What is an Instant of Time? Car on racetrack example:

1. How long did a car spend at any one location? 2. Each position is linked to a time, but how long did that time last? 3. You could say an “instant”, but how long is that?

CHAPTER – 5 A Mathematical Model of Motion 5.1 Graphing Motion in One Dimension Position – Time Graphs Position – Time GraphsExample: Football running back motion (displacement) diagram at 1 second intervals. Plot the Position/Time Graph

4. If an instant lasts for a finite amount of time, then, because the car would be at the same position during that time, the car would be at rest. But, a moving object (car) cannot be at rest;  an instant is not a finite period of time. 5. This means that an instant of time lasts “0” seconds.

Using a Graph to Find Out Where and When (Pick various locations)

Graphing the Motion of Two or More Objects A = running back C = center B = linebackerD = defensive back

From Graphs to Words and Back Again Keep in mind that when t=0, the position of the object does not necessarily have to be zero.

Uniform Motion Definition of uniform motion = Means that equal displacements occur during successive equal time intervals.

What does the “slope” of the pos/time graph give us? Velocity

rise Δy Y f - Y i rise Δy Y f - Y i slope = run = Δx = x f - x i Δd d f - d i Δd d f - d i slope = v = Δt = t f - t i

For Objects With Diff. Velocities

Using an Equation to Find out Where and When Δd d f - d i Δd d f - d i Average Velocity = v = Δt = t f – t i d = d i + vt

Example Problem (PP-11) A car starts 200m west of the town square and moves with a constant velocity of 15m/s towards the east. a) Where will the car be 10 min later? b) When will the car reach the town square?

a) draw sketch d = d i + vt d = d i + vt d = -200m + (15ms)(600s) d = -200m m d = 8800m b) d = d i + vt 0 = -200m + (15m/s)t 200m = (15m/s)t 13.3s = t

Example-2 (PP 12) A car starts 200m west of the town square and moves with a constant velocity of 15m/s towards the east. At the same time a truck was 400m east of the town square moving west at a constant velocity of 12m/s. Find the time and place where the car meets the truck.

Draw a sketch. d = d i + vt

5.2 Graphing Velocity in One Dimension, Determining Instantaneous Velocity Q: When an object is not moving with uniform motion, the object is said to be…? A: accelerating

Position-Time Graph Uniform Motion

Position-Time Graph Acceleration

Instantaneous Velocity = ? How fast an object is moving at a particular instant in time, ie: how fast is it moving “Right Now”

What is the instantaneous velocity at 2s? What is the instantaneous velocity at 4s?

Instantaneous Velocity The Instantaneous Velocity is equal to the “Slope” of the tangent line of a position/time graph at any particular time. Draw previous graph and calculate the instantaneous velocity at 2s & 4s.

Velocity-Time Graphs 2 planes Plane-A & Plane-B v B is a constant 75m/s v A is constantly increasing (constant “a”) Draw sketch on board Q: At the point of intersection, will the planes crash? A: ??? Not enough information given, the graph merely indicates the planes have the same velocity at that point.

Displacement from a Velocity-Time Graph Q: What does the area under a V-T graph represent? A: Δd, displacement.

5.3 Acceleration Determining Average Acceleration Average acceleration, “a”, is equal to the slope (rise/run, Δv/Δt) of a velocity-time graph.

Constant and Instantaneous Acceleration If there is a constant slope on a Velocity- Time Graph then there is also a constant acceleration, any point “a” is the same. Acceleration is simply the slope of the line.

Instantaneous Acceleration of a Velocity-Time Graph, Curved Line Draw graph on board.

Q: What is the instantaneous acceleration at 2s? How could it be determined? A: Draw a tangent line at 2s then calculate the slope.

