Concept preview

What you will be able to do

1. 01Formulate motion integrals: velocity is an antiderivative of acceleration and position is an antiderivative of velocity, so $s(b)=s(a)+\displaystyle\int_a^b v(t)\,dt$ and $v(b)=v(a)+\displaystyle\int_a^b a(t)\,dt$ by the net change theorem.
2. 02Compute displacement as the signed integral $\displaystyle\int_a^b v\,dt$ and total distance as $\displaystyle\int_a^b |v|\,dt$, and explain precisely why they differ when velocity changes sign.
3. 03Set up a total-distance integral correctly by finding where $v=0$, splitting the interval, and integrating $|v|$ piecewise (flipping the sign on intervals where the particle moves backward).

The connection begins with the net change theorem, which is just the Fundamental Theorem of Calculus read as a statement about accumulation: for a function $F$ with derivative $F'$, $\int_a^b F'(t)\,dt=F(b)-F(a).$ The definite integral of a rate of change over $[a,\,b]$ equals the net change in the original quantity. Since velocity is the derivative of position, $v=s'$, setting $F=s$ gives $\int_a^b v(t)\,dt=s(b)-s(a)$. That difference — where the particle ends minus where it started — is the displacement.