# Absolute value and distance traveled

I don't know what to do again. I can get the displacement (but don't know what it means) but I can't get the distance traveled.

The velocity function is given for a particle moving along a lone. Find the displacement and the distance traveled by the particle during the given time interval.

$v(t) = 3t - 5$, $0 \leq t \leq 3$.

I know that the anti derivative is what I want and that is

$\frac {3}{2} t^2 - 5t$

I know that the function is negative from 0 - 3 so what do I do?

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Write $|v(t)|$ as a piecewise function: $$|v(t)|= \cases{5-3t, & 0\le t\le 5/3 \cr 3t-5, & 5/3\le t\le 3 }$$ Then, split the integral $\int_0^3 |v(t)|\,dt$ into two pieces.

$$\int_0^3 |v(t)|\,dt =\int_0^{5/3} |v(t)|\,dt +\int_{5/3}^3 |v(t)|\,dt =\int_0^{5/3} 5-3t\,dt +\int_{5/3}^3 3t-5\,dt,$$ and evaluate.

Essentially, find the set over which $v$ is positive and the set over which $v$ negative; then integrate $|v|$ over these sets separately. On the set where $v$ is negative, you'd integrate $-v$ and over the set where $v$ is positive, you'd integrate $v$.

The difference between displacement and total distance should be clear: here the point travels to the left the first $5/3$ seconds then travels to the right from $t=5/3$ to $t=3$. The displacement is the difference between the final and initial position of the point. Since the point traveled left and then right, this will be less than the total distance traveled (which is the sum of the distance traveled left with the distance traveled right).

For example if you move along a line left $4$ units then right $5$ units, the displacement is $1$ and the total distance traveled is $9$.

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If you know where the function is negative and positive (where it changes direction - either by plotting it or other means), then you can write the displacement as two integrals.

You would make the integral positive from 0 to 3 (by simply throwing a negative outside) then and leave the second integral alone (since it is positive already.) If you showed your work so far, it would be easier to see where you went astray on this step.

By the way, if your equation is $0 = 3t - 5$, then $t = \dfrac{5}{3}$ not $3$ sec - which means that on a velocity-time graph, the velocity is instantaneously zero and is changing between negative and positive.

Edit: It is easy to see the difference between distance and displacement with an example. Say you start at $0$ meters and travel $10$ meters to the right (positive) and then $10$ meters back to the left (negative). Your total displacement would be $0$ meters (since you started at zero and ended at zero, whereas the total distance you traveled would obviously be $10 + 10$ meters = $20$ meters.

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