II. Applications
    A. Police
        1. In a car accident, Police can find the distance a car traveled using acceleration, time and velocity to recover the car
        2. When bullets are fired in a crime, the distance of the bullet can be calculated to recover the bullet for evidence
    B. Military
        1. In firing a shell, the distance the shell will travel due to the velocity the gun discharges to result in an accurate shot.
        2. In dropping bombs, using a combination of the equations, the exact place that the bomb will land can be calculated.

        3. In recovering ejected pilots and wrecked planes the kinematic equations are used to discover their locations.
    C. Sports
        1. The distance a football will travel due to its hang time can be calculated to know the best way to punt the ball.
        2. In baseball the angle by which a hitter can get the most distance can be calculated to show the batter how to hit the ball and the pitcher how to throw it.
III.  Meaning of Variables

X=Distance in meters   V=Final Velocity in meters per second   Vo=Initial Velocity in meters per second   A=Acceleration in meters squared per second  T=Time in seconds
              (Note: A sub letter of  h or v represents the horizontal or vertical aspect, respectively, of that variable)
IV.  Important Concepts
    A. In every launched, dropped, or otherwise airborne and motorless object, there are horizontal and vertical velocities that are independent of one another.
        1. The horizontal velocity does not change while the object is in the air unless it is acted upon by another object or motor.
        2. The vertical velocity is constantly acted upon by gravity.  When an object is rising it has and acceleration of -9.8 m/s^2 and while it is falling it has an acceleration of 9.8 m/s^2.  The object rises for have of its time in the air and falls for half of its time in the air.
    B. Objects have a tendency to remain still until they are acted upon
     C. All four kinematic equations can be used in place of one another.
    D. The kinematic equations can only be used if there is an acceleration of some sort.  If there is not, the formula for velocity (V=X/T) is used to find distance, time and velocity.
    E. A launched motorless object has the same velocity at the time it hits the ground as when it was launched, assuming level ground, no wind and no airborne collisions.
V. Related formulas
                                                For a launched object:
Horizontal velocity(Vh)=cosine of the angle it is launched from x the original velocity
Horizontal velocity=distance/time
Vertical velocity(Vv)=sine of the angle it is launched from x the original velocity
                                                         The First two Kinematic Equations
V=Vo+AT
X=1/2(V+Vo)T

VI. Practice Problems
    1.A car traveling at 30m/s accelerates at 2.14m/s^2.  How far does it travel if it accelerates for 10 seconds?
                                                        X=VoT+1/2(AT^2)

                                    X=30m/s(10s)+1/2(2.14m/s^2)(10^2s)
                                    X=407m

    2.A shell is fired at a 34° angle with an initial velocity of 34m/s. Solve for the highest point the shell reaches(Xv), the distance it travels(Xh), and the time it is in the air.
                                                Vh=cos(34°) x 34m/s=24.19m/s
                                                Vv=sin(34°) x 34m/s=19.01m/s

                              V=Vo+AT                                                    Vh=Xh/T
                          0m/s=19.01m/s+-9.8m/s^2(1/2T)          24.19m/s=Xh/3.88s
                          1/2T=1.94s                                                  Xh=93.86m
                               T=3.88s

                                                    Xv=Vo(1/2T)+1/2(A(1/2T)^2)
                                                    Xv=19.01m/s(1.94)+1/2(-9.8m/s^2(1.94^2))
                                                    Xv=18.44m

    3.  If a car traveling at 95m/s and then accelerates at 10 meters per second over a distance of 9000 meters.  What is the car's final velocity.
                                                V^2=Vo^2+2AX
                                                V^2=95^2m/s+2(10m/s^2)(9000m)
                                                V^2=189025m/s
                                                   V=434.77m/s

Click here to see the first two kinematic equations.
Click here to learn about graphing.
Click here to see the Physics Home Page.