Understanding Motion

A High School Guide to Speed and Acceleration

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Position, Distance, and Displacement

To talk about motion, we first need to agree on where things are.

Key Terms

Position ($x$): An object's specific location, like a point on a map.

Distance: The total length of the path you travel. It's a "scalar" - just a number.

Displacement ($\Delta x$): The straight-line change in position. It's a "vector" - it has direction!

A (Start) B (End) Displacement Distance

Speed vs. Velocity

One tells you "how fast," the other tells you "how fast and where."

Speed (Scalar)

How fast an object is moving.

\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]

Velocity (Vector)

The rate of change of position.

\[ \vec{v} = \frac{\Delta \vec{x}}{\Delta t} \]

Race Track Example

A car completes one lap on a 1-mile track. Its average speed might be 120 mph, but its average velocity is 0 mph because it ended up where it started (displacement = 0)!

Acceleration: The Rate of Change

"Putting the pedal to the metal" is just one type of acceleration.

\[ \vec{a} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_{final} - \vec{v}_{initial}}{t} \]

Speeding Up

Positive acceleration.

Slowing Down

Negative acceleration (deceleration).

Changing Direction

Even at constant speed!

The Kinematic Equations

Your toolkit for solving problems with constant acceleration.

\(v = v_0 + at\)
\(\Delta x = v_0 t + \frac{1}{2}at^2\)
\(v^2 = v_0^2 + 2a\Delta x\)
\(\Delta x = \frac{1}{2}(v_0 + v)t\)
\(\Delta x = vt - \frac{1}{2}at^2\)
Where: $v_0$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $t$ = time, $\Delta x$ = displacement

Using the Formulas: An Example

A car, starting from rest, accelerates at a constant 2 m/s² for 5 seconds.

1. What is its final velocity?

Known: \(v_0=0, a=2, t=5\). Find: \(v\).

\[v = v_0 + at \rightarrow v = 0 + (2)(5) = 10 \text{ m/s}\]

2. How far did it travel?

Known: \(v_0=0, a=2, t=5\). Find: \(\Delta x\).

\[\Delta x = v_0 t + \frac{1}{2}at^2 \rightarrow \Delta x = (0)(5) + \frac{1}{2}(2)(5)^2 = 25 \text{ m}\]

Graphs of Motion

A picture is worth a thousand numbers.

Position vs. Time

Position Time Slope = Velocity No Motion Curve = Acceleration

Velocity vs. Time

Velocity Time Slope = Acceleration Area = Displacement

Problem Set: Group 1

Problem 1.1

A sprinter runs 100 meters in 10 seconds. What is her average speed?

Problem 1.2

A bike starting from rest accelerates at 1 m/s² for 4 seconds. What is its final velocity?

Solutions: Group 1

Problem 1.1: Sprinter's Speed

Formula: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)

Calculation: \( \text{Speed} = \frac{100 \text{ m}}{10 \text{ s}} \)

Answer: 10 m/s

Problem 1.2: Bike's Velocity

Given: \(v_0 = 0\) (from rest), \(a = 1\) m/s², \(t = 4\) s

Formula: \(v = v_0 + at\)

Calculation: \(v = 0 + (1 \text{ m/s²})(4 \text{ s})\)

Answer: 4 m/s

Problem Set: Group 2

Problem 2.1

A train traveling at 30 m/s applies its brakes and slows to 10 m/s in 5 seconds. What is its acceleration?

Problem 2.2

A ball is dropped from a tall building. How far does it fall in 3 seconds? (Use g = 9.8 m/s² and ignore air resistance).

Solutions: Group 2

Problem 2.1: Train's Acceleration

Given: \(v_0 = 30\) m/s, \(v = 10\) m/s, \(t = 5\) s

Formula: \(a = (v - v_0) / t\)

Calculation: \(a = (10 \text{ m/s} - 30 \text{ m/s}) / 5 \text{ s}\)

Answer: -4 m/s² (deceleration)

Problem 2.2: Falling Ball

Given: \(v_0 = 0\) (dropped), \(a = g = 9.8\) m/s², \(t = 3\) s

Formula: \(\Delta x = v_0 t + \frac{1}{2}at^2\)

Calculation: \(\Delta x = (0)(3) + \frac{1}{2}(9.8)(3)^2 \)

Answer: 44.1 meters

Problem Set: Group 3

Problem 3.1

A car accelerates from rest at 4 m/s². How much distance does it cover before it reaches a speed of 20 m/s?

Problem 3.2

A stone is thrown upwards with an initial velocity of 19.6 m/s. What is its maximum height? (Use g = 9.8 m/s²).

Solutions: Group 3

Problem 3.1: Car's Distance

Given: \(v_0 = 0\), \(a = 4\) m/s², \(v = 20\) m/s

Formula: \(v^2 = v_0^2 + 2a\Delta x\)

Calculation: \((20)^2 = (0)^2 + 2(4)\Delta x \rightarrow 400 = 8\Delta x\)

Answer: \(\Delta x = 50\) meters

Problem 3.2: Stone's Maximum Height

Given: \(v_0 = 19.6\) m/s, \(v = 0\) (at max height), \(a = -g = -9.8\) m/s²

Formula: \(v^2 = v_0^2 + 2a\Delta x\)

Calculation: \((0)^2 = (19.6)^2 + 2(-9.8)\Delta x \)

Answer: \(\Delta x = 19.6\) meters

Key Concepts in Review

Remember these key visual connections!

Position vs. Time Graph

The slope of this graph tells you the object's velocity.

Position Time Slope = Velocity Curve = Acceleration

Velocity vs. Time Graph

Here, the slope is acceleration, and the area is displacement.

Velocity Time Slope = Acceleration Area = Displacement

Real-World Examples

Let's see this in action!

Braking Car

Stopping Distance

A car at 15 m/s brakes with an acceleration of -5 m/s². How far until it stops?

  • Use: \(v^2=v_0^2+2a\Delta x\)
  • Calculation: \(0^2 = 15^2 + 2(-5)\Delta x\)

Answer: \(\Delta x = 22.5\) m

Dropped Phone

Height = 20m

You drop your phone from a 20m balcony. How long until it hits the ground? (g=9.8 m/s²)

  • Use: \(\Delta x=v_0t+\frac{1}{2}at^2\)
  • Calculation: \(20 = \frac{1}{2}(9.8)t^2 \rightarrow t^2 = 4.08\)

Answer: \(t \approx 2.02\) s

You've Mastered the Basics!

You're now ready to tackle more complex motion problems.