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Purpose: Find the relationship between the position (x) of an object as it rolls down an incline and the time taken (t) to reach each position. t2,x2 Picture represents equal position intervals. If so, what t1,x1 would happen to the time? t3,x3 Procedure: 1) Set up the ramp. 2) Control the ramp angle and the What’s required: mass of the ball. 3) Perform 3 trials for each of the 5 1) Data Table 2) Original Graph positions along the ramp. 3) Modified graph. 4) Derivation of Equation. 5) Answer Questions #1-3 4) Plot Position vs. time and then modify. (Plot time on x axis) 5) Derive an equation for the relationship. One could speak of the average velocity of the object in the graph, but since the object started very slowly and steadily increased its speed, the term average velocity has little meaning. x x t t What happens as you shrink the time interval t over which you calculate the average velocity? x x x x x x t t t t t t As one shrinks the interval, t to zero, the secant becomes a tangent; the slope of the tangent is the average velocity at this instant, or simply the instantaneous velocity at that clock reading. x x t t Instantaneous velocity (v1, v2, vf, vi, etc.) the velocity at a given point (instant) in time. It is equal to the slope of the tangent to the x vs. t graph. Acceleration x Δv v Δt t t Δv Slope = = acceleration Δt The slope of a v vs. t graph is the acceleration. Acceleration Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity (magnitude or direction). Acceleration values are expressed in units of velocity/time. Typical acceleration units include the following: m/s/s mi/hr/s km/hr/s Changing speed, constant direction Acceleration exists Constant speed, changing direction Acceleration exists Changing speed, constant direction Acceleration exists Direction of Motion Action/Sign of Sign of Velocity Acceleration New Graphs Written Description Motion Map (x vs t) (v vs t) (a vs t) Starting Dir (v) (a) point x x t t2 • Position α time2 • y = mx2 + b • Final Equation: x (m) = 2.5 m/s2 x time2 (s2) • How does the position vs time graph differ from the previous unit’s lab? There is no longer a direct relationship between position and time. • Explain how the position vs. time graph shows the motion exhibited by the ball? The slope is no longer constant, therefore the velocity is changing. CONSTANT ACCELERATION Galileo (1564 – 1642) 1 displacement 3 5 7 1 second = 1 displacement 9 2 second = 1 + 3 displacement = 2 2 3 second = 1 + 3 + 5 displacement = 32 4 second = 1 + 3 + 5 + 7 displacement = 42 5 second = 1 + 3 + 5 + 7 + 9 displacement = 52 Therefore, displacement a time2 x Time (s) Position (m) Instantaneous velocity Group Ave. Velocity (m/s) (m/s) 0 0 1 2.5 1.5 5.63 2 10 2.5 13.6 3.0 22.5 0 3.6 t (s) Read Directions: 1. Draw tangent lines to the curve below at all but the last point. 2. Find the slope of each tangent line. Then slope of the tangent line is the instantaneous velocity at that point in time Record the value of the velocity in the table above. 3. Average the slope values within your lab group for each point. Use Graphical analysis to obtain a velocity vs. time graph using your groups average values. Draw a picture of the graph on another sheet of paper. Label axis w/ units and record stats. Compare Graphs Same original x graph t • What is the difference between a X vs t2 graph & the V vs t graph. • Derive the equation for both. v x t2 t 4. Write the mathematical model for this 5. The equation for the curve is x=2.5m/s^2(t^2). graph underneath your V vs t graph. How does the slope of your velocity vs. time graph compare with the slope of the curve above. Y = mX + b Y = mX2 + b vf = 5 m/s2 (time) + vi Final Equation: X = 2.5 m/s2 (time2) Since slope = ave. accel. So, for X vs t2 graph slope = ½ a vf = a Δ t + vi or vf = vi + a Δ t vf - vi X = ½ at2 a= Δt Motion Graphs x v a t t t x 0 Motion Graphs x v a t t t x 0 Starting Point Direction Velocity Acceleration 0 + forward + V (speeding up) +constant Acceleration Example #1 V (m/s) a (m/s/s) 0 +5 +5 +5 +10 +5 15 +5 Δv from +5 to +10 m/s requires a +5 m/s/s acceleration! Motion Graphs x v a t t t x 0 Starting Point Direction Velocity Acceleration 0 + forward + V (slowing down) -constant Acceleration Example #2 V (m/s) a (m/s/s) + 10 -5 + 5 -5 0 -5 Δv from +10 to + 5 m/s requires a - 5 m/s/s acceleration! Motion Graphs x v a t t t x 0 Starting Point Direction Velocity Acceleration above backward -V (speeding up) -constant Acceleration Example #3 V (m/s) a (m/s/s) 0 -5 -5 -5 -10 -5 -15 -5 Δv from -5 to -10 m/s requires a -5 m/s/s acceleration! a negative acceleration does not mean the velocity is slowing down, it has to do with direction! Motion Graphs a v x t t t x 0 Starting Point Direction Velocity Acceleration above backward - V (slowing down) +constant Acceleration Example #4 V (m/s) a (m/s/s) - 10 +5 - 5 +5 0 +5 Δv from -10 to -5 m/s requires a +5 m/s/s acceleration! More Conventions… Speeding Up + v, + a Same sign - v, - a Slowing Down + v, - a Opposite sign - v, + a Motion Graphs x v a t t t x 0 Speed is not acceleration As the ball rolls down this hill a) Its speed increases and its acceleration decreases. b) Its speed decreases and acceleration increases. c) Both increase d) Both remain constant e) Both decrease Graphing Acceleration B v A C t D A: speeding up + velocity ↑ + acceleration B: const motion const. velocity 0 acceleration C: slowing down + velocity ↓ - acceleration D: speeding up - velocity ↑ - acceleration Rearranging Equation #1 vf - vi a= Slope of an V vs t graph Δt a Δt = vf - vi vf = vi + a Δt Deriving the Acceleration Equations 2 • modify x vs t graph x t x Equation is: x = 2.5 m/s/s x time2 t2 So, for x vs t2 graph m = ½ a Therefore, Δx = ½ at2 + vit Deriving the Acceleration Equations 3 Through a complex derivation – when you combine equations #1 and #2 you get: vf2 = vi2 + 2aΔx Constant Acceleration Equations Missing variable ΔX vf = vi + a Δt Vf ΔX = ½ at2 + vit Vf2 = vi 2 + 2aΔx t a ΔX = ½ (v f+vi)t All the equations in one place constant velocity uniform acceleration x v v a t t x vt v f at v 0 x v0t at 1 2 2 v v 2ax 2 f 2 0 1. A bicyclist has an acceleration of 2 m/s2. If he starts from rest, determine his velocity and position at t = 5s. 2. A car is traveling at a speed of 30 m/s when the brakes are suddenly applied, causing a constant deceleration of 4 m/s2. Determine the time required to stop the car and the distance traveled before stopping. Answers: 1.10 m/s, 25 m 2. 7.5 s, 112.5 m Early one morning Farmer Joiner tries his hand at milking a new breed of cow. He chooses one named Hackensack. (Yes he’s milking Hackensack, the new Jersey). However, Hackensack fails to realize how much pull Farmer Joiner has and reacts oddly. Every time Dr. J milks her, she has trouble with her moo. It sounds like M- m-m-ooo (a sort of udder stutter, you might say). Finally she can’t take it any longer. a) Starting from rest, she accelerates at 2 m/s2 for 4 seconds out the barn door. b) She then runs at a constant velocity for 5 minutes, c) until Dr. J succeeds in throwing a rope around her neck, after which she slows down to a stop at a rate of 1 m/s2. When the ordeal is over, Dr. J admits that it was a mooving experience indeed. Qualitatively sketch the following graphs for Hackensack’s trip. A. position vs. time graph B. velocity vs. time graph C. acceleration vs. time graph D. motion map. An automobile accelerates from rest at 2 m/s2 for 20 s. The speed is then held constant for 20 s. Then there is an acceleration of - 3 m/s2 until the automobile stops. Find: a. the distance traveled between the 4th and 5th seconds. b. the total distance traveled. c. the total time traveled Answers: a. 9.00 m b. 1466.53 m c. 53.33 s Ella Vader is driving her flashy Porsche down the highway at 30 m/s. She sees a deer ahead and 2.0 s after she spots the deer she hits the brakes. It takes her 7.5 more seconds to stop. a) How far did Ella travel before stopping? b) What was Ella’s acceleration while she slowed down? Terms for Constant Acceleration Test Instantaneous Velocity: definition find from x vs t (tangent line, slope) Acceleration: definition (a = Δv / Δt) find from a v vs t (slope) For accelerated Motion: x vs t, v vs t, a vs t graphs motion map Graphs & Tracks Exercise Conceptual Questions – Acceleration & Velocity Constant Acceleration Equations vf = vi + at Δx = ½at2 + vit vf2 = vi2 + 2aΔx

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posted: | 11/6/2012 |

language: | English |

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