I. Kinematics
    1. General notes
        a. Assign (or at least think about) a positive direction
        b. Be sure you have appropriate and consistent units
    2. Displacement
        a. Vector quantity - needs a direction
        b. Net change in position
        c. Distance may be the magnitude (size)
        d. In some instances the distance and the displacement are very different
            i. Example
    3. Velocity
        a. Vector quantity - needs a direction
        b. Average velocity = total displacement/total time      v = x / t
        c. May also look at instantaneous velocities - particularly initial and final
        d. Uniform velocity means no changes, all instantaneous are the same (and the same as the average)
        e. Example
    4. Acceleration
        a. Vector quantity - needs a direction
            i. Either + or - is sufficient at this point
        b. Means the velocity is changing
        c. Acceleration = change in velocity / change in time a = (v_f - v_o) / t
        d. Example
    5. Kinematic equations
        a. Put different combinations of information about motion together
        b. See table 2.3 on page 36 in your text
        c. Other notes about using the equations
        d. There is generally more than one way to solve motion problems
            i. The more information you have, the more choices you have
        e. Some examples
 

II. Free Fall
    1. Acted on only by gravity
    2. Uniformly Accelerated motion, so the same motion equations apply
    3. Assign a positive direction
    4. a is known - it's the acceleration due to gravity
        a. a = g = 9.8 m/s²
        b. the direction of g is always straight downward
    5. Three basic cases or situations - each will have specific implied information
        a. dropped
        b. thrown downward
        c. thrown upward
    6. May be non-symmetrical
        a. see example 2.6 (page 42) in textbook
        b. another example
 

Examples:

1) Assume you are running on a circular track and the diameter of the track is 2.00 m.
        a) If you run halfway around the track, what are the distance and the displacement traveled?

                Distance = ½ circumference = 3.14 m

                Displacement = diameter = 2.00 m
                        (the straight-line distance from one point on a circle to another point on the same circle)

        b) If you run all the way around the track, what are the distance and the displacement traveled?

                Distance = circumference = 6.28 m

                Displacement = 0      (You are back at your starting point, therefore no displacement has taken place)
 

2. It took a person 1.00 minute to travel 95.0 m. What was his average velocity?

        v = x / t = 95.0 m / 60.0 s = 1.58 m/s
            (Direction of the velocity is the same as the direction of the displacement, + in this case.)

3. A car speeds up from rest to 120 m/s in 10.0 seconds. What was its acceleration?

            a = (v_f - v_o) / t = (120 m/s - 0) / 10.0 s = 12.0 m/s²

4. Notes about equations:
 

Equation	Notes about use
x = vavg t	Use for uniform velocity; if the velocity is not uniform, use in conjunction with the next equation
vavg = ½(vf + vo)	Use to find vavg when the velocity is not uniform
vf = vo + a t	No displacement (x); use when you don't know and aren't trying to find x
x = vo t + ½ a t2	No final velocity (vf); use when you don't know and aren't trying to find vf
vf2 = vo2 + 2 a x	No time (t); use when you don't know and aren't trying to find t

5. Using the equations

A truck accelerates from 0 to 28.0 m/s in 10.0 seconds. Find

    a) its acceleration

        v_f = v_o + a t; solve for acceleration;  a = (v_f - v_o) / t = (28.0 m/s - 0) / 10.0 s = 2.80 m/s²

    b) its displacement

        x = v_o t + ½ a t² = 0 + ½ (2.80 m/s²)(10.0 s)² = 140 m

            or use v_f² = v_o² + 2 a x and solve for x

            or use x = v_avg t with v_avg = ½(v_f + v_o) to find x

        Try it!  All will give the same result (within some minor differences if you have done any rounding).
 

6. Free Fall - Dropped

Implied Information a = g; v_o = 0; let downward be positive

    A ball is dropped from a height of 10.0 m above the ground.

        a) How long does it take to hit the ground?

                y = v_ot + ½ a t² with v_o = 0 and solving for t

                t = [2 y / g]^½ = [2 (10.0 m) / 9.8 m/s²]^½ = 1.43 s

        b) How fast is it going just before it hits the ground?

                v_f = v_o + g t = 0 + (9.8 m/s²)(1.43 s) = 14.0 m/s

                or v_f² = v_o² + 2 a x = 0 + 2 (9.8 m/s²)(10.0 m)
                            v_f = 14.0 m/s

7. Free Fall - Thrown downward

Implied Information a = g; v_o not = 0, otherwise the same approach; let downward be positive

    A ball is thrown downward with an initial velocity of 5.00 m/s from a height of 10.0 m above the ground.

        a) How long does it take to hit the ground?

                y = v_ot + ½ a t² with v_o 0 and solving for t will give a quadratic equation so solve for v_f first

        (Using the quadratic equation will give two roots, you can discard the negative root and the positive root is 1.01 s.)

        b) How fast is it going just before it hits the ground?

                v_f² = v_o² + 2 a x = (5.00 m/s)² + 2 (9.8 m/s²)(10.0 m)
                            v_f = 14.9 m/s

        Then back to part a:

                v_f = v_o + g t and t = (v_f - v_o)/g = (14.9 m/s - 5.00 m/s)/9.8 m/s² = 1.01 s

8. Free Fall - thrown upward

Implied Information a = g; v_o  not = 0; v at the maximum height = 0; let upward be positive; if symmetrical, then v_f =-v_o and t_up = t_down which also gives t_total = 2 t_up

    A ball is thrown straight upward with an initial velocity of 10.0 m/s.

            a) How high does it go?

                    v_f² = v_o² + 2 g y and solve for y, v_f = 0 since y is at maximum height

                    y = (v_f² - v_o²)/(2g) = [0 - (10.0 m/s)²]/[2(-9.8 m/s²)] = 5.10 m

            b) What is the total time in flight?

                    t_total = 2 t_up and t_up = (v_f - v_o)/g = (0 - 10.0 m/s)/(-9.8 m/s²) = 1.02 s

                            so t_total = 2 t_up = 2(1.02 s) = 2.04 s

9. Free Fall - non symmetrical

    A ball is thrown straight upward with an initial velocity of 8.00 m/s and caught on its way back down by another person 1.50 m above the thrower. (Let upward be positive.)

    a) What is the maximum height the ball reaches?

            v_f² = v_o² + 2 g y and solve for y, v_f = 0 since y is at maximum height

            y = (v_f² - v_o²)/(2g) = [0 - (8.00 m/s)²]/[2(-9.8 m/s²)] = 3.27 m

    b) How long is the ball in flight?

        t_total = t_up + t_down (but t_up not = t_down since it is not symmetrical)

            t_up = (v_f - v_o) / g = (0 - 8.00 m/s) / (-9.8 m/s²) = 0.816 s (v_f = 0 since it's at the maximum height)

            For t_down use y = v_ot + ½ a t² with v_o = 0 since it starts at the maximum height
                y = 3.27 m - 1.50 m (the maximum height - height where it's caught) = 1.77 m

            t_down = [2 y / g]^½ = [2 (-1.77 m)/ -9.8 m/s²]^½ = 0.600 s

        Finally, t_total = t_up + t_down = 0.816 s + 0.600 s = 1.416 s

You could also use y = v_ot + ½ a t² and solve the quadratic equation; y = 1.50 m. The two roots you obtain are t = 0.216 s and t = 1.42 s. The 0.216 s represents the time the ball reaches 1.50 m above it starting point on the way up and so we need the second root for the way back down.