Why Velocity’s Sign Matters
A velocity integral gives displacement: the net change in position, including direction. Total distance traveled is different. If a particle moves forward and then reverses direction, the backward part reduces its displacement but still adds to its total distance.
To find total distance from a velocity function, identify when velocity is positive or negative. Those sign changes mark times when the particle switches direction. The total distance is found by adding the distance traveled on each resulting time interval.
Velocity, Direction, and Distance
Velocity includes both speed and direction. Where \(v(t)>0\), the particle moves in the positive direction; where \(v(t)<0\), it moves in the negative direction. Where \(v(t)=0\), it is instantaneously stationary. The absolute value \(|v(t)|\) is the particle’s speed, which is nonnegative in either direction.
For a time interval on which velocity is nonnegative, \(|v(t)|=v(t)\). On an interval where velocity is nonpositive, \(|v(t)|=-v(t)\). These identities also hold at times when velocity is zero. Therefore, once the velocity’s sign is known on each interval, the absolute-value integral can be evaluated using ordinary integrals with the appropriate signs.
In typical AP problems, \(v\) is continuous and its zeros can be found algebraically or from given information. A zero inside the interval is a candidate point for splitting the integral. It is important to check the sign on both sides: not every zero is a sign change. For example, a velocity function may touch zero and remain negative on both sides.
A Derivation of the Total-Distance Formula
Let \(x(t)\) be the particle’s position, so that \(x'(t)=v(t)\). Suppose first that the interval is divided into subintervals on which velocity has a constant sign. On any one of them, position is monotonic: it increases if velocity is positive and decreases if velocity is negative. Thus the distance traveled on that subinterval is the absolute value of the change in position.
Let its endpoints be \(u\) and \(w\), and suppose velocity has one sign throughout it, allowing zeros at its endpoints.
The position change is \(x(w)-x(u)=\int_u^w x'(t)\,dt=\int_u^w v(t)\,dt\).
Because position is monotonic on this subinterval, its distance traveled is \(|x(w)-x(u)|=\left|\int_u^w v(t)\,dt\right|\).
If \(v\geq 0\), the distance is \(\int_u^w v(t)\,dt=\int_u^w |v(t)|\,dt\). If \(v\leq 0\), it is \(-\int_u^w v(t)\,dt=\int_u^w |v(t)|\,dt\).
Adding over all sign-consistent subintervals gives the total distance \(\int_a^b |v(t)|\,dt\).
This reasoning explains why a sign change requires a split: a single signed integral across a reversal can cancel motion in opposite directions. The formula using \(|v|\) prevents that cancellation.
A Reliable Calculation Method
For a continuous velocity function on \([a,b]\), use the following process. Find its zeros in the interval, including endpoints if useful for describing the sign pattern. Use the zeros inside the interval as possible split points. Test the sign of \(v\) between consecutive zeros, then integrate \(v\) where it is nonnegative and \(-v\) where it is nonpositive. Add all the resulting nonnegative distances.
A sign chart is often helpful. The sign at a test point between consecutive zeros determines the sign throughout that interval when there are no other zeros there and the function is continuous. A zero where signs on either side agree does not represent a reversal, although splitting there is still valid if the integral is evaluated consistently.
Worked Examples
Worked Example: A Velocity That Changes Sign Once
A particle has velocity \(v(t)=t^2-4\) for \(0\leq t\leq 3\). Find its total distance traveled.
First find the zeros:
Only \(t=2\) lies in \([0,3]\). Test the intervals: \(v(0)=0^2-4=-4<0\), while \(v(3)=3^2-4=5>0\). Thus velocity is negative on \([0,2)\) and positive on \((2,3]\). The particle reverses direction at \(t=2\).
Integrate speed by splitting at that time:
Check by using the position antiderivative \(F(t)=\frac{t^3}{3}-4t\), for which \(F'(t)=t^2-4=v(t)\). On \([0,2]\), the position change is \(F(2)-F(0)=-\frac{16}{3}\), so the distance there is \(\frac{16}{3}\). On \([2,3]\), it is \(F(3)-F(2)=\frac{7}{3}\), so that distance is \(\frac{7}{3}\). Their sum is \(\frac{23}{3}\), as calculated.
