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A cart on a track

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A cart on a straight track has velocity $v (t) = t^{2} - 4 t + 3$ metres per second for $0 \leq t \leq 3$ . What total distance, in metres, does it travel in that time, counting motion in each direction? Give the answer as a fraction in lowest terms.

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Hint

Distance integrates the speed $∣ v (t) ∣$ , so split the interval where $v$ changes sign.

Solution

$v (t) = (t - 1) (t - 3)$ , so $v$ is positive on $[0, 1]$ and negative on $[1, 3]$ .
With $F (t) = \frac{t ^{3}}{3} - 2 t^{2} + 3 t$ , we get $F (0) = 0$ , $F (1) = \frac{1}{3} - 2 + 3 = \frac{4}{3}$ and $F (3) = 9 - 18 + 9 = 0$ .
The cart moves forward $F (1) - F (0) = \frac{4}{3}$ and back $F (1) - F (3) = \frac{4}{3}$ .
The net displacement is $0$ , but the total distance is $\frac{4}{3} + \frac{4}{3} = \frac{8}{3}$ .

Answer: $\frac{8}{3}$

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