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CalculusBy Educator Today

C34 · Unit C8 Applications of integration

Displacement and total distance

CoreReview editionNot yet human-reviewed
Time
About 45 minutes, untimed
Course framework reference
AP 8.2
Builds on
Part of

Progress on this lesson counts in every pathway listed.

Understand the idea

Integrating velocity gives displacement; integrating speed |v| gives total distance. Locate every velocity sign change and split the interval before removing the absolute value. Position at the final time adds the displacement to initial position. An acceleration integral gives velocity change and likewise requires an initial velocity for an absolute value of velocity.

Worked example

For v(t) equals t minus 2 on [0,5], displacement is [t squared /2 minus 2t] sub 0 to the power 5 equals 2.5. Distance is minus ∫ sub 0 squared (t minus 2)dt+∫ sub 2 to the power 5 (t minus 2)dt equals 2+4.5 equals 6.5. The distance exceeds the magnitude of displacement because the direction changes.

Explore the mathematics

Displacement and total distance

Separate signed displacement from total distance.

Setting 3 of 5

Scroll sideways to see the full diagram.

Predict whether displacement and total distance agree.xy12Marked input 1; displayed curve height 0Velocity v(t)=t−1 on [0,2]

Predict before continuing

Predict whether displacement and total distance agree.

Read the complete explanation
  1. Predict whether displacement and total distance agree.
  2. Velocity is negative until t=1 and positive afterward.
  3. The signed integral is 0, but distance integrates |v|.
  4. Distance is 1/2+1/2=1; displacement is 0. Split at each sign change.
See the idea: Displacement and total distance
See the idea: Displacement and total distanceThe velocity changes sign at t=2. The negative triangle has area magnitude 2 and the positive one 4.5. Displacement is −2+4.5=2.5; distance is 2+4.5=6.5.xy0-31.25-1.252.50.53.752.2554

The velocity changes sign at t=2. The negative triangle has area magnitude 2 and the positive one 4.5. Displacement is −2+4.5=2.5; distance is 2+4.5=6.5.

Discuss: explain what each label represents, then draw a related example with different values.

A common misconception

Total distance cannot be negative. Taking the absolute value after integrating may hide backtracking.

Investigate and explain

Draw a signed area diagram under velocity and physically trace the motion along a number line.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.