Positive and Negative Acceleration V 1 speed increase/decrease/constant V 2 velocity +/- V 3 acceleration +/-/0 V 4 V 5 V 6

When v+ and a+, speed increases(+) When v+ and a-, speed decreases(+) When v- and a-, speed increases(-) When v- and a+, speed decreases(-)

Calculating Velocity from Acceleration v = v o + at

Example Problem Hines Ward is running for a touchdown at 4m/s. He accelerates for 3s. His velocity entering the end zone is 7m/s. What is his acceleration? v = v o + at 7m/s = 4m/s +a(3s) 3m/s = a(3s) 1m/s 2 = a

Displacement Under Constant Acceleration d = d i + ½(v f + v i )t d = d i + v i t + 1/2at 2 v 2 = v i 2 + 2a(d f – d i )

5.4 Free Fall Acceleration Due to Gravity 1. Drop a flat sheet of paper 2. Drop a crumpled piece of paper. 3. Drop a tennis ball. Q: Is there a difference in the acceleration of each of the objects above? A: NO All “free falling” objects accelerate with a magnitude of 9.8m/s 2 towards the center of the Earth. Acceleration due to gravity “g” = -9.8m/s 2

Example: Drop A Rock After 1s it is falling at ____m/s After 2s it is falling at ____m/s After 3s it is falling at ____m/s After 4s it is falling at ____m/s  Each second during free fall the rock will Δ its velocity by -9.8m/s.

Drop a Rock Diagram DIAGRAM During each equal successive time interval the rock will fall a greater distance b/c a = “-g” Q: Could this diagram also apply to a rock thrown upward? A: YES, the diagram would look the same. Q: Why? A: Once the rock leaves the hand, the only force acting on the rock is gravity “g”.

Diagram/Example Diagram of a ball thrown upward with a velocity of 49m/s. Show the velocity changes for 10 seconds, 5 seconds up & 5 seconds down. DRAW SKETCH

Time Velocity 0s 49m/s 1s 39.2m/s 2s29.4m/s 3s19.6m/s 4s9.8m/s 5s0m/s 6s-9.8m/s 7s-19.6m/s 8s-29.4m/s 9s-39.2m/s 10s-49m/s

Example Problem If you throw a rock upward with an initial velocity of 35m/s: a) what is its velocity after 1,2,3,4,5 sec? b) what is its position after 1,2,3,4,5 sec? c) how long will it take to reach its maximum height? d) what is its maximum height? e) how long will it be in the air?

Example Thrown Upward

Quadratic Equation y = ax 2 + bx + c When y = 0 0 = ax 2 + bx + c Solving for “x” - b ± √(b 2 – 4ac) - b ± √(b 2 – 4ac) x = 2a

A ball is thrown upward with an initial velocity of 35m/s, how long will the ball be in the air? Equation to be used…? d = d i + v i t + 1/2at 2 d = ½(a)t 2 + v i t + d i 0 = ½(-9.8m/s 2 )t 2 + (35m/s)t = (-4.9m/s 2 )t 2 + (35m/s)t Solve for “t”

- b + √(b 2 – 4ac) - b + √(b 2 – 4ac) x = 2a -35m/s + √﴾(35m/s) 2 – 4(-4.9m/s)(0)﴿ -35m/s + √﴾(35m/s) 2 – 4(-4.9m/s)(0)﴿ t = 2(-4.9m/s 2 ) -35m/s + √1225m 2 /s 2 -35m/s + √1225m 2 /s 2 t = -9.8m/s 2 -35m/s + 35m/s 0__ -35m/s + 35m/s 0__ t = -9.8m/s 2 = -9.8m/s 2 = 0s OR →

- b - √(b 2 – 4ac) - b - √(b 2 – 4ac) x = 2a -35m/s - √﴾(35m/s) 2 – 4(-4.9m/s)(0)﴿ t = 2(-4.9m/s 2 ) -35m/s - √1225m 2 /s 2 -35m/s - √1225m 2 /s 2 t = -9.8m/s 2 -35m/s - 35m/s -70m/s -35m/s - 35m/s -70m/s t = -9.8m/s 2 = -9.8m/s 2 t = 7.143s