Prerequisites
Before studying this topic, make sure you understand:
- Motion in a Straight Line - basic definitions
- Kinematic Equations - for uniform acceleration
Why Graphical Analysis?
Graphs provide a visual representation of motion that helps us:
- Understand the nature of motion at a glance
- Extract quantities like velocity and acceleration
- Calculate displacement from the area under curves
- Solve problems where algebraic methods are complex
Position-Time (x-t) Graph
What the Graph Tells Us
| Feature | Physical Meaning |
|---|---|
| Slope | Velocity |
| Shape | Type of motion |
| y-intercept | Initial position |
Common x-t Graph Shapes
1. Horizontal Line (slope = 0)
x │ ___________
│
└─────────────────► t
- Motion: Object at rest
- Velocity: $v = 0$
2. Straight Line with Positive Slope
x │ /
│ /
│ /
└─────────────────► t
- Motion: Uniform velocity (constant speed)
- Velocity: $v = \text{constant} > 0$
3. Parabola Opening Upward
x │ /
│ /
│ __/
└─────────────────► t
- Motion: Uniformly accelerated ($a > 0$)
- Velocity: Increasing
4. Parabola Opening Downward
x │ ___
│ / \
│ / \
└─────────────────► t
- Motion: First accelerating up, then decelerating
- Velocity: First increasing, then decreasing
Slope Interpretation
At any point on an x-t graph:
$$v = \frac{dx}{dt} = \text{slope of tangent}$$- Steeper slope = higher velocity
- Horizontal tangent = momentarily at rest
Interactive Demo: Visualize Velocity-Time Graphs
Explore the relationship between position, velocity, and time graphically.
Velocity-Time (v-t) Graph
What the Graph Tells Us
| Feature | Physical Meaning |
|---|---|
| Slope | Acceleration |
| Area under curve | Displacement |
| y-value | Instantaneous velocity |
Common v-t Graph Shapes
1. Horizontal Line
v │ ___________
│
└─────────────────► t
- Motion: Uniform velocity
- Acceleration: $a = 0$
2. Line with Positive Slope
v │ /
│ /
│ /
└─────────────────► t
- Motion: Uniformly accelerated
- Acceleration: $a = \text{constant} > 0$
3. Line with Negative Slope (above x-axis)
v │ \
│ \
│ \
└─────────────────► t
- Motion: Uniformly decelerating
- Acceleration: $a = \text{constant} < 0$ (retardation)
Calculating Displacement from v-t Graph
$$\text{Displacement} = \int_{t_1}^{t_2} v \, dt = \text{Area under v-t curve}$$
Area Sign Convention
- Area above time axis = positive displacement
- Area below time axis = negative displacement
- Total displacement = algebraic sum of areas
Example: Area Calculation
For a v-t graph showing:
- Triangle above axis (base 4s, height 20 m/s)
- Triangle below axis (base 2s, depth 10 m/s)
Displacement:
- Area above = $\frac{1}{2} \times 4 \times 20 = 40$ m
- Area below = $\frac{1}{2} \times 2 \times 10 = 10$ m (negative)
- Net displacement = 40 - 10 = 30 m
Distance: 40 + 10 = 50 m
Acceleration-Time (a-t) Graph
| Feature | Physical Meaning |
|---|---|
| Area under curve | Change in velocity |
| y-value | Instantaneous acceleration |
Converting Between Graphs
From x-t to v-t
- Find slope at each point on x-t graph
- Plot slope values as v on v-t graph
From v-t to x-t
- Find area under v-t curve up to each time
- Plot cumulative area as x on x-t graph
From v-t to a-t
- Find slope at each point on v-t graph
- Plot slope values as a on a-t graph
Summary Table
| Graph | Slope Gives | Area Gives |
|---|---|---|
| x-t | Velocity | — |
| v-t | Acceleration | Displacement |
| a-t | — | Change in velocity |
Practice Problems
Problem 1
A v-t graph is a straight line from (0, 10) to (5, 30). Find: (a) Acceleration (b) Displacement in 5 seconds
Solution: (a) $a = \frac{30 - 10}{5 - 0} = $ 4 m/s²
(b) Area = Trapezium area = $\frac{1}{2}(10 + 30) \times 5 = $ 100 m
Problem 2
An x-t graph is a parabola: $x = 2t^2 + 3t$ (x in m, t in s). Find velocity and acceleration at t = 2s.
Solution: $v = \frac{dx}{dt} = 4t + 3$ At t = 2s: $v = 4(2) + 3 = $ 11 m/s
$a = \frac{dv}{dt} = 4$ m/s² (constant)