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Total distance traveled from a velocity function with an absolute value

A particle moves along a straight line with velocity v(t)=∣t−3∣v(t) = |t - 3| (in meters per second), where tt is measured in seconds. Consider the time interval [0,6][0, 6].

Find the total distance traveled by the particle over the time interval [0,6][0, 6].

Total distance traveled from a velocity function with an absolute value graph
Hint

Since v(t)v(t) changes sign of its expression inside the absolute value at t=3t=3, split the integral ∫06∣t−3∣ dt\int_0^6 |t-3|\,dt into two pieces where v(t)v(t) can be written without absolute value bars, then integrate each piece separately and add the results.

Step-by-step solution

Since v(t)=∣t−3∣v(t) = |t-3|, the velocity is never negative, so total distance equals total displacement here: ∫06∣t−3∣ dt\int_0^6 |t-3|\,dt. Split at t=3t=3, the point where t−3t-3 changes sign: for 0≤t≤30 \le t \le 3, ∣t−3∣=3−t|t-3| = 3-t, and for 3≤t≤63 \le t \le 6, ∣t−3∣=t−3|t-3| = t-3.

Compute each piece: ∫03(3−t) dt=[3t−t22]03=9−4.5=4.5\int_0^3 (3-t)\,dt = \left[3t - \frac{t^2}{2}\right]_0^3 = 9 - 4.5 = 4.5. Similarly, ∫36(t−3) dt=[t22−3t]36=(18−18)−(4.5−9)=4.5\int_3^6 (t-3)\,dt = \left[\frac{t^2}{2} - 3t\right]_3^6 = (18-18) - (4.5-9) = 4.5.

Adding the two pieces gives the total distance traveled: 4.5+4.5=94.5 + 4.5 = 9 meters.

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