Kinematics Study of an object’s motion without regard to the forces causing the motion.

Kinematics asks three questions at a particular instant in time  Where is the object?  Position: distance and direction (“displacement”) from a given reference point. (Ex: 4 meters East of the wall)  How fast is the object moving and in what direction?  Instantaneous Velocity: The rate at which the object changes position. (Ex: 4 meters per second East.)  How fast is the object speeding up, slowing down or changing direction?  Acceleration: The rate at which velocity changes. (Ex: 4 meters per second each second East).

Two Types of Quantities in Physics  Scalars: Have “magnitude” (amount with units) only.  Ex: Speed: 4 meters per second  Vectors: Have magnitude and direction.  Ex: Velocity: 4 meters per second East  Identify the following as a vector or scalar:  A mass: 6 kilograms (6 kg)  A distance: 6 meters (4m)  A displacement: 6 meters (6m) East  A position: 6 meters (6m) East of the wall.

Table of Kinematic Quantities QUESTIONSCALARVECTORS.I. UNIT When? Where? How Fast? Changing Velocity?

Table of Kinematic Quantities QUESTIONSCALARVECTOR (vector symbols are in boldface) S.I. UNIT When?Time (t) Second (s) Where?Distance (d)Position (x, y) Displacement (d) Meter (m) How Fast?Instantaneous Speed (v) Instantaneous Velocity (v) m/s Changing Velocity? Acceleration (a) m/s/s, or m/s 2

Average Speed and Velocity (Always applies to a TRIP) Average Speed = (distance of trip)/(time of trip) V AV = d/t Average Velocity = (displacement of trip)/(time of trip) V AV = d/t (Note: The boldface type means the quantity ia a vector)

How does average walking pace vary from person to person? Each person walks naturally along the route specified: d = ______ ft. The time for the walker is measured by his/her partner with a stopwatch. The average speed is then: v AV = d/t Convert from feet per second to miles per hour. If the trip’s endpoint is the same as the starting point, what is each person’s average velocity?

How do we convert from feet per second to miles per hour?

Three Cases of Kinematic Motion Definition of Acceleration: a = Δv/ Δt In each case we are given: Initial velocity (v i ) and acceleration (a) We’ll look at three cases: – Case 1: Constant velocity Motion: v i ≠ 0, a = 0 – Case 2: Accelerating from Rest: v i = 0, a ≠ 0 – Case 3: Accelerating from Nonzero initial velocity: v i ≠ 0, a ≠ 0 In each case, we want to know: – Velocity at a particular time (t). How Fast? – Distance traveled after a particular time (t). How Far? – We plot v vs. t by knowing v i and a.

Case 1: Constant Velocity Motion Car moving at constant velocity: v i ≠ 0, a = 0 Ex: v i = 4 m/s, Plot v vs. t What does the slope of the v vs. t graph mean? What does the area under the v vs. t graph mean? Plot d vs. t t (s)v (m/s)d (m)

Case 2: Accelerating from Rest: v i =0, a ≠ 0 A car accelerates from rest at a constant rate of 2 m/s 2 Plot v vs. t What does the slope of the v vs. t graph mean? What does the area under the v vs. t graph mean? Can we find v and d using algebra? Plot d vs. t t (s)v (m/s)d (m) 0V i =

How do free falling objects differ in acceleration due to gravity? Measure the distance (d) from the ceiling to the floor. Drop various objects and time how long it takes each one to hit the floor (t). Describe each object: Ping-Pong, wood, copper tube, black rubber, steel, brass, lead. Calculate the acceleration d = ½ at 2 or: a = 2d/t 2 for each ball.

