4.1 Kinematics: Speed, Velocity, Acceleration & Graphs
Key Takeaways
- Scalar quantities like distance and speed describe magnitude only, whereas vector quantities like displacement, velocity, and acceleration describe both magnitude and direction.
- Speed measures the rate of total distance covered over time ($v = d/t$), while velocity measures the rate of position change or displacement over time ($\vec{v} = \Delta \vec{x}/t$).
- Acceleration ($a = (v_f - v_i)/t$) represents the rate of change of velocity; zero acceleration indicates constant velocity, not necessarily a state of rest.
- On a position-time ($d$-$t$) graph, the slope represents velocity, whereas on a velocity-time ($v$-$t$) graph, the slope represents acceleration and the area beneath the curve represents displacement.
4.1 Kinematics: Speed, Velocity, Acceleration & Graphs
Kinematics is the subfield of physics focused on describing how objects move. On the GED Science test, kinematics questions test your ability to calculate motion parameters using formulas, distinguish between scalar and vector quantities, and extract information from motion graphs.
Scalar vs. Vector Quantities
Every physical measurement in kinematics falls into one of two fundamental categories: scalars or vectors.
| Feature | Scalar Quantity | Vector Quantity |
|---|---|---|
| Definition | Measured by magnitude (numerical value and unit) only | Measured by both magnitude AND direction |
| Direction Matter? | No | Yes |
| Distance / Displacement | Distance ($d$): Total path length traveled (e.g., $50\text{ meters}$) | Displacement ($\Delta \vec{x}$): Straight-line change in position from start to finish (e.g., $10\text{ meters North}$) |
| Speed / Velocity | Speed ($v$): Rate of distance covered ($v = d/t$) | Velocity ($\vec{v}$): Speed in a specified direction ($\vec{v} = \Delta \vec{x}/t$) |
| Acceleration | N/A | Acceleration ($\vec{a}$): Rate of change of velocity over time |
The Distance vs. Displacement Trap
Imagine a track athlete running exactly one full $400\text{-meter}$ lap around a circular track and returning to the exact starting line in $50\text{ seconds}$:
- Distance covered ($d$): $400\text{ meters}$.
- Displacement ($\Delta \vec{x}$): $0\text{ meters}$ (since the final position equals the initial position).
- Average Speed: $v = \frac{d}{t} = \frac{400\text{ m}}{50\text{ s}} = 8\text{ m/s}$.
- Average Velocity: $\vec{v} = \frac{\Delta \vec{x}}{t} = \frac{0\text{ m}}{50\text{ s}} = 0\text{ m/s}$.
GED Tip: If an object returns to its exact starting point, its overall displacement and average velocity are zero, even if its speed and distance are non-zero!
Core Motion Formulas & Calculations
To solve kinematics calculation items on the GED exam, you must be comfortable manipulating three core mathematical relationships.
1. Speed Formula
- Units: Meters per second ($\text{m/s}$), kilometers per hour ($\text{km/h}$), or miles per hour ($\text{mph}$).
- Algebraic Rearrangements: $d = v \cdot t$ and $t = \frac{d}{v}$.
2. Velocity Formula
- Units: $\text{m/s}$ along with a direction (e.g., $+5\text{ m/s}$ or $5\text{ m/s East}$). Positive and negative signs frequently denote opposite directions (e.g., $+v$ for right/up, $-v$ for left/down).
3. Acceleration Formula
- Units: Meters per second squared ($\text{m/s}^2$).
- Understanding $\text{m/s}^2$: An acceleration of $3\text{ m/s}^2$ means that the object's velocity increases by $3\text{ m/s}$ every single second.
- Deceleration / Negative Acceleration: If an object slows down, its final velocity $v_f$ is less than its initial velocity $v_i$, yielding a negative acceleration value ($a < 0$).
Worked Examples
Worked Example 1: Calculating Average Speed
Problem: A delivery truck travels $120\text{ kilometers}$ during its first $2\text{ hours}$ of travel, stops for $0.5\text{ hours}$ for lunch, and then travels an additional $90\text{ kilometers}$ in $1.5\text{ hours}$. What is the average speed of the truck for the entire journey?
