5.2 · Net change, displacement, and total distance

Displacement preserves velocity signs

For velocity $\displaystyle v(t)$, displacement from $\displaystyle a$ to $\displaystyle b$ is $\displaystyle \int_a^bv(t)\,dt$. Final position equals initial position plus this signed change. A negative result describes a net movement in the negative direction.

Total distance is $\displaystyle \int_a^b|v(t)|\,dt$. Find zeros and sign changes of velocity, then split the integral so each segment contributes positively.

$$\mathrm{distance}=\int_a^b|v(t)|\,dt$$

A zero of velocity is a candidate direction change

If $\displaystyle v(t)=t-1$ on $\displaystyle [0,3]$, the direction changes at $\displaystyle 1$: velocity is negative before and positive after. The two signed areas have different signs, but both contribute to distance.

Not every zero changes direction. If velocity touches zero and retains its sign, no absolute-value reversal is needed there. A sign chart makes the distinction explicit.

For an amount, subtract outflow from inflow

If $\displaystyle I(t)$ is inflow and $\displaystyle O(t)$ is outflow, the amount changes at rate $\displaystyle I(t)-O(t)$. Accumulate the net rate and add the initial amount. Do not integrate the initial amount a second time.

An amount can increase while its rate decreases, provided the net rate remains positive. When looking for a maximum amount, inspect zeros of net rate and endpoint amounts, just as in a derivative-based extremum problem.