Topic 8.2 Notes – Connecting Position, Velocity, and Acceleration of Functions Using Integrals

Position, Velocity, and Acceleration in Rectilinear Motion

Rectilinear motion means motion along a straight line, described as functions of time $t$.

We use:

$$v(t) = s'(t)$$

$$a(t) = v'(t) = s''(t)$$

So:

• Position $s(t)$ tells you where the particle is.
• Velocity $v(t)$ tells you how fast and in what direction it’s moving.
• Acceleration $a(t)$ tells you how velocity is changing.

Integrals reverse those relationships:

• $\int v(t)\,dt = s(t) + C$
• $\int a(t)\,dt = v(t) + C$

And with definite integrals, you get accumulated change over an interval.

The three graphs below show how these ideas line up for the same moving object.

Position, velocity, and acceleration for the same motion

Look vertically across the panels. The slope of the position graph matches the velocity graph. The slope of the velocity graph matches the acceleration graph. The area under the acceleration curve gives change in velocity. The area under the velocity curve gives change in position.

Lock this in:

• Slope = derivative
• Area = accumulated change

That idea runs this entire topic.

Displacement vs Total Distance Traveled

These two are constantly tested together.

Displacement

Displacement is net change in position:

$$\text{Displacement} = s(b) - s(a) = \int_a^b v(t)\,dt$$

Important:

• It uses the signed area under the velocity curve.
• If velocity is negative, that area subtracts.
• Displacement can be zero even if the particle moved.

If a particle goes right 5 units and left 5 units, displacement is 0.

Total Distance Traveled

Distance measures the entire path length:

$$\text{Distance} = \int_a^b |v(t)|\,dt$$

Key differences:

• Always positive.
• Uses the absolute value of velocity.
• You must split the integral at points where $v(t)=0$ if the sign changes.

Example idea:

If $v(t)=t-2$ on $[0,4]$, velocity changes sign at $t=2$.

You would compute:

• $\int_0^2 |t-2|\,dt$
• $\int_2^4 |t-2|\,dt$

On FRQs, students lose points by forgetting to split at sign changes. Always check where velocity equals zero.

Using Integrals to Move Between a, v, and s

Most questions give you one function and initial conditions.

From Velocity to Position

If you’re given $v(t)$:

• Displacement on $[a,b]$:
  $\int_a^b v(t)\,dt$
• Position function:
  $$s(t) = \int v(t)\,dt + C$$

Use a condition like $s(1)=4$ to solve for $C$.

Remember:

• Indefinite integral → family of functions.
• Definite integral → numerical change.

From Acceleration to Velocity

If given $a(t)$:

$$v(t) = \int a(t)\,dt + C$$

Use an initial velocity like $v(0)=3$ to find $C$.

Then integrate again to get position.

This layering matters:

• Acceleration accumulates into velocity.
• Velocity accumulates into position.

On calculator-active problems, you may not even need a formula. You might just evaluate definite integrals numerically to get velocity or displacement values.

Reading Motion from Graphs

AP loves graph interpretation.

From a Velocity Graph

• Area = displacement
• Total area (absolute value) = distance
• Above x-axis → moving right
• Below x-axis → moving left
• Slope of graph → acceleration

If velocity and acceleration have the same sign, the particle is speeding up. Opposite signs mean slowing down.

From an Acceleration Graph

• Area under curve = change in velocity
• Positive area → velocity increases
• Negative area → velocity decreases

Sometimes you build velocity step-by-step by accumulating areas over subintervals.