Step-by-Step Solution:
- Identify total distance ($d_{\text{total}}$):
- Identify total elapsed time ($t_{\text{total}}$):
- Apply average speed formula:
Worked Example 2: Calculating Acceleration
Problem: A sports car accelerates along a straight test track from an initial velocity of $10\text{ m/s}$ to a final velocity of $35\text{ m/s}$ in a time span of $5\text{ seconds}$. Calculate the car's average acceleration.
Step-by-Step Solution:
- Identify given variables: $v_i = 10\text{ m/s}$, $v_f = 35\text{ m/s}$, $t = 5\text{ s}$.
- Apply acceleration formula:
Interpreting Motion Graphs
The GED Science test relies heavily on line graphs to evaluate your understanding of kinematics. The most critical step in graph analysis is reading the Y-axis label first to identify whether you are looking at a Position-Time ($d$-$t$) graph or a Velocity-Time ($v$-$t$) graph.
1. Position-Time ($d$-$t$) Graphs
On a position-time graph, the slope of the line represents velocity ($\text{Slope} = \frac{\Delta y}{\Delta x} = \frac{\text{Position}}{\text{Time}} = v$).
- Horizontal Flat Line ($\text{Slope} = 0$): Position is unchanged over time. The object is stationary / at rest ($v = 0\text{ m/s}$).
- Straight Sloped Line (Constant Non-Zero Slope): Position changes at a uniform rate. The object moves with constant velocity (zero acceleration).
- Positive Slope: Object moves forward / away from origin.
- Negative Slope: Object moves backward / toward origin.
- Curved Line (Changing Slope): Velocity is changing over time. The object is accelerating.
- Curve steepening upwards: Speeding up.
- Curve flattening out: Slowing down.
2. Velocity-Time ($v$-$t$) Graphs
On a velocity-time graph, the slope of the line represents acceleration ($\text{Slope} = \frac{\Delta y}{\Delta x} = \frac{\text{Velocity}}{\text{Time}} = a$). The area under the line represents the total displacement.
- Horizontal Flat Line at $y > 0$ ($\text{Slope} = 0$): Velocity is constant. Acceleration is zero ($a = 0\text{ m/s}^2$).
- Horizontal Line at $y = 0$: Velocity is zero. The object is at rest.
- Straight Sloped Line Upward ($\text{Positive Slope}$): Velocity is increasing at a steady rate. Constant positive acceleration.
- Straight Sloped Line Downward ($\text{Negative Slope}$): Velocity is decreasing at a steady rate. Constant negative acceleration (deceleration).
- Line Crossing the X-Axis ($y = 0$): The object changes its direction of motion (velocity changes sign from positive to negative or vice versa).
Graphical Comparison Summary
| Motion State | Position-Time ($d$-$t$) Graph | Velocity-Time ($v$-$t$) Graph |
|---|---|---|
| At Rest | Horizontal line at current position | Horizontal line directly on the X-axis ($v=0$) |
| Constant Velocity | Straight diagonal line (constant slope) | Horizontal flat line above X-axis |
| Constant Acceleration | Curved parabolic line | Straight diagonal line (constant slope) |
Common GED Distractors & Exam Pitfalls
- Confusing Graph Axes: A horizontal line on a $d$-$t$ graph means at rest, but a horizontal line on a $v$-$t$ graph means moving at constant speed. Always check the Y-axis label before answering graph questions.
- Confusing Negative Acceleration with Moving Backward: On a velocity-time graph, a negative slope means slowing down (if velocity is positive). Moving backward occurs only when the line drops below the X-axis into negative velocity values.
- Unit Mismatches: Ensure time units match before calculating (e.g., converting minutes to seconds if speed is given in $\text{m/s}$).
A cyclist travels along a straight road covering 15 kilometers in 30 minutes, stops for 15 minutes to repair a flat tire, and then completes the remaining 15 kilometers in 45 minutes. What is the cyclist's average speed for the entire 90-minute period?
Examining a velocity-time graph of a subway train reveals a straight diagonal line sloping downward from a velocity of 24 m/s at t = 0 s to a velocity of 0 m/s at t = 8 s. What is the acceleration of the train during this interval?
A runner starts at a marked position on a straight track, runs 100 meters East in 12 seconds, turns directly around, and walks 40 meters West toward the start line in 8 seconds. What is the magnitude of the runner's average velocity for the entire 20-second interval?