Worked Example: Two Direction Changes
Suppose \(v(t)=(t-1)(t-3)\) on \(0\leq t\leq 4\). Find the total distance traveled.
The zeros are \(t=1\) and \(t=3\), so test the intervals \([0,1]\), \([1,3]\), and \([3,4]\). At \(t=0\), \(v(0)=3>0\). Check the factor signs: for \(1<t<3\), \(t-1>0\) and \(t-3<0\), so their product is negative. In particular, at \(t=2\), \(v(2)=(2-1)(2-3)=-1<0\), so the velocity is negative between the roots. For \(t=4\), \(v(4)=3>0\). The signs are therefore positive, negative, positive.
Expand the velocity and find an antiderivative:
Integrate \(v\) on the positive intervals and \(-v\) on the negative interval:
The endpoint values are \(F(0)=0\), \(F(1)=\frac13-2+3=\frac43\), \(F(3)=9-18+9=0\), and \(F(4)=\frac{64}{3}-32+12=\frac43\). Hence
Check each contribution using position changes: the changes are \(\frac43\), \(-\frac43\), and \(\frac43\). Their absolute values sum to \(4\), confirming the result. The sign test at \(t=2\) is essential; multiplying the factors gives a negative value there.
Worked Example: A Zero Without a Sign Change
Let \(v(t)=-(t-1)^2\) for \(0\leq t\leq 3\). Find the total distance traveled and explain what happens at \(t=1\).
The velocity is zero when \(-(t-1)^2=0\), so \(t=1\). For every \(t\neq1\), \((t-1)^2>0\), and therefore \(v(t)<0\). The velocity is negative on both sides of \(t=1\): the particle does not reverse direction there. It is instantaneously stationary at that one time.
Since \(v\leq0\) throughout, speed is \(-v(t)=(t-1)^2\). Thus
For a check, a position antiderivative is \(F(t)=-\frac{(t-1)^3}{3}\), since \(F'(t)=-(t-1)^2=v(t)\). The position change is \(F(3)-F(0)=-\frac83-\frac13=-3\), whose absolute value is \(3\). Because velocity is nonpositive throughout, this equals the total distance.
Common Mistake: Using Displacement as Distance
A frequent error is to calculate \(\int_a^b v(t)\,dt\) and report its value as total distance, even when velocity changes sign. For instance, let \(v(t)=t-1\) on \([0,2]\). Its signed integral is
That zero is the displacement, not the distance. The velocity is negative for \(0\leq t<1\) and positive for \(1<t\leq2\), so the distance must be split at \(t=1\):
The correction is to integrate \(|v(t)|\), or equivalently to split at the sign change and make each interval’s contribution nonnegative. A related mistake is assuming every zero is a reversal. The third example shows why the signs on both sides of a zero must be checked.
Why This Method Is Useful
Velocity describes motion direction as well as rate of position change. Integrating signed velocity answers how far the particle ends from its starting position, with direction accounted for. Integrating speed answers how much path the particle has covered. Keeping those questions distinct is especially important whenever a particle reverses direction.
For a final check, each interval’s distance contribution should be nonnegative. If an interval produces a negative contribution, the sign of the integrand was not handled correctly. One can also check each part by evaluating the corresponding position change and taking its absolute value.
Check Your Understanding
For each question, state the velocity sign pattern before evaluating any distance integral.
- For \(v(t)=t-2\) on \([0,4]\), where does the velocity change sign, and what integral gives total distance?
- For \(v(t)=4-t^2\) on \([0,3]\), find the total distance traveled.
- Explain why a zero of \(v\) does not necessarily indicate a direction change.
- For \(v(t)=(t-2)^2\) on \([0,4]\), does the particle reverse direction at \(t=2\)? Explain using signs.
- In one sentence, distinguish \(\int_a^b v(t)\,dt\) from \(\int_a^b |v(t)|\,dt\).
