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Total Distance Traveled — Definition, Formula & Examples

Total distance traveled is the cumulative length a particle covers along its path, regardless of direction. Unlike displacement, it counts every bit of motion — even when the particle reverses.

For a particle with velocity function v(t)v(t) on the interval [a,b][a, b], the total distance traveled is ∫ab∣v(t)∣ dt\int_a^b |v(t)|\, dt. This quantity is always non-negative and equals the sum of the distances covered on each sub-interval where the particle moves without changing direction.

Key Formula

Total Distance=∫ab∣v(t)∣ dt\text{Total Distance} = \int_a^b |v(t)|\, dt
Where:
  • v(t)v(t) = Velocity of the particle at time t
  • aa = Start of the time interval
  • bb = End of the time interval

How It Works

To find total distance traveled, you integrate the absolute value of the velocity function. In practice, this means you first find where v(t)=0v(t) = 0 to locate direction changes. Then you split the integral at those zeros, take the absolute value of each piece, and add the results. If v(t)≥0v(t) \ge 0 on an interval, the integral gives distance directly; if v(t)<0v(t) < 0, you negate the integral on that interval to make it positive.

Worked Example

Problem: A particle moves along a line with velocity v(t) = t² − 4 for 0 ≤ t ≤ 3. Find the total distance traveled.
Find where velocity is zero: Set v(t) = 0 and solve.
t2−4=0  ⟹  t=2(only t=2 lies in [0,3])t^2 - 4 = 0 \implies t = 2 \quad (\text{only } t=2 \text{ lies in } [0,3])
Integrate on each sub-interval: On [0, 2], v(t) < 0, so |v(t)| = 4 − t². On [2, 3], v(t) > 0, so |v(t)| = t² − 4.
∫02(4−t2) dt=[4t−t33]02=8−83=163\int_0^2 (4 - t^2)\, dt = \left[4t - \frac{t^3}{3}\right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}
Evaluate the second integral: Integrate v(t) on [2, 3] where it is positive.
∫23(t2−4) dt=[t33−4t]23=(9−12)−(83−8)=−3+163=73\int_2^3 (t^2 - 4)\, dt = \left[\frac{t^3}{3} - 4t\right]_2^3 = \left(9 - 12\right) - \left(\frac{8}{3} - 8\right) = -3 + \frac{16}{3} = \frac{7}{3}
Add the pieces: Total distance is the sum of the two non-negative results.
163+73=233\frac{16}{3} + \frac{7}{3} = \frac{23}{3}
Answer: The total distance traveled is 233\dfrac{23}{3} units (approximately 7.67 units).

Why It Matters

AP Calculus free-response questions frequently ask for total distance traveled alongside displacement. Confusing the two costs points. This concept also appears in physics whenever you need the odometer reading rather than the net change in position.

Common Mistakes

Mistake: Computing the plain integral ∫abv(t) dt\int_a^b v(t)\, dt without absolute value signs.
Correction: That gives displacement, not total distance. You must split the integral at zeros of v(t)v(t) and take the absolute value of each piece so that motion in both directions counts as positive.

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