Motion Graphs

Position-time, velocity-time, and acceleration-time graphs

Motion Graphs

Introduction

Graphs are powerful tools for visualizing and analyzing motion. The three main types are:

Position-time graphs ( $x$ vs $t$ )
Velocity-time graphs ( $v$ vs $t$ )
Acceleration-time graphs ( $a$ vs $t$ )

Position-Time Graphs

Interpreting Slope

The slope of a position-time graph gives the velocity.

$\text{slope} = \frac{\Delta x}{\Delta t} = v$

Key features:

Horizontal line → zero velocity (at rest)
Positive slope → positive velocity (moving forward)
Negative slope → negative velocity (moving backward)
Steeper slope → greater speed
Curved line → changing velocity (acceleration)

Interpreting Curvature

Straight line: constant velocity ( $a = 0$ )
Curve upward: increasing velocity (positive acceleration)
Curve downward: decreasing velocity (negative acceleration)

Reading the Graph

y-value: position at that time
Slope: velocity at that time
Change in slope: acceleration

Velocity-Time Graphs

Interpreting Slope

The slope of a velocity-time graph gives the acceleration.

$\text{slope} = \frac{\Delta v}{\Delta t} = a$

Key features:

Horizontal line → zero acceleration (constant velocity)
Positive slope → positive acceleration
Negative slope → negative acceleration (deceleration)
Steeper slope → greater acceleration

Interpreting Area

The area under a velocity-time graph gives the displacement.

$\text{area} = \int v \, dt = \Delta x$

For constant velocity (rectangle): $\Delta x = v \cdot t$

For constant acceleration (trapezoid or triangle): $\Delta x = \frac{1}{2}(v_0 + v)t$

Sign matters:

Area above time axis → positive displacement
Area below time axis → negative displacement

Reading the Graph

y-value: velocity at that time
Slope: acceleration at that time
Area under curve: displacement

Acceleration-Time Graphs

Interpreting Area

The area under an acceleration-time graph gives the change in velocity.

$\text{area} = \int a \, dt = \Delta v$

For constant acceleration: $\Delta v = a \cdot t$

Reading the Graph

y-value: acceleration at that time
Area under curve: change in velocity
Horizontal line: constant acceleration

Relationships Between Graphs

| 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) |

Common Motion Patterns

Constant Velocity

Position-time: straight line (slope = velocity)
Velocity-time: horizontal line
Acceleration-time: zero (horizontal at $a = 0$ )

Constant Acceleration

Position-time: parabola (curved)
Velocity-time: straight line (slope = acceleration)
Acceleration-time: horizontal line

Speeding Up vs. Slowing Down

Speeding up: velocity and acceleration have the same sign

Moving right and accelerating right: $v > 0$ , $a > 0$
Moving left and accelerating left: $v < 0$ , $a < 0$

Slowing down: velocity and acceleration have opposite signs

Moving right and accelerating left: $v > 0$ , $a < 0$
Moving left and accelerating right: $v < 0$ , $a > 0$

Problem-Solving Tips

Identify the type of graph (position, velocity, or acceleration)
Look at the y-value for the quantity itself
Calculate slope to find the derivative quantity
Calculate area to find the integral quantity
Check signs carefully (positive/negative)
Draw the graph if only given data

📚 Practice Problems

1Problem 1easy

❓ Question:

A position-time graph is a straight line passing through $(0, 5)$ m and $(10, 25)$ m. What is the velocity?

💡 Show Solution

Given information:

Position at $t = 0$ s: $x_0 = 5$ m
Position at $t = 10$ s: $x = 25$ m
Graph is a straight line (constant velocity)

Find: Velocity

Solution: The velocity is the slope of the position-time graph.

$v = \frac{\Delta x}{\Delta t} = \frac{x - x_0}{t - t_0}$

$v = \frac{25 - 5}{10 - 0} = \frac{20}{10} = 2 \text{ m/s}$

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.

2Problem 2medium

❓ Question:

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.

💡 Show Solution

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.

3Problem 3medium

❓ Question:

💡 Show Solution

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.

4Problem 4hard

❓ Question:

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?

💡 Show Solution

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

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.

5Problem 5easy

❓ Question:

A velocity-time graph shows a horizontal line at $v = 15$ m/s from $t = 0$ to $t = 8$ s. Find the displacement and acceleration.

💡 Show Solution

Given information:

Velocity: $v = 15$ m/s (constant)
Time interval: $0$ to $8$ s
Graph is horizontal line

Find:

Displacement
Acceleration

Part 1: Displacement Displacement = area under velocity-time graph

Since velocity is constant, the area is a rectangle: $\text{Area} = \text{base} \times \text{height}$ $\Delta x = t \times v = 8 \times 15 = 120 \text{ m}$

Part 2: Acceleration Acceleration = slope of velocity-time graph

Since the line is horizontal: $a = \frac{\Delta v}{\Delta t} = \frac{0}{8} = 0 \text{ m/s}^2$

Answers:

Displacement: 120 m
Acceleration: 0 m/s² (moving at constant velocity)

Key insight: Horizontal line on $v$ - $t$ graph means constant velocity (zero acceleration).

6Problem 6hard

❓ Question:

💡 Show Solution

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

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.

7Problem 7medium

❓ Question:

A velocity-time graph shows a straight line from $(0, 0)$ to $(5, 20)$ m/s. Find: (a) the acceleration, (b) the displacement during this time.

💡 Show Solution

Given information:

Initial velocity at $t = 0$ : $v_0 = 0$ m/s
Final velocity at $t = 5$ s: $v = 20$ m/s
Graph is a straight line (constant acceleration)

Part (a): Find acceleration Acceleration = slope of velocity-time graph

$a = \frac{\Delta v}{\Delta t} = \frac{v - v_0}{t - 0}$

$a = \frac{20 - 0}{5 - 0} = \frac{20}{5} = 4 \text{ m/s}^2$

Part (b): Find displacement Displacement = area under velocity-time graph

The graph forms a triangle with:

Base = $5$ s
Height = $20$ m/s

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

$\Delta x = \frac{1}{2} \times 5 \times 20 = \frac{1}{2} \times 100 = 50 \text{ m}$

Alternative method for (b): Use kinematic equation: $\Delta x = \frac{1}{2}(v_0 + v)t$

$\Delta x = \frac{1}{2}(0 + 20)(5) = \frac{1}{2}(100) = 50 \text{ m}$

Answers:

(a) Acceleration: 4 m/s²
(b) Displacement: 50 m

Check: Using $v = v_0 + at$ : $20 = 0 + 4(5) = 20$ ✓

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