Results of Free Falling Balls ( d = 2.765m) Accepted value: g = 9.81 m/s 2 = acceleration due to gravity. ObjectMass (g)Acceleration (a) (m/s 2 ) Ping-Pong2.3 Wood2.5 Rubber8.9 Copper Tube12.1 Small Steel28.2 Brass Mass50.0 Lead93.5

Results of Free Falling Balls ( d = 2.765m) Accepted value: g = 9.81 m/s 2 = acceleration due to gravity. ObjectMass (g)Acceleration (a) (m/s 2 ) Ping-Pong2.3 Wood2.5 Rubber8.9 Copper Tube12.1 Small Steel28.2 Brass Mass50.0 Lead93.5

How can we estimate error due to our reaction time?

How High Is This Cliff? 1.You’re at the top of a high cliff, overlooking a river. How can you find out its height? 2.It takes 10 seconds for a dropped rock to hit the river below. How high is the cliff? 3.How Fast is the rock traveling after 3.5 seconds? 4.How Far has the rock traveled after 3.5 seconds?

Case 3: Accelerating from an initial velocity: v i ≠ 0, a ≠ 0 How can we find equations for v and d as “functions” of time (t)? From definition of acceleration, we get:  v = v i + a (Δt) From area under v vs. t graph, we get:  d = v i (Δt) + ½ a (Δt) 2

Case 3: Accelerating from an initial velocity: v i ≠ 0, a ≠ 0 A car accelerates from v i = 4 m/s at a constant rate of 2 m/s 2. Fill out the following table and plot v vs. t and d vs. t v = v i + a (Δt) d = v i (Δt) + ½ a (Δt) 2 t (s)v (m/s)d (m) 0V i =

Summarizing Kinematic Cases Δt = time interval v i = velocity at start of time interval v = velocity at end of time interval d = distance traveled at end of time interval Case 1: Constant velocity: v i ≠ 0; a = 0  v = v i, d = v i (Δt) Case 2: Accelerating from Rest: v i = 0; a ≠ 0  v = a (Δt), d = ½ a (Δt) 2 Case 3: Accelerating from an initial velocity v i ≠ 0; a ≠ 0  v = v i + a (Δt) (“Kinematic Equation 1”)  d = v i (Δt) + ½ a (Δt) 2 (“Kinematic Equation 2”)

Summarizing Kinematic Cases (setting initial time at: t 0 =0) Case 1: Constant velocity: v i ≠ 0; a = 0  v = v i, d = v i t Case 2: Accelerating from Rest: v i = 0; a ≠ 0  v = a t, d = ½ a t 2 Case 3: Accelerating from an initial velocity v i ≠ 0; a ≠ 0  v = v i + a t (“Kinematic Equation 1”)  d = v i t + ½ a t 2 (“Kinematic Equation 2”)

Summarizing Kinematic Cases General Time Interval Case 1: Constant velocity: v 0 ≠ 0; a = 0  v = v i, d = v i (Δt) Case 2: Accelerating from Rest: v i = 0; a ≠ 0  v = a (Δt), d = ½ a (Δt) 2 Case 3: Accelerating from an initial velocity v i ≠ 0; a ≠ 0  v = v i + a (Δt) (“Kinematic Equation 1”)  d = v i (Δt) + ½ a (Δt) 2 (“Kinematic Equation 2”) Time interval starting at t 0 = 0 Case 1: Constant velocity: v i ≠ 0; a = 0  v = v i, d = v i t Case 2: Accelerating from Rest: v i = 0; a ≠ 0  v = a t, d = ½ a t 2 Case 3: Accelerating from an initial velocity v i ≠ 0; a ≠ 0  v = v i + a t (“Kinematic Equation 1”)  d = v i t + ½ a t 2 (“Kinematic Equation 2”)

Relative Motion Person A stands on the sidewalk. Given: A sees B drive by at 40 mi/h East. A sees C drive by at 60 mi/h East. Questions: According to B  What is A’s velocity?  What is C’s velocity? According to C  What is A’s velocity?  What is B’s velocity? Suppose B throws a ball sideways out the window.  How does B see the ball move?  How does A see the ball move?