1. For uniform motion (constant velocity), the equation
v = ∆d/∆t or v = (df – di)/ ∆t
can be used to describe the motion
∆d = v ∆t can only be used if the velocity is constant.
For uniform accelerated motion (changing velocity, constant acceleration), the equations
vf = v0 + a t
vavg = (v0 + vf ) / 2
d = v0 t + ½ a t 2
vf2 – v02 = 2 a d

2. For constant velocity, v = ∆d/∆t will give you the velocity at any time (since the velocity is constant)
For changing velocity, v = ∆d/∆t will give you the average velocity for a time interval
NOT the change in velocity
NOT the final velocity

3. Kinematics Equations that Make Sense For Uniform Acceleration
Average velocity: vav = ∆d / ∆t SI unit: m/s
vav = (df - di) / t
df = di + vav t
Average acceleration: aav = ∆v / ∆t SI unit: m/s/s = m/s2
aav = (vf - vi) / t
vf = vi + aav t
Note: if the time intervals are very small we call these quantities instantaneous

4. Using Split Times! Position (m) 5 10 15 20 25 Split Time (s) 1.6 2.4
3.0
3.5
4.0
Av. Velocity (m/s)
3.1
2.1
1.7
1.4
1.2(5)
Determine the average velocity for each distance interval
Determine average velocity of the object over the time recorded
vav = df - di / t
= 25m - 0m / 14.5 s
= 1.7 m / s
Determine the average acceleration over the time recorded
a = vf - vi / t
= (1.25 m/s - 3.1m/s) / 14.5 s
= m / s2
Note: the a is negative because the change in v is negative!!

5. Constant Motion aav = (10 - 10)m/s / (5 - 0)s
On the d-t graph at any point in time … vav = ∆d / ∆t
On the v-t graph at any point in time… aav = vf - vi / t
On the a-t graph the area between the line and the x-axis is….
aav = ( )m/s / (5 - 0)s
vav = (50 - 0)m / (5 - 0)s
Area of rectangle = b x h
aav = 0 m/s2
Area = 5s x 0 m/s2 = 0 m/s
vav = 10 m/s
Looking at the area between the line and the x-axis….
The area thus represents….
The slope is constant on this graph so the velocity is constant
∆v = aav ∆t
Area of rectangle = b x h
Area = 5s x 10 m/s = 50 m
Change in velocity
Which is of course displacement

6. Changing Motion aav = (20 - 0)m/s / (5 - 0)s aav = 4 m/s2
On the d-t graph at any point in time … vav = ∆d / ∆t
On the v-t graph at any point in time… aav = vf - vi / t
On the a-t graph the area between the line and the x-axis is….
aav = (20 - 0)m/s / (5 - 0)s
The slope is constantly increasing on this graph so the velocity is increasing at a constant rate
Area of rectangle = b x h
aav = 4 m/s2
Area = 5s x 4 m/s2 = 20 m/s
Looking at the area between the line and the x-axis….
The slope of a tangent line drawn at a point on the curve will tell you the instantaneous velocity at this position
The area thus represents….
Area of triangle = 1/2 (b x h)
Change in velocity
Area = 1/2 (5s x 20 m/s) = 50 m
Which is of course displacement

7. The 3 Kinematic equations
There are 3 major kinematic equations than can be used to describe the motion in DETAIL. All are used when the acceleration is CONSTANT.

8. Kinematic #1

9. Kinematic #1
Example: A boat moves slowly out of a marina (so as to not leave a wake) with a speed of 1.50 m/s. As soon as it passes the breakwater, leaving the marina, it throttles up and accelerates at 2.40 m/s/s.
a) How fast is the boat moving after accelerating for 5 seconds?
What do I know?
What do I want?
vo= 1.50 m/s
v = ?
a = 2.40 m/s/s
t = 5 s
13.5 m/s

10. Kinematic #2
b) How far did the boat travel during that time?
37.5 m

11. Does all this make sense?
13.5 m/s
1.5 m/s
Total displacement = = 37.5 m = Total AREA under the line.

12. Interesting to Note A = HB
Most of the time, xo=0, but if it is not don’t forget to ADD in the initial position of the object.
A=1/2HB

13. Kinematic #3 What do I know? What do I want? vo= 12 m/s x = ?
Example: You are driving through town at 12 m/s when suddenly a ball rolls
out in front of your car. You apply the brakes and begin decelerating at
3.5 m/s/s.
How far do you travel before coming to a complete stop?
What do I know?
What do I want?
vo= 12 m/s
x = ?
a = -3.5 m/s/s
V = 0 m/s
20.57 m

14. Common Problems Students Have
I don’t know which equation to choose!!!
Equation
Missing Variable x v t