Total Distance Traveled

A particle that turns twice

A particle moves along a line with velocity $v=t^2-4t+3$v = t² − 4t + 3 from t = 0 to t = 4. Factorized, v = (t − 1)(t − 3), so v is positive before t = 1, negative between t = 1 and t = 3, and positive again after t = 3. The particle moves forward, turns at t = 1, moves backward, and turns again at t = 3.

$v=t^2-4t+3$v = t² − 4t + 3 from t = 0 to 4. It crosses the axis at the dots, t = 1 and t = 3, where the particle turns. Between them the velocity is negative and the shaded piece lies below the axis.

The signed integral is displacement

An antiderivative of v is $s=\frac{t^3}{3}-2t^2+3t$s = t³/3 − 2t² + 3t, which is the position if the particle starts at s = 0. Then $s\left(1\right)=\frac{1}{3}-2+3=\frac{4}{3}$s(1) = 1/3 − 2 + 3 = 4/3, s(3) = 9 − 18 + 9 = 0 and $s\left(4\right)=\frac{64}{3}-32+12=\frac{4}{3}$s(4) = 64/3 − 32 + 12 = 4/3.

So the particle goes out to $\frac{4}{3}$4/3, comes back to 0 at t = 3, and goes out to $\frac{4}{3}$4/3 again. The integral of v from 0 to 4 is $s\left(4\right)-s\left(0\right)=\frac{4}{3}$s(4) − s(0) = 4/3, the displacement: the trip back has canceled one of the trips out.

Integrate the speed

The total distance counts every stretch as ground covered, whichever way the particle was going. That is the integral of the speed |v|, which is never negative, so no stretch can cancel another.

The graph of |v| is the graph of v with every piece below the axis reflected above it. The area under it is the total distance.

The velocity $v=t^2-4t+3$v = t² − 4t + 3 in gold and the speed |v| in plain; the two run together wherever v is positive. Between t = 1 and 3 the speed is the velocity turned over, so all of the shaded area under |v| is above the axis, and it adds to 4.

In practice: split at the turns

|v| has no single antiderivative to use across the turns, so work in pieces. Solve v = 0 to find the turning times, integrate v between consecutive turns, and add the sizes of the answers.

Here v = 0 at t = 1 and t = 3. The pieces are $s\left(1\right)-s\left(0\right)=\frac{4}{3}$s(1) − s(0) = 4/3, $s\left(3\right)-s\left(1\right)=-\frac{4}{3}$s(3) − s(1) = −4/3 and $s\left(4\right)-s\left(3\right)=\frac{4}{3}$s(4) − s(3) = 4/3. The total distance is $\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4$4/3 + 4/3 + 4/3 = 4; with the signs kept, $\frac{4}{3}-\frac{4}{3}+\frac{4}{3}=\frac{4}{3}$4/3 − 4/3 + 4/3 = 4/3 is the displacement.

Distance is never less than the size of the displacement, and the two are equal only when the particle never turns back.

A ball thrown upward with $s=20t-5t^2$s = 20t − 5t² and v = 20 − 10t. At t = 3 it is at s = 15 on the way down, with v = −10. Its velocity is zero at t = 2, where it turns at s = 20. From t = 0 to 4 its displacement is s(4) − s(0) = 0, but it rises 20 and falls 20, a distance of 40. Drag t to 2 to see the turn.

A car that comes back

A car has v = 10 − 2t meters per second. Then v = 0 at t = 5, and $s=10t-t^2$s = 10t − t². From t = 0 to 5 the car moves s(5) − s(0) = 50 − 25 = 25 meters forward; from 5 to 10 it moves s(10) − s(5) = 0 − 25 = −25, that is 25 meters back.

Over the ten seconds its displacement is 0 and its distance is 25 + 25 = 50 meters.

The usual mistakes

Integrating v straight through. That gives the displacement, $\frac{4}{3}$4/3, not the distance, 4.

Solving a = 0 for the turns. a = 0 where the velocity stops changing: here a = 2t − 4 = 0 at t = 2, where v = −1, so the particle is still moving backward. A turn happens only where v = 0.

Adding the pieces with their signs. $\frac{4}{3}-\frac{4}{3}+\frac{4}{3}$4/3 − 4/3 + 4/3 is the displacement; the distance drops the minus sign.

Saying a return trip has zero distance. A particle that comes back to its start has displacement 0, but it covered ground both ways.

A survey vehicle on a tether

In the application below, a vehicle on a tether has $v=1.5n^2-12n+18$v = 1.5n² − 12n + 18 meters per minute, which is zero at n = 2 and n = 6. One integral gives where it ends up; the tether the winch moves is the distance, from three pieces added without their signs.