Kinematic equations

See kinematics.

Constant acceleration

We can only use these where $a$ is constant, usually in projectile motion exercises.

$$v=v_0+a_{0}t$$ $$s=s_0+v_{0}t+\frac{1}{2}{a}_0{t}^2$$ $$v^2=v_0^2+2a_0(s-s_0)$$ $$s=\frac{(v+v_0)t}{2}$$

Acceleration as a function of time

Where $a=f(t)$. Recall $a=f(t)=\frac{dv}{dt}$. If we separate the variables and integrate:

$${\int }_{v_0}^{v}dv={\int }_0^{t}f(t)dt$$ $$v=v_0+{\int }_0^{t}f(t)dt$$

We use similar logic for the rest of the formulas, but I won’t hash it out again.

$$s=s_0+{\int }_0^{t}vdt$$

Acceleration as a function of velocity

Where $a=f(v)$.

$$t={\int }_0^{t}dt={\int }_{v_0}^v\frac{dv}{f(v)}$$

This above equation gives us $t$ as a function of $v$. We would then need to solve for $v$ as a function of $t$, then integrate $v=\frac{ds}{dt}$ to get $s$ as a function of $t$. We have another approach instead: where we use $vdv=f(v)ds$. This won’t give us $s$ with reference to $t$, however.

$$s=s_0+{\int }_{v_0}^v\frac{vdv}{f(v)}$$

Acceleration as a function of displacement

Where $a=f(s)$.

$${\int }_{v_0}^{v}vdv={\int }_{s_0}^{s}f(s)ds\to v^2=v_0^2+2{\int }_{s_0}^{s}f(s)ds$$

Next, we solve for $v$ and set it equal to $g(s)$, then substitute.

$$t={\int }_{s_0}^s\frac{ds}{g(s)}$$

Addendum

Often times when using the formulas for acceleration as a function of time, velocity, displacement, the integration just gets extremely tedious. So we typically integrate numerically, graphically, or using a computer, if necessary.