The Derivative as a “Motion Detector”

Good day, students! Today, we’re going to connect the abstract world of calculus to something we see every day: things that move. Whether it’s a car on the highway, a ball thrown in the air, or a planet orbiting a star, we can describe their motion precisely using the derivative.

Part 1: The Foundation – Position, Velocity, and Acceleration

First, let’s define our key players. We will describe motion along a straight line (1-dimensional).

1. Position, $s(t)$: This function tells us where an object is at any given time $t$. Its units are typically meters (m), feet (ft), miles (mi), etc.

   • Example: $$s(t)=t^{2}-4t+3$$ means at time $t$, the object is at position $$t^{2}-4t+3$$ .

2. Velocity, $v(t)$: This tells us how fast and in what direction the object is moving. It’s the rate of change of position.

   • Key Insight: The instantaneous velocity is the derivative of the position function with respect to time.

     $$v(t)=s^\prime(t)=\frac{ds}{dt}$$​

   • Interpretation:

     • If v(t)>0, the object is moving forward (in the positive direction).

     • If v(t)<0, the object is moving backward (in the negative direction).

     • If v(t)=0, the object is instantaneously at rest (it’s changing direction or pausing).

3. Acceleration, a(t): This tells us how the velocity is changing. It’s the rate at which the object is speeding up or slowing down.

   • Key Insight: The instantaneous acceleration is the derivative of the velocity function, and therefore the second derivative of the position function.

     $$a(t)=v^\prime(t)=\frac{dv}{dt}=s^\dprime(t)=\frac{d^{2}s}{dt^{2}}$$​

   • Interpretation:

     • If a(t)>0, the velocity is increasing (speeding up in the positive direction or slowing down in the negative direction).

     • If a(t)<0, the velocity is decreasing (slowing down in the positive direction or speeding up in the negative direction).

     • Crucial Point: The sign of acceleration alone does not tell you if the object is speeding up or slowing down. You must compare it to velocity!

Part 2: The “How-To” – Solving Motion Problems

Let’s work through a classic example.

The Scenario: A ball is thrown vertically upward from a platform. Its height (in meters) above the ground after $t$ seconds is given by the position function:

$$s(t)=-4.9t^{2}+19.6t+24.5$$

Question 1: Find the velocity and acceleration functions.

• Velocity is the derivative of position:

  $$v(t)=s^\prime(t)=\frac{d}{dx}(-4.9t^{2}+19.6t+24.5)=-9.8t+19.6$$

• Acceleration is the derivative of velocity:

  $$a(t)=v^prime(t)=\frac{d}{dx}(-9.8t+19.6)=-9.8$$

Interpretation: The acceleration is a constant $$-9.8m/s^{2}$$. This is the acceleration due to gravity, as we would expect! The negative sign indicates it’s acting downward.

Question 2: What is the initial velocity?

The initial velocity is the velocity at time t=0.

$$v(0)=-9.8(0)+19.6=19.6m/s$$

The ball starts with an upward velocity of 19.6 m/s.

Question 3: When does the ball reach its maximum height?

Think about what happens at the very top of the ball’s flight. For an instant, it stops moving upward and hasn’t started moving downward yet. Its velocity is zero.

$$v(t)=0$$ then $$-9.8+19.6=0 \longrightarrow -9.8t=-19.6 \longrightarrow 9.8t=19.6$$

therefore t=2seconds

The ball reaches its maximum height after 2 seconds.

Question 4: What is the maximum height?

We now know the ball is at its highest point at t=2 seconds. We find the height by plugging this time into the original position function.

$$s(2)=-4.9(2)^{2}+19.6(2)+24.5=-19.6+}39.2+24.5=44.1m$$

The maximum height is 44.1 meters.

Question 5: When does the ball hit the ground?

The ball hits the ground when its height, s(t), is zero.

We can solve this quadratic equation using the quadratic formula.

$$t=\frac{-19.6\pm\sqrt{(19.6)^{2}-4(-4.9)(24.5)}}{2(-4.9)}=\frac{-19.6\pm29.4}{-9.8}$$

This gives two solutions: t=−1 and t=5. Since time cannot be negative, we take $t=5$ seconds. The ball hits the ground after 5 seconds.

Question 6: How fast is the ball moving when it hits the ground?

We found the time of impact (t=5), so we plug this into the velocity function, not the position function.

$$v(t)=-9.8(5)+19.6=-49+19.6=-29.4m/s$$

The velocity is -29.4 m/s. The negative sign is crucial! It tells us the ball is moving downward at 29.4 m/s when it impacts.