Discover how calculus describes motion in the real world through an intuitive ball-throwing example that connects displacement, velocity, and acceleration.
Imagine you throw a ball straight up into the air from the ground. What happens? The ball:
Goes up
Slowing down due to gravity
Reaches a peak
Velocity becomes zero
Falls back down
Speeding up due to gravity
The Big Question:
How can we describe this motion mathematically? And what does calculus have to do with it?
In this lesson, we'll start by measuring the ball's height at different times, just like a scientist would. Then, we'll discover that there's a beautiful mathematical equation hiding behind our measurements — and calculus is the key to unlocking it.
Let's say we throw a ball upward and measure its height above the ground at different times. Here's what we recorded:
| Time $t$ (seconds) | Height $s$ (meters) |
|---|---|
| 0 | 0.0 |
| 1 | 15.1 |
| 2 | 20.4 |
| 3 | 15.9 |
| 4 | 1.6 |
Starting point
At $t = 0$, ball is at ground level
Highest point
At $t = 2$ s, ball reaches 20.4 m
Coming down
Height decreases after peak
Quick Questions:
When we plot these points on a graph, something interesting emerges:
Observations:
What this tells us:
Key Insight:
The curved graph tells us that the ball's speed is changing over time. If the speed were constant, we'd have a straight line!
Let's answer an important question: How fast is the ball moving when it hits the ground?
We can figure this out using physics you already know, without any calculus. Here's the plan:
From our table:
Maximum height = 20.4 m
Velocity at peak = 0 m/s
At the highest point, the ball stops moving upward for an instant before falling back down.
Height at peak = 20.4 m
Height at ground = 0 m
Distance fallen = 20.4 m
Given:
Initial velocity at peak: $v_i = 0$ m/s
Acceleration due to gravity: $a = 9.8$ m/s²
Displacement: $\Delta x = 20.4$ m
Using the equation of motion:
Direction: Downward → velocity = −20 m/s
Beautiful Symmetry!
If we threw the ball up at 20 m/s, it hits the ground at 20 m/s. What goes up must come down — at the same speed!
Here's where it gets interesting. It turns out there's an exact mathematical equation that describes the ball's height at any time:
What each part means:
$20t$
Initial upward motion (thrown at 20 m/s)
$-4.9t^2$
Effect of gravity pulling down
Let's verify:
At $t = 1$:
$s(1) = 20(1) - 4.9(1)^2 = 15.1$ ✓
At $t = 2$:
$s(2) = 20(2) - 4.9(2)^2 = 20.4$ ✓
At $t = 3$:
$s(3) = 20(3) - 4.9(3)^2 = 15.9$ ✓
Important Realization:
The table was just samples from this continuous equation. The equation tells us the height at any time, not just at whole seconds!
We can find this by setting height equal to zero:
The ball hits the ground at approximately $t = 4.08$ seconds.
Before we dive into the calculus method, let's understand what velocity and acceleration really mean in mathematical terms.
The rate of change of displacement
Velocity tells us how quickly the position (displacement) is changing with respect to time.
Mathematical Definition:
In words: velocity is the derivative of displacement with respect to time
In this lesson:
$s(t) = 20t - 4.9t^2$
Displacement function
Therefore:
$v(t) = s'(t) = 20 - 9.8t$
Velocity function
The rate of change of velocity
Acceleration tells us how quickly the velocity is changing with respect to time.
Mathematical Definition:
In words: acceleration is the derivative of velocity with respect to time
In this lesson:
$v(t) = 20 - 9.8t$
Velocity function
Therefore:
$a(t) = v'(t) = -9.8$
Acceleration function
Since acceleration is the derivative of velocity, and velocity is the derivative of displacement, we can say:
Acceleration is the second derivative of displacement
For our ball example:
Key Insight
Calculus gives us a precise mathematical language to describe motion. When we say "velocity is the rate of change of position," we're really saying "velocity is the derivative of displacement." This isn't just fancy terminology — it's a powerful tool that lets us calculate these quantities directly from equations!
Now that we understand velocity and acceleration as derivatives, let's see how calculus provides an elegant shortcut compared to the physics method we used earlier!
Key Question:
What does the gradient (slope) of the height-time graph represent?
Answer: Velocity!
Step 1: Start with displacement
Step 2: Differentiate to get velocity
Step 3: Find velocity at impact ($t = 4.08$ s)
🎯 Exact Match!
Physics Method:
v = −20.0 m/s
Multiple steps, used $v^2 = u^2 + 2as$
Calculus Method:
v = −20.0 m/s
One step, just differentiate!
The Power of Calculus
Calculus didn't give us a different answer — it gave us the same answer, faster and more elegantly!
"Before calculus, we jump through hoops. With calculus, we just differentiate."
Now let's see the beautiful connection between displacement, velocity, and acceleration through their graphs:
Displacement
$s(t) = 20t - 4.9t^2$
Graph: Parabola (curved)
Shows where the ball is
Velocity
$v(t) = 20 - 9.8t$
Graph: Straight line
Shows how fast and direction
Acceleration
$a(t) = -9.8$
Graph: Horizontal line
Constant gravity
How they're connected:
Notice:
Displacement (Parabola)
Curved graph showing position over time
Velocity (Straight Line)
Derivative of displacement
Acceleration (Horizontal)
Derivative of velocity
The Derivative Chain:
Apply what you've learned with these practice problems.
Test your understanding with these questions.
"Calculus wasn't invented to make maths harder — it was invented to explain motion."
| Quantity | Graph Shape | Meaning |
|---|---|---|
| Displacement | Parabola | Where the ball is |
| Velocity | Straight line | How fast & direction |
| Acceleration | Horizontal line | Gravity (constant) |
"The derivative is just velocity hiding inside an equation."