AP Calculus AB and BC

Position vs Displacement

Q: Is displacement the same as distance travelled?

No. Displacement integrates v(t) and can be negative or zero, while total distance integrates \lvert v(t) \rvert and counts movement in both directions.

Position is where the particle is at a single instant. Displacement is the change in position across an interval, found by integrating velocity, so it can be zero even though the particle moved the whole time. Add the displacement to the starting position to get the final position.

Position

Use when: The question asks where the particle is at a given time, or for a value such as $s(4)$ .

Displacement

Use when: The question asks for the net change in position over an interval, or how far the particle ended up from where it started.

Side by side

	Position	Displacement
What it measures	Location at one instant	Net change in location over an interval
Written as	$s(t)$ or $x(t)$	$s(b) - s(a)$
How you find it	$s(a) + \int_a^t v(u)\,du$	$\int_a^b v(t)\,dt$
Depends on	One time value	Two endpoints
Needs an initial condition	Yes, or the position is undetermined	No, the constant cancels
Common trap	Reporting it when the question asked for the change	Reporting it as the final position

Position is a value and displacement is a difference, which is the whole distinction. The Fundamental Theorem states the link as $\int_a^b v(t)\,dt = s(b) - s(a)$ , so integrating a velocity function never hands you a location. It hands you the gap between two locations.

That is why finding a position from a velocity function requires the initial condition: $s(b) = s(a) + \int_a^b v(t)\,dt$ . A free response part that supplies $s(0) = 2$ along with $v(t)$ and then asks for $s(5)$ is testing exactly this line, and answering with the integral alone loses the point.

Zero displacement, moving particle

A particle that leaves $x = 1$ , travels out to $x = 7$ , and comes back has displacement $0$ even though it never stopped except for the single instant it turned around. Zero displacement means it finished where it started, not that it stood still. The total distance travelled over that trip is $12$ .

Frequently asked questions

Is displacement the same as distance travelled?

No. Displacement integrates $v(t)$ and can be negative or zero, while total distance integrates $\lvert v(t) \rvert$ and counts movement in both directions.

How do I get position from velocity?

Integrate velocity to get the change, then add the given starting position: $s(t) = s(0) + \int_0^t v(u)\,du$ . Without an initial condition you can only report displacement.

Can displacement be negative?

Yes. A negative value means the particle finished to the left of, or below, where it started, which is information a distance answer throws away.

In the CED: Unit 4: Contextual Applications, Unit 8: Applications of Integration