What does the area under an acceleration-displacement curve represent?

Considering the equation where,

$$\frac {1}{2} \left (v^2_f - v^2_i \right) = \int_0^s ads\,$$

What does the left-hand side of the equation actually represent? Is there an intuitive explanation similar to how the area under an a-t curve yields velocity and the area under a v-t curve yields displacement?

• Welcome to stack exchange. The answer depends on the system you are currently interested in. E.g. it could be the kinetic energy per unit mass. Thus, it would be great if you could provide more information. – Semoi Jan 19 '20 at 16:13
• The equation was given without much elaboration in Hibbeler's book on Dynamics. Perhaps it was kept general since, as you said, its meaning will depend on the system. Still, knowing that it could represent more than one thing, such as "the kinetic energy per unit mass", is quite helpful to building intuition. – Octavius Jan 19 '20 at 16:34
• This is one of the four usual motion equations. Typically shown in this form: $$v_f^2=v_i^2+2a(s_f-s_i)$$ – Steeven Jan 19 '20 at 16:40

As Newton says $$F=ma$$ we have $$a=F/m$$. Now force times distance is work and, for a body experiencing only the force $$F$$, the amount of work done $$W=\int_{\rm start}^{\rm end} F\,ds$$ is the change in kinetic energy
$$W= \Delta {\rm K.E.} =1/2 mv_{\rm end}^2 - 1/2 mv_{\rm start}^2$$.
• @Octavius Yes, correctly understood. The area under the curve of an $a$-$s$ graph is the value of your left-hand-side expression $\frac 12(v_f^2-v_i^2)$. Multiplying that area with $m$ means multiplying $\frac 12(v_f^2-v_i^2)$ with $m$. That gives you exactly the expression Mike shows here - which equals the total work done by the (total) force that causes the acceleration. – Steeven Jan 19 '20 at 16:44