up to freefall

content/physics/kinematics.md

@@ -48,5 +48,40 @@ a(t)=\tfrac{\mathrm{d}v}{\mathrm{d}t}=\lim_{\Delta t \to 0}{\tfrac{\Delta v}{\De
 $$
 Please take calculus/study calculus if you want a neuron or two to function during the course of this, well, course.
 
+## Motion at Constant Acceleration
+During this section and any point you use these formulas, always take $t_0$ to be equivalent to $0=t_1$. This allows for the following line of reasoning:
+$$
+\bar{v}=\frac{x-x_0}{t-t_0}=\frac{x-x_0}{t}
+$$
+This clearly makes sense because $x_0$ is taken to be the initial position of the particle in 1D, and $t_0$ is taken to be simply $0$, when the clock starts. Therefore acceleration must be a constant like so:
+$$
+a=\frac{v-v_0}{t}
+$$
+Firstly, it is important to note that all variables without subscripts are taken to be definitions of that variable completely. In contrast, when you saw that $v$ had a little *bar* on top of it, $\bar{v}$ meant to be the ***average*** of $v$. Moving on, if you are required to find the velocity of an object after elapsed time $t$:
+$$
+v=v_0+at 
+$$
+> [!NOTE]
+> **Example:** You are given that the acceleration of a motorcycle is constantly $4.0 \mathrm{m}/\mathrm{s}^2$. You need to find the speed of that motorcycle (note that we haven't given you any sense of position, $\therefore$ did not ask for velocity) after elapsed time $t=6.0\mathrm{s}$. The motorcycle starts from total rest, and when $t_1=t_0=0$. 
+
+> [!faq]- Answer
+> $4.0*6.0=24\mathrm{m}/\mathrm{s}$
+
+Next, calculating the position:
+$$
+x=x_0+\bar{v}t 
+$$
+Because of the symmetric and uniform growth of the velocity over time:
+$$
+\bar{v}=(\frac{v_0+v}{2})t=x_0+t(\frac{v_0+v_0+at}{2})=x_0+t(\frac{2v_0+at}{2})
+$$
+This finally results in the formula for position in these scenarios:
+$$
+x=x_0+tv_0+\tfrac{1}{2}at^2
+$$
+
+## Objects in free fall
+Just remember that $g=9.80\mathrm{m}/\mathrm{s}^2$
+
 
 #physics