Total Distance Traveled Using Definite Integrals

Distance vs. Displacement

One of the most important distinctions in AP Calculus motion problems is the difference between displacement and total distance traveled.

If an object has velocity , then integrating velocity over a time interval gives its net change in position, or displacement.

Displacement can be positive, negative, or zero. It tells you where the object ends up relative to where it started.

Total distance traveled is different. Distance measures how much ground the object actually covers, regardless of direction. Because of that, negative velocity cannot cancel positive velocity.

The absolute value is the key. Whenever velocity is negative, we use its magnitude rather than allowing that portion of the motion to subtract from the total.

Why the Absolute Value Matters

Suppose an object moves forward for part of a trip and then reverses direction. Its velocity might look like this:

If you calculate over the entire interval, the negative portion subtracts from the positive portion. That gives displacement—not total distance.

For total distance, we need:

On an AP Calculus problem, this usually means finding the times when and splitting the integral at those times.

The Step-by-Step Method

Start with the velocity function. Make sure you know the time interval over which the distance is being measured.
Find where the velocity is zero. Solve for all times in the interval. These are possible direction changes.
Determine the sign of velocity on each resulting time interval. A positive velocity means motion in one direction; a negative velocity means motion in the opposite direction.
Split the integral at every velocity zero where the sign changes.
Make every contribution positive. Integrate where velocity is positive and where velocity is negative.
Add the distances. The result is the total distance traveled.

Worked Examples

Example 1: Velocity Stays Positive

An object moves along a straight path with velocity for . Find the total distance traveled.

Step 1: Because is always positive on the interval, the object never reverses direction.

Step 2: Distance is therefore simply the integral of velocity:

Step 3: Evaluate the definite integral:

Result: The object travels units.

Example 2: Velocity Changes Sign

An object has velocity for . Find the total distance traveled.

Step 1: Find when the velocity is zero:

Step 2: Determine the sign of velocity. For , velocity is positive. For , velocity is negative. Therefore, the object changes direction at .

Step 3: Split the distance calculation at the direction change:

Step 4: Because velocity is negative on the second interval, reverse its sign when calculating distance:

Step 5: Evaluate:

Result: The total distance traveled is units.

Example 3: Multiple Direction Changes

An object has velocity for . Find the total distance traveled.

Step 1: Find all zeros of the velocity:

Step 2: The velocity changes sign at each of these times, so divide the interval into three pieces.

Step 3: Determine the sign of velocity on each interval. The velocity is positive on , negative on , and positive again on .

Step 4: Write the total distance as the sum of the magnitudes of the three integrals:

Result: After evaluating the three pieces, the total distance is units.

Example 4: Distance from a Position Function

Suppose the position of a particle is , where . Find the total distance traveled.

Step 1: Total distance is determined from velocity, so differentiate position:

Step 2: Find where the particle changes direction by solving :

Step 3: Determine the sign of velocity on each interval and split the integral at the direction-change time.

Step 4: Calculate the total distance:

Result: The particle travels units.

A Faster AP Exam Strategy

On a timed AP Calculus question, you do not always need to explicitly write the absolute-value integral. A reliable strategy is to make a quick sign chart for velocity.

Then translate that sign chart directly into positive distance contributions. This prevents the most common mistake: accidentally calculating displacement when the question asks for total distance.

Distance and displacement can be compared directly

If an object never changes direction, then distance and the absolute value of displacement are the same. But if the object reverses direction, total distance is larger.

For example, an object could travel 10 meters forward and then 6 meters backward. Its displacement is only 4 meters, but its total distance is 16 meters.

Common Mistakes

Using one integral of velocity for total distance

The integral gives displacement. It gives total distance only when velocity never becomes negative on the interval. If velocity changes sign, you must split the integral and account for the negative portion.

Forgetting to check where velocity is zero

The zeros of velocity are the places where direction changes can occur. Always solve before setting up a total-distance calculation.

Assuming every zero of velocity means a direction change

A velocity of zero does not automatically mean the particle reverses direction. Check the sign of velocity on both sides of the zero. If the sign stays the same, there is no direction change and you do not need to split the distance integral there.

Reporting displacement when the question asks for distance

A negative contribution from velocity is perfectly valid for displacement but cannot make total distance smaller. Distance is always nonnegative.

Practice Problems

Question 1

A particle has velocity for . Find the total distance traveled.

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Question 2

A particle has velocity for . Find the total distance traveled.

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Question 3

A particle has velocity for . Find the total distance traveled.

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Question 4 (Position Function)

A particle's position is for . Find the total distance traveled.

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