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

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

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

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 = - 0.13 m / s2 Note: the a is negative because the change in v is negative!!

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 = (10 - 10)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

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

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.

Kinematic #1

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

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

Does all this make sense? 13.5 m/s 1.5 m/s Total displacement = 7.50 + 30 = 37.5 m = Total AREA under the line.

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

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

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