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## What is an Instant of Time?

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**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 – 5A Mathematical Model of Motion**• 5.1 Graphing Motion in One Dimension Position – Time Graphs Example: 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 = linebacker D = 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**riseΔyYf - Yi**• slope = run = Δx = xf - xi Δddf - di slope = v = Δt = tf - ti**Using an Equation to Find out Where and When**Δddf - di • Average Velocity = v = Δt = tf – ti • d = di + 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 = di + vt d = -200m + (15ms)(600s) d = -200m + 9000m d = 8800m • d = di + 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 = di + 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**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 • vB is a constant 75m/s • vA 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 AccelerationDetermining 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**LineDraw 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**• V1 speed increase/decrease/constant • V2 velocity +/- • V3 acceleration +/-/0 • V4 • V5 • V6**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 = vo + 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 = vo + at • 7m/s = 4m/s +a(3s) • 3m/s = a(3s) • 1m/s2 = a**Displacement Under Constant Acceleration**• d = di + ½(vf + vi)t • d = di + vit + 1/2at2 • v2 = vi2 + 2a(df – di)**5.4 Free FallAcceleration 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/s2 towards the center of the Earth. • Acceleration due to gravity “g” = -9.8m/s2**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 distanceb/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 2s 29.4m/s 3s 19.6m/s 4s 9.8m/s 5s 0m/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?**Quadratic Equation**• y = ax2 + bx + c • When y = 0 • 0 = ax2 + bx + c • Solving for “x” - b ± √(b2 – 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 = di + vit + 1/2at2 • d = ½(a)t2 + vit + di • 0 = ½(-9.8m/s2)t2 + (35m/s)t + 0 • 0 = (-4.9m/s2)t2 + (35m/s)t • Solve for “t”**- b + √(b2 – 4ac)**• x = 2a • -35m/s + √﴾(35m/s)2 – 4(-4.9m/s)(0)﴿ • t = 2(-4.9m/s2) • -35m/s + √1225m2/s2 • t = -9.8m/s2 • -35m/s + 35m/s 0__ • t = -9.8m/s2 = -9.8m/s2 = 0s • OR →**- b - √(b2 – 4ac)**• x = 2a -35m/s - √﴾(35m/s)2 – 4(-4.9m/s)(0)﴿ • t = 2(-4.9m/s2) • -35m/s - √1225m2/s2 • t = -9.8m/s2 • -35m/s - 35m/s-70m/s • t = -9.8m/s2 = -9.8m/s2 • t = 7.143s