Distance traveled is the total length of the path a particle covers over a time interval, found by integrating speed: ∫|v(t)|dt for motion on a line, or ∫√((dx/dt)² + (dy/dt)²)dt for parametric or vector-valued motion in the plane (FUN-8.B.2).
Distance traveled measures how much ground a moving particle actually covers, counting every step forward AND backward. If you walk 5 meters right and 3 meters left, your distance traveled is 8 meters, even though you only ended up 2 meters from where you started. That 2 meters is displacement, and the AP exam loves checking whether you know the difference.
In calculus terms, the CED (FUN-8.B.2) says the definite integral of speed gives total distance traveled, while the definite integral of velocity gives displacement. For motion along a line, speed is |v(t)|, so distance = ∫|v(t)|dt. For planar motion defined parametrically or with vector-valued functions, speed is the magnitude of the velocity vector, so distance = ∫√((dx/dt)² + (dy/dt)²)dt. The absolute value (or the magnitude) is the whole point. It stops backward motion from canceling out forward motion the way it does in a displacement integral.
Distance traveled lives in Topic 9.6, Solving Motion Problems Using Parametric and Vector-Valued Functions, in Unit 9 (BC only). It directly supports learning objective AP Calc 9.6.A, determining values for positions and rates of change in planar motion, and essential knowledge FUN-8.B.2, which draws the exact line you need to memorize. Velocity integrates to displacement; speed integrates to distance traveled. It's also one of the cleanest examples of a big AP Calc idea, that integrating a rate of change accumulates total change. And here's a connection worth noticing. The planar distance-traveled formula is identical to the arc length formula for a parametric curve, because the distance a particle travels IS the length of the path it traces.
Keep studying AP Calculus Unit 9
Visual cheatsheet
view galleryDisplacement (Unit 9)
Displacement is the net change in position, ∫v(t)dt, where backward motion subtracts from forward motion. Distance traveled uses |v(t)| so nothing cancels. Distance is always greater than or equal to |displacement|, with equality only when the particle never changes direction.
Speed (Unit 9)
Speed is what you actually integrate to get distance traveled. On a line it's |v(t)|, and in the plane it's the magnitude of the velocity vector, √((dx/dt)² + (dy/dt)²). If you can write the speed function, you're one definite integral away from the answer.
Velocity Vector (Unit 9)
For planar motion, the velocity vector ⟨dx/dt, dy/dt⟩ does double duty. Integrate the vector itself component by component and you get displacement. Take its magnitude first, then integrate, and you get distance traveled. The order of operations is the entire difference.
Arc Length of Parametric Curves (Unit 9)
Distance traveled along a parametric path and arc length use the same integral, ∫√((dx/dt)² + (dy/dt)²)dt. A particle's distance traveled is literally the arc length of the curve it traces, as long as it doesn't retrace its path.
Particle motion shows up on the FRQ section almost every year, and distance traveled is a standard part. The 2018, 2019, and 2021 exams all featured FRQ Q2 motion problems built around a velocity function v(t), where you needed to integrate |v(t)| for total distance but ∫v(t)dt plus an initial position for location. The 2023 BC FRQ Q2 moved this to a parametric curve, where dx/dt was given and you had to build the speed integrand yourself. Multiple choice tests the setup too, asking what you integrate to find distance traveled for parametric motion (answer: speed, the magnitude of velocity). These are usually calculator-active, so your job is correct setup and notation. Write the integral with the absolute value or square root visible, evaluate on your calculator, and label units. Forgetting the absolute value is the single most common way to lose these points.
Displacement is the net change in position, ∫v(t)dt, and it can be negative or zero even when the particle moved a lot. Distance traveled is the total path length, ∫|v(t)|dt, and it's never negative. Quick gut check. Run a lap around a track and your displacement is 0, but your distance traveled is 400 meters. On the exam, the question's wording tells you which integral to set up, so read for 'total distance' versus 'position' or 'displacement.'
Distance traveled is the definite integral of speed over the time interval, while displacement is the definite integral of velocity (FUN-8.B.2).
For motion along a line, distance traveled = ∫|v(t)|dt, and the absolute value prevents backward motion from canceling forward motion.
For parametric or vector-valued motion in the plane, distance traveled = ∫√((dx/dt)² + (dy/dt)²)dt, which is the same formula as parametric arc length.
Distance traveled is always greater than or equal to the absolute value of displacement, with equality only if the particle never changes direction.
On calculator-active FRQs, show the correct integral setup with the absolute value or square root written out, then evaluate numerically.
Distance traveled in planar motion is tested in Topic 9.6 under learning objective AP Calc 9.6.A on the BC exam.
Distance traveled is the total length of the path a particle covers over a time interval. You find it by integrating speed, which is ∫|v(t)|dt for motion on a line or ∫√((dx/dt)² + (dy/dt)²)dt for parametric motion in the plane.
No. Displacement is the net change in position (∫v(t)dt) and can be negative or zero, while distance traveled (∫|v(t)|dt) counts all motion and is never negative. A particle that moves right 5 units and left 5 units has displacement 0 but distance traveled 10.
Velocity is negative when the particle moves backward, so ∫v(t)dt lets backward motion cancel forward motion. Taking |v(t)| makes every interval of motion count positively, which is what total distance means.
Integrate the speed, √((dx/dt)² + (dy/dt)²), over the time interval. For example, with x = 3t² and y = 2t³ you'd compute ∫₀² √((6t)² + (6t²)²)dt, usually on your calculator since these FRQs are calculator-active.
Both, in different forms. Distance traveled for straight-line motion (∫|v(t)|dt) appears on AB and BC, as in the 2018, 2019, and 2021 FRQs. The parametric and vector-valued version in Topic 9.6 is BC only, like the 2023 BC FRQ Q2.