Motion Revisited: Accumulating Velocity

You’ll be able to

Recover displacement by integrating velocity over an interval
Compute total distance traveled by integrating the speed |v(t)|
Find position from velocity and an initial condition

Integration reverses the motion derivatives

In Unit 4 you differentiated position to get velocity; integration runs it backward. The displacement over [a, b] is the accumulated velocity, ∫[a to b] v(t) dt = s(b) − s(a) (the Net Change Theorem). Likewise integrating acceleration recovers velocity. Integration is how you rebuild position and velocity from the rates that describe the motion.

Displacement vs. total distance — with integrals

Integrating velocity directly gives displacement, where motion in opposite directions cancels. To get total distance traveled you integrate the speed, ∫[a to b] |v(t)| dt, so every bit of motion counts as positive. In practice you find where v changes sign, split the interval there, and add the absolute values of the pieces — the integral version of the Unit 4 idea.

Position from velocity and a starting point

To find the actual position at a time, you need an initial condition. Since s(t) = s(a) + ∫[a to t] v(τ) dτ, the position at time t is the starting position plus the accumulated displacement since a. Without the initial position you can find only the change in position, not where the particle actually is.

Motion by integration

displacement = ∫[a to b] v(t) dt · total distance = ∫[a to b] |v(t)| dt · s(t) = s(a) + ∫[a to t] v(τ) dτ

Velocity integrates to displacement; speed |v| integrates to total distance. An initial condition pins down the position.

Worked example

A particle has velocity v(t) = t − 3 (m/s) for 0 ≤ t ≤ 5. Find its displacement and its total distance traveled.

1.Displacement is ∫[0 to 5] (t − 3) dt = [t²/2 − 3t] from 0 to 5 = (25/2 − 15) − 0 = −2.5 m.
2.For total distance, find where v = 0: t − 3 = 0 at t = 3, where velocity changes sign.
3.On [0, 3], v < 0: ∫[0 to 3] (t − 3) dt = (9/2 − 9) = −4.5, absolute value 4.5.
4.On [3, 5], v > 0: ∫[3 to 5] (t − 3) dt = (25/2 − 15) − (9/2 − 9) = (−2.5) − (−4.5) = 2. Add absolute values: 4.5 + 2.

Answer: Displacement = −2.5 m (net, 2.5 m in the negative direction); total distance = 4.5 + 2 = 6.5 m. The two differ because the particle reverses direction at t = 3.

Watch out

To get total distance you must integrate |v(t)|, not v(t). Split the interval at every time velocity changes sign and add the absolute values — integrating v directly gives only the net displacement, where opposite motions cancel.

Checkpoint

The displacement of a particle over [a, b] is given by:

Checkpoint

A particle starts at position s(0) = 4 with velocity v(t). What is its position at t = 3?

On the exam

Watch the wording: "displacement" or "position change" means ∫v dt, while "total distance traveled" means ∫|v| dt with the interval split at sign changes. And "where is the particle" always needs the initial position added to the integral.

Answer the 2 checkpoints as you read.