To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function.
-## Kinematics
-
-$${dV \over dt} = {\operatorname{change in volume} \over \operatorname{respect to time}}$$
-
-**position $x$** - distance from origin or fixed point
-**displacement $s$** - change in position from starting point (vector)
-**velocity $v$** - change in position with respect to time
-**acceleration $a$** - change in velocity
-**speed** - magnitude of velocity
-
-$$v_{\operatorname{avg}}={\Delta x \over \Delta t}={{x_2 - x_1} \over {t_2 - t_1}}$$
-$$\operatorname{speed}_{\operatorname{avg}}={\Delta v \over \Delta t}$$
-
--- /dev/null
+# Kinematics
+
+$${dV \over dt} = {\operatorname{change in volume} \over \operatorname{respect to time}}$$
+
+**position $x$** - distance from origin or fixed point
+**displacement $s$** - change in position from starting point (vector)
+**velocity $v$** - change in position with respect to time
+**acceleration $a$** - change in velocity
+**speed** - magnitude of velocity
+
+$$v_{\operatorname{avg}}={\Delta x \over \Delta t}={{x_2 - x_1} \over {t_2 - t_1}}$$
+$$\operatorname{speed}_{\operatorname{avg}}={\Delta v \over \Delta t}$$
+
+## Constant acceleration
+
+| | no |
+| - | -- |
+| $v=u+at$ | $s$ |
+| $s=ut + {1 \over 2} at^2$ | $v$ |
+| $v^2 = u^2 + 2as$ | $t$ |
+| $s= {1 \over 2}(u+v)t$ | $a$ |
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