Displacement And Distance During Reversals

Why Reversals Change the Calculation

If an object moves along a straight line, its velocity records both how quickly it moves and which direction it moves. A positive velocity indicates motion in the chosen positive direction; a negative velocity indicates motion in the opposite direction.

When velocity keeps the same sign, displacement and distance are closely related. When velocity changes sign, the object reverses direction: its motion in one direction partly cancels its motion in the other when we calculate displacement. Distance does not cancel in this way. It counts the whole path travelled.

The key is to distinguish signed accumulation from unsigned accumulation. Integrating velocity gives signed displacement. Integrating the magnitude of velocity gives total distance.

Definition: Let \(x(t)\) be the position of an object moving on a line, and let \(v(t)=x'(t)\) be its velocity. On a time interval \([a,b]\), the displacement is \(x(b)-x(a)\), and the total distance travelled is the length of the path the object covers during that interval.

Displacement and Total Distance

Suppose \(v\) is continuous on \([a,b]\), and \(x'(t)=v(t)\). The Fundamental Theorem of Calculus gives the displacement:

\text{displacement} = x(b)-x(a) = \int_a^b v(t)\,dt.

The sign of velocity matters in this integral. A positive part contributes positively; a negative part contributes negatively. The resulting displacement is the net change in position, not necessarily the distance travelled.

For total distance, each part of the motion contributes a nonnegative amount. The speed is \(|v(t)|\), so the corresponding formula is:

\text{total distance} = \int_a^b |v(t)|\,dt.

These formulas use the same velocity function but answer different questions. Displacement depends on the initial and final positions. Total distance depends on all the motion between those times.

Key distinction: Displacement is a signed net change, so use \(\int_a^b v(t)\,dt\). Total distance is unsigned, so use \(\int_a^b |v(t)|\,dt\).

How to Locate and Handle a Reversal

A reversal occurs when the object changes direction. For a continuous velocity function, this means that velocity changes sign. To find possible reversals, solve \(v(t)=0\) and check the sign of \(v\) on either side of each solution. A zero alone is not enough: velocity might touch zero and keep the same sign.

Once the interval is divided at its reversals, velocity has one sign on each resulting piece. The signed integral on a piece is that piece’s displacement. Its distance contribution is the absolute value of that displacement. Adding the signed contributions gives net displacement; adding their absolute values gives total distance.

Find candidate times.
Solve \(v(t)=0\) in the interval and include any breakpoints where the velocity formula changes.

Check the signs.
Test velocity between consecutive candidate times. Keep as reversals only the times where the sign changes.

Calculate signed changes.
Integrate \(v(t)\) on each piece. These values add to the full-interval displacement.

Calculate distance.
Take the absolute value of each piece’s displacement and add those nonnegative amounts.

Why Splitting at Reversals Works

Here is a justification of the distance formula when there are finitely many reversals. Partition the interval at the sign-changing times, writing the resulting subintervals as \([t_{i-1},t_i]\). On each one, velocity has a constant sign, so position moves in just one direction. The position change on that piece is

x(t_i)-x(t_{i-1}) = \int_{t_{i-1}}^{t_i} v(t)\,dt.

Because \(v\) does not change sign on the piece, the absolute value of this position change is the distance covered on it. Equivalently, \(|v|\) equals either \(v\) throughout the piece or \(-v\) throughout the piece. Therefore the distance on that piece is

\int_{t_{i-1}}^{t_i} |v(t)|\,dt = \left| \int_{t_{i-1}}^{t_i} v(t)\,dt \right|.

Adding the pieces gives the total distance. The signed integrals add to the net position change, while their absolute values add to the full path length:

\text{displacement} = \sum_i \int_{t_{i-1}}^{t_i} v(t)\,dt, \qquad \text{total distance} = \sum_i \left|\int_{t_{i-1}}^{t_i} v(t)\,dt\right|.

This also explains why taking the absolute value only after integrating over the entire interval generally does not give distance: cancellation between opposite directions has already occurred.

Worked Examples

Worked Example: A Single Reversal

An object has velocity \(v(t)=t-2\) metres per second for \(0\le t\le5\) seconds. Find its displacement and total distance over the interval. Also find its final position if \(x(0)=10\) metres.

First locate the velocity zero:

t-2=0 \quad\Longrightarrow\quad t=2.

For \(0\le t<2\), \(t-2<0\); for \(2<t\le5\), \(t-2>0\). Thus the object reverses direction at \(t=2\) seconds. An antiderivative of \(v(t)\) is \(F(t)=\frac{t^2}{2}-2t\). The displacement on the whole interval is

\int_0^5 (t-2)\,dt = F(5)-F(0) = \left(\frac{25}{2}-10\right)-0 = \frac{5}{2}\text{ m}.

The displacement from \(0\) to \(2\) is

F(2)-F(0) = (2-4)-0 = -2\text{ m}.

The displacement from \(2\) to \(5\) is

F(5)-F(2) = \frac{5}{2}-(-2) = \frac{9}{2}\text{ m}.

The object travels \(2\) metres in the first piece and \(\frac{9}{2}\) metres in the second. Therefore,

\text{total distance} = 2+\frac{9}{2} = \frac{13}{2}\text{ m}.

