...Sergio Hernandez Jose Roque Paul Zuniga October 7th, 2012 Laboratory Report: Acceleration on an Incline Purpose: 0 Use a Motion Detector to measure the speed and acceleration of a cart rolling down an incline. 1 Determine the mathematical relationship between the angle of an incline and the acceleration of the cart. 2 Determine the value of free fall acceleration, g, by extrapolating the acceleration vs. sine of track angle graph. 3 Determine if an extrapolation of the acceleration vs. sine of track angle is valid. Materials: * Computer * Vernier computer interface. * Logger Pro. * Vernier Motion Detector. * Dynamics cart. * Meter stick. * Ramp. * Books. Procedure: 1. Connect the Motion Detector to the DIG/SONIC 1 channel of the interface. 2. Place a single book under one end of a 1 – 3 m long board or track so that it forms a small angle with the horizontal. Adjust the points of contact of the two ends of the incline, so that the distance, x, in Figure 1 is between 1 and 3 m. 3. Place the Motion Detector at the top of an incline. Place it so the cart will never be closer than 0.4 m. 4. Open the file “04 g On An Incline” from the Physics with Vernier folder. 5. Hold the cart on the incline about 0.5 m from the Motion Detector. 6. Click to begin collecting data; release the cart after the Motion Detector starts to click. Get your hand out of the Motion Detector path quickly. You may have to adjust the position...
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...P03: Acceleration on an Incline (Acceleration Sensor) |Concept |DataStudio |ScienceWorkshop (Mac) |ScienceWorkshop (Win) | |Linear motion |P03 Acceleration.ds |(See end of activity) |(See end of activity) | |Equipment Needed |Qty |Equipment Needed |Qty | |Acceleration Sensor (CI-6558) |1 |Dynamics Cart (inc. w/ Track) |1 | |Angle Indicator (inc. w/ Track) |1 |Meter stick |1 | |Base and Support Rod (ME-9355) |1 |1.2 m Track System (ME-9429A) |1 | What Do You Think? When a sled accelerates down a snow-covered hill, on what does its acceleration depend? You may want to consider the height of the hill, the slope of the hill and the mass of the sled. How does its acceleration depend on the variable(s) you selected? Take time to answer the ‘What Do You Think?’ question(s) in the Lab Report section. Background A cart on an incline will roll down the incline as it is pulled by gravity. The direction of the acceleration due to gravity is straight down as shown in the diagram. The component of the acceleration due to...
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...Lab #4: Kinematics - Velocity and Acceleration Introduction: The purpose of this lab is to discover and understand the relationships between position, velocity, and acceleration. Additionally, constant/uniform acceleration due to the force of gravity will be examined to find possible mathematical relationships to position and velocity. Velocity and acceleration are changes in position and velocity, respectively, with regards to time. This change can be shown mathematically in calculus derivatives: EQ 1. EQ 2. As dt decreases in value, the instantaneous velocity and acceleration can also be found. Furthermore, if constant acceleration is established, two basic relations between distance, velocity, and the constant acceleration can be found: EQ 3. EQ 4. In any environment near Earth, the acceleration in the vertical direction is constant at a value of g=9.8m/s2 towards the center of Earth or often written as g=-9.8m/s2. In such an environment there is no natural acceleration in the horizontal direction, thus the horizontal motion is analyzed independently of the vertical motion. Thus it can be established that the general form of a position curve for a projectile would follow an inverse parabola shape and the maximum height occurs when vertical velocity is zero. By calculus derivation, it can also be found that the velocity graph would display a linear line with a negative slope. Procedure: This lab consists of two separate but related experiments...
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...Acceleration vs. Time * On the first graph, the man walks slowly to the house from the origin. On the Position-Time graph, the line is a positive consistent rise. This is because his position is going in a positive direction as well as the time is going in a consistent positive direction. On the Velocity-Time graph, the line is straight across at 2 m/s because the velocity does not change because of the consistent speed of the man walking to the house. Since the velocity is constant the acceleration is zero. On the Acceleration-Time graph, the line is flat and straight across at the 0 m/s line because the man does not accelerate. He just walks at a consistent pace to the house. This is called constant speed because there is no variation in his speed. * On the second example, the man is sleeping then wakes up and runs toward the house constantly speeding up as he goes. On the Position-Time graph, there is a positive upward curved line. This is because both are moving in a positive direction but because he is running, the position is rising faster than the time. This upward curve indicates an increase in velocity. On the Velocity-time graph, the line is a straight consistent rise. This is caused because the man is running so the velocity is rising throughout the graph, as is the position. A positive slope indicates a changing velocity which is a positive acceleration. On the Acceleration-Time graph, the line constantly rising because the man is running, constantly speeding...
