# Kinematic Equations

1. Oct 24, 2004

### Tom McCurdy

Since a good amount of the posts refer to problems involving kinematics I figured this may be helpful

$$V_0 =$$ Inital Velocity
$$V =$$ Final Velocity
$$t=$$ time
$$a=$$ acceleration
$$x=$$ distance

$$x=V_0t+\frac{1}{2}at^2$$

$$V=V_0+at$$

$$V^2=V_0^2+2ax$$

$$x= \frac{1}{2}(V_0+V)*t$$

Last edited: Oct 24, 2004
2. Oct 24, 2004

### HallsofIvy

Staff Emeritus
Assuming that you mean "x is the distance traveled in time t" and that a is a constant, then
$$x= V_0t+ \frac{1}{2}t^2$$
NOT
$$x= V_0t*\frac{1}{2}t^2$$

3. Oct 24, 2004

### Tom McCurdy

Very True... it does equal plus