Since \(x(0)=10\) metres, the final position is \(x(5)=10+\frac{5}{2}=\frac{25}{2}\) metres. Check: the object first moves from \(10\) metres to \(8\) metres, then moves forward by \(\frac{9}{2}\) metres to \(\frac{25}{2}\) metres. The changes give displacement \(-2+\frac{9}{2}=\frac{5}{2}\) metres and distance \(2+\frac{9}{2}=\frac{13}{2}\) metres, as calculated.

Worked Example: Two Reversals

Let \(v(t)=(t-1)(t-3)\) metres per second on \(0\le t\le4\) seconds. Find the displacement and total distance.

The candidate reversal times are \(t=1\) and \(t=3\). The velocity is positive on \([0,1)\), negative on \((1,3)\), and positive on \((3,4]\). Both zeros are reversals. Expand the velocity and use the antiderivative

v(t)=t^2-4t+3, \qquad F(t)=\frac{t^3}{3}-2t^2+3t.

The signed changes on the three pieces are

\int_0^1 v(t)\,dt=F(1)-F(0)=\frac{4}{3}-0=\frac{4}{3}\text{ m}, $$ $$ \int_1^3 v(t)\,dt=F(3)-F(1)=0-\frac{4}{3}=-\frac{4}{3}\text{ m}, $$ $$ \int_3^4 v(t)\,dt=F(4)-F(3)=\frac{4}{3}-0=\frac{4}{3}\text{ m}.

Adding the signed changes gives displacement \(\frac{4}{3}-\frac{4}{3}+\frac{4}{3}=\frac{4}{3}\) metres. Adding their absolute values gives total distance \(\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4\) metres. Check: differentiating \(F\) gives \(F'(t)=t^2-4t+3=(t-1)(t-3)=v(t)\), and the three interval changes sum to \(F(4)-F(0)=\frac{4}{3}\) metres.

Worked Example: A Zero That Is Not a Reversal

Suppose \(v(t)=(t-1)^2(t-3)\) metres per second on \(0\le t\le4\) seconds. Find displacement and total distance. In particular, determine whether both velocity zeros are reversals.

The zeros are \(t=1\) and \(t=3\). The squared factor \((t-1)^2\) is positive on both sides of \(t=1\), while \(t-3\) is negative up to \(t=3\). Thus velocity is negative on both sides of \(t=1\): that zero is not a reversal. At \(t=3\), velocity changes from negative to positive, so \(t=3\) is a reversal.

Expand and integrate:

v(t)=t^3-5t^2+7t-3, \qquad F(t)=\frac{t^4}{4}-\frac{5t^3}{3}+\frac{7t^2}{2}-3t.

The displacement from \(0\) to \(3\) is \(F(3)-F(0)=-\frac{9}{4}\) metres. The displacement from \(3\) to \(4\) is \(F(4)-F(3)=\frac{43}{12}\) metres. Therefore the full displacement is

-\frac{9}{4}+\frac{43}{12} = -\frac{27}{12}+\frac{43}{12} = \frac{4}{3}\text{ m}.

The total distance adds the magnitudes of the two changes:

\frac{9}{4}+\frac{43}{12} = \frac{27}{12}+\frac{43}{12} = \frac{35}{6}\text{ m}.

Check: \(F'(t)=t^3-5t^2+7t-3=(t-1)^2(t-3)\). Also, the signed piece changes sum to \(F(4)-F(0)=\frac{4}{3}\) metres, while their magnitudes sum to \(\frac{35}{6}\) metres. The zero at \(t=1\) does not require a separate distance piece because the direction does not change there.

A Common Mistake

A tempting but incorrect method is to integrate velocity over the whole interval and then take the absolute value to get distance. In the first example, this would give

\left|\int_0^5 (t-2)\,dt\right| = \left|\frac{5}{2}\right| = \frac{5}{2}\text{ m}.

That is the magnitude of the displacement, not the total distance. The first \(2\) metres of travel were in the negative direction and the later \(\frac{9}{2}\) metres were in the positive direction. The correct total is \(2+\frac{9}{2}=\frac{13}{2}\) metres. Taking the absolute value after integration cannot restore motion that has already cancelled.

Practical check: Total distance must be at least the magnitude of displacement. If your computed distance is smaller, a sign or interval-splitting error has occurred.

Interpreting the Results

Displacement describes the change from the starting position to the ending position, and can be positive, negative, or zero. Total distance is nonnegative and records all motion, including motion that is later undone by a reversal. If there is no reversal and velocity has one sign throughout the interval, then total distance equals the magnitude of displacement. If the object reverses, the two quantities generally differ.

For a model, both answers can be useful. Displacement answers how far the final position is from the initial position in the signed coordinate system. Distance answers how much path the object travelled, regardless of direction. State which quantity is being requested before choosing the integral.

Check Your Understanding

For each question, distinguish net change from accumulated travel and check whether velocity actually changes sign.

If velocity is negative throughout an interval, how are displacement and total distance related?
For \(v(t)=t-4\) on \(0\le t\le6\), at what time does the object reverse direction?
Why is a time \(t_0\) satisfying \(v(t_0)=0\) not necessarily a reversal?
What does \(\left|\int_a^b v(t)\,dt\right|\) measure, and why can it differ from total distance?
In a calculation split at reversals, which quantities should be added with their signs, and which should be added by magnitude?

Displacement And Distance During Reversals

What You'll Learn

Why Reversals Change the Calculation

Displacement and Total Distance

How to Locate and Handle a Reversal

Why Splitting at Reversals Works