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Position-time, velocity-time, and acceleration-time graphs
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Graphs are powerful tools for visualizing and analyzing motion. The three main types are:
A position-time graph is a straight line passing through m and m. What is the velocity?
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The slope of a position-time graph gives the velocity.
Key features:
The slope of a velocity-time graph gives the acceleration.
Key features:
The area under a velocity-time graph gives the displacement.
For constant velocity (rectangle):
For constant acceleration (trapezoid or triangle):
Sign matters:
The area under an acceleration-time graph gives the change in velocity.
For constant acceleration:
| If you have... | To get... | Operation |
|---|---|---|
| Position-time | Velocity-time | Take slope (derivative) |
| Velocity-time | Acceleration-time | Take slope (derivative) |
| Velocity-time | Position-time | Find area (integral) |
| Acceleration-time | Velocity-time | Find area (integral) |
Speeding up: velocity and acceleration have the same sign
Slowing down: velocity and acceleration have opposite signs
Given information:
Find: Velocity
Solution: The velocity is the slope of the position-time graph.
Answer: The velocity is 2 m/s (constant).
Interpretation: The object moves in the positive direction at a steady 2 m/s. Since the slope is constant (straight line), there is no acceleration.
A velocity vs. time graph shows a straight line starting at v = 10 m/s at t = 0 and ending at v = 30 m/s at t = 4 s. (a) What is the object's acceleration? (b) What is the displacement during this time interval? (c) Describe what the area under the v-t curve represents.
Solution:
(a) Acceleration from v-t graph: Acceleration = slope of v-t graph a = Δv/Δt = (30 - 10)/(4 - 0) = 20/4 = 5.0 m/s²
(b) Displacement: Displacement = area under v-t curve The area is a trapezoid with parallel sides 10 m/s and 30 m/s, height 4 s.
Area = ½(v₁ + v₂)t = ½(10 + 30)(4) = ½(40)(4) = 80 m
(c) Meaning of area under v-t curve: The area under a velocity-time graph represents the displacement of the object. For motion in one direction, this equals the distance traveled.
A velocity-time graph shows a horizontal line at m/s from to s. Find the displacement and acceleration.
Given information:
Find:
Part 1: Displacement Displacement = area under velocity-time graph
An object's position is given by the equation x(t) = 2t³ - 6t² + 4t + 5, where x is in meters and t is in seconds. (a) Find the velocity function v(t). (b) Find the acceleration function a(t). (c) At what time(s) is the object momentarily at rest?
Solution:
(a) Velocity function: v(t) = dx/dt = d/dt(2t³ - 6t² + 4t + 5) v(t) = 6t² - 12t + 4 m/s
(b) Acceleration function: a(t) = dv/dt = d/dt(6t² - 12t + 4) a(t) = 12t - 12 m/s²
(c) When object is at rest: Object at rest when v(t) = 0 6t² - 12t + 4 = 0 Divide by 2: 3t² - 6t + 2 = 0
Using quadratic formula: t = [6 ± √(36-24)]/6 = [6 ± √12]/6 = [6 ± 2√3]/6
t = (6 + 2√3)/6 ≈ 1.58 s and t = (6 - 2√3)/6 ≈ 0.42 s
The object is momentarily at rest at t ≈ 0.42 s and t ≈ 1.58 s.
A velocity-time graph shows a straight line from to m/s. Find: (a) the acceleration, (b) the displacement during this time.
Given information:
Part (a): Find acceleration Acceleration = slope of velocity-time graph
Part (b): Find displacement Displacement = area under velocity-time graph
The graph forms a triangle with:
Alternative method for (b): Use kinematic equation:
Answers:
Check: Using : ✓
Since velocity is constant, the area is a rectangle:
Part 2: Acceleration Acceleration = slope of velocity-time graph
Since the line is horizontal:
Answers:
Key insight: Horizontal line on - graph means constant velocity (zero acceleration).