out of place an object is from its starting position. Unlike distance, which measures the total path length, displacement only considers the straight-line difference between start and end points.

• Represented mathematically as: $\Delta x = x - x_0$ where $x$ is the final position and $x_0$ is the initial position
• Displacement is a vector quantity with both magnitude and direction
• An object that returns to its starting point has zero displacement, regardless of how far it traveled
• In multiple dimensions, displacement can be calculated using vector components: $\Delta \vec{r} = (x - x_0)\hat{i} + (y - y_0)\hat{j} + (z - z_0)\hat{k}$

Average Velocity and Acceleration

Calculation of Averages

When analyzing motion over a time interval, we often need to calculate average values rather than tracking every instantaneous change. Average calculations provide a simplified overview of an object's behavior during a specific time period.

Average values consider only the net change between the beginning and end of the interval, effectively smoothing out any variations that occurred during that time. This approach is particularly useful when detailed moment-by-moment analysis isn't necessary or when we want to characterize overall motion trends.

Average Velocity Formula

Average velocity represents how quickly an object's position changes over a given time interval. It provides the rate of displacement, not the rate of distance traveled.

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

• The direction of average velocity matches the direction of displacement
• Average velocity can be zero even when an object is moving (if it returns to its starting point)
• Units are typically meters per second (m/s)
• Average velocity differs from average speed, which is the total distance divided by time

Average Acceleration Formula

Average acceleration describes how an object's velocity changes over time. It indicates both how quickly the speed changes and how the direction of motion changes.

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

• Calculated by dividing the change in velocity by the time interval
• Represents the rate of velocity change in a particular direction
• Units are typically meters per second squared (m/s²)
• Can be positive (speeding up) or negative (slowing down) relative to the direction of motion

Acceleration Conditions

Acceleration occurs whenever there is a change in an object's velocity. This change can involve the speed, direction, or both.

• An object accelerates when it speeds up or slows down (change in speed)
• An object accelerates when it changes direction, even at constant speed
• In circular motion at constant speed, acceleration is always present because the direction constantly changes
• The acceleration vector points in the direction of the velocity change, not necessarily in the direction of motion

Instantaneous Kinematics

Limit of Average Values

Instantaneous values tell us exactly what's happening at a specific moment in time, rather than over an interval. These values are found by taking the limit of average values as the time interval approaches zero.

• Instantaneous velocity: $\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{x}}{\Delta t} = \frac{d\vec{r}}{dt}$
• Instantaneous acceleration: $\vec{a} = \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t} = \frac{d\vec{v}}{dt}$
• These limits lead to derivatives in calculus, connecting position, velocity, and acceleration
• For component-wise analysis: $v_x = \frac{dx}{dt}$ and $a_x = \frac{dv_x}{dt}$

Time-Dependent Functions

In many physics problems, we can express position, velocity, and acceleration as functions of time. These functions allow us to determine the state of motion at any instant.

• Position functions $x(t)$ describe where an object is at time $t$
• Velocity functions $v(t) = \frac{dx}{dt}$ describe how fast and in what direction an object is moving at time $t$
• Acceleration functions $a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$ describe how the velocity is changing at time $t$
• We can find displacement by integrating velocity: $\Delta x = \int_{t_1}^{t_2} v(t) dt$
• We can find velocity change by integrating acceleration: $\Delta v = \int_{t_1}^{t_2} a(t) dt$

Practice Problem 1: Displacement Calculation

Solution

To solve this problem, we need to find the net displacement vector by adding all individual displacements.

Step 1: Break down the displacements into components.

• East: 50 m in the positive x-direction: (50 m, 0 m)
• North: 30 m in the positive y-direction: (0 m, 30 m)
• West: 20 m in the negative x-direction: (-20 m, 0 m)

Step 2: Add the x-components and y-components separately.

• Net x-displacement: 50 m + 0 m + (-20 m) = 30 m
• Net y-displacement: 0 m + 30 m + 0 m = 30 m

Step 3: Find the magnitude and direction of the resultant displacement.

• Magnitude: $|\vec{d}| = \sqrt{(30 \text{ m})^2 + (30 \text{ m})^2} = \sqrt{1800 \text{ m}^2} = 42.4 \text{ m}$
• Direction: $\theta = \tan^{-1}\left(\frac{30 \text{ m}}{30 \text{ m}}\right) = 45°$ (northeast)

The car's total displacement is 42.4 meters at 45° northeast from its starting point.

Practice Problem 2: Average and Instantaneous Velocity

Solution

(a) To find the average velocity between $t = 2$ s and $t = 3$ s:

Step 1: Calculate the positions at the given times.

• At $t = 2$ s: $x(2) = 3(2)^2 - 2(2) + 5 = 3(4) - 4 + 5 = 12 - 4 + 5 = 13$ m
• At $t = 3$ s: $x(3) = 3(3)^2 - 2(3) + 5 = 3(9) - 6 + 5 = 27 - 6 + 5 = 26$ m

Step 2: Calculate the displacement.

• $\Delta x = x(3) - x(2) = 26 \text{ m} - 13 \text{ m} = 13 \text{ m}$

Step 3: Calculate the time interval.

• $\Delta t = 3 \text{ s} - 2 \text{ s} = 1 \text{ s}$

Step 4: Calculate the average velocity.

• $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{13 \text{ m}}{1 \text{ s}} = 13 \text{ m/s}$

(b) To find the instantaneous velocity at $t = 2$ s:

Step 1: Find the velocity function by differentiating the position function.

• $v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 2t + 5) = 6t - 2$

Step 2: Substitute $t = 2$ s into the velocity function.

• $v(2) = 6(2) - 2 = 12 - 2 = 10 \text{ m/s}$

Therefore, the average velocity between $t = 2$ s and $t = 3$ s is 13 m/s, and the instantaneous velocity at $t = 2$ s is 10 m/s.