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...Acceleration Due to Gravity Introduction In this lab you will measure the acceleration due to gravity near the earth’s surface with two experiments: first, by determining the time for a steel ball to fall a known vertical distance (free fall), and then second, by measuring the velocity of a cart at various points as it glides down a slightly inclined and nearly frictionless air track (slow fall). Equipment Part 1: Free-Fall • Free-fall apparatus (steel plate, drop mechanism) • Electronic Timer • Steel Ball Part 2: Slow-Fall • Air Track • Electronic Timer (may be different brand/model than in Part 1) • Gliding Car • Laser Photogate Background: Free Fall Acceleration Under the constant acceleration of gravity near the Earth’s surface, g, the vertical position, y, of a falling object is related to the time it has fallen by 1 y = y 0 + v 0 t " gt 2 2 where y0 and v0 are the initial position and velocity, respectively. The distance fallen after a time, t, has elapsed is: ! 1 y 0 " y = gt 2 " v 0 t 2 If you release the object from rest, v0 = 0, the equation simplifies to ! y0 " y = 1 2 gt 2 By varying the distance the ball drops and measuring the corresponding transit times, we can determine the acceleration of gravity from a best fit line to a linear graph of the experimental data. ! ! Procedure: Free-Fall Acceleration A diagram of the experimental apparatus is shown in Figure 1. When the ball loses contact with the release mechanism, the timer starts counting. It stops...
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...starts from rest and reaches a speed of 5 m/s after travelling with uniform acceleration in a straight line for 2 s. Calculate the acceleration of the body. 2. A body starts from rest and moves with uniform acceleration of 2m/s2 in a straight line. a. Calculate the velocity after 5s. b. Calculate the distance travelled in 5s. c. Find the time taken for the body to reach 100m from its starting point. 3. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff. 4. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car. 5. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall? 6. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled. 7. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon. 8. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance which the sled travels? 9. A bike accelerates...
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...Acceleration of free falling metal ball Brian Ahn November 11, 2013 Raw Data <Table of time taken for ball to reach ground in different heights> Height (m) ± 0.005 m | Time taken (s) ± 0.01 s | | Trial 1 | Trial 2 | Trial 3 | 0.5 | 0.35 | 0.35 | 0.33 | 1.0 | 0.45 | 0.43 | 0.40 | 1.5 | 0.53 | 0.44 | 0.47 | 2.0 | 0.56 | 0.56 | 0.53 | 2.5 | 0.72 | 0.64 | 0.72 | 3.0 | 0.75 | 0.76 | 0.79 | 3.5 | 0.77 | 0.84 | 0.78 | Processing Data <Table of average time taken for ball to reach ground in different heights> Height (m) ± 0.005 | Average time taken (s) | 0.5 | 0.34 ± 0.01 | 1.0 | 0.43 ± 0.03 | 1.5 | 0.48 ± 0.05 | 2.0 | 0.55 ± 0.02 | 2.5 | 0.69 ± 0.04 | 3.0 | 0.77 ± 0.02 | 3.5 | 0.80 ± 0.04 | Average of the time taken is (t1 + t2 + t3) ÷ 3 Uncertainty for the time taken is (tmax– tmin) ÷ 2 Average of the time taken in height 1 is (0.45 + 0.43 + 0.40) ÷ 3 = 0.42667 ≅ 0.43 (to two significant figures) Uncertainty for the time taken in height 1 is (0.45 – 0.40) ÷ 2 = 0.025 ≅ 0.03 (to one significant figure) <Table of average time taken squared by dropping an object in different heights> Height (m) ± 0.005 | Average time taken squared (s2) | 0.5 | 0.12 ± 0.007 | 1.0 | 0.18 ± 0.03 | 1.5 | 0.23 ± 0.05 | 2.0 | 0.30 ± 0.02 | 2.5 | 0.48 ± 0.05 | 3.0 | 0.59 ± 0.03 | 3.5 | 0.64 ± 0.06 | Uncertainty for average time taken squared is 2 × (percentage uncertainty) × (time taken) 2 Uncertainty...
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...Motion of Freely Falling Objects Objectives: * To study the motion of free falling objects * To determine the acceleration due to gravity, g. * To derive quantities from the slope and intercept of graphs * To determine if the motion of a falling object changes by varying its mass Theory: The free fall is a known example or the most common example of a uniformly accelerated movement, with an acceleration a = -9.8m/s2 (vertical axis pointing vertically upward). If you choose the vertical axis pointing vertically downward, the acceleration is taken as + 9.8m/s2. The kinematic equations for a rectilinear movement under the acceleration of gravity are the same as any movement with constant acceleration: (1) v = vi - gt velocity as function of time. (2) y - yi = ½(vi + v)t displacement as function of time (3) y - yi = vit - ½gt2 displacement as function of time (4) v2 = vi2 -2g(x - xi) velocity as function of displacement The sub index i denotes initial quantities, g the gravity acceleration and t, the time. But for this type of motion, the displacement of the object as a function of time is described mathematically as: (5) ∆y=Vot + ½gt2 where Vo is the initial velocity of the object. If object if just drop velocity, Equation (5) becomes (6) ) ∆y=½gt2 Methodology Procedure: * A digital balance was used to measure the masses of the small and big steel balls. ...
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...While rolling level a ball does not roll with or against the vertical force of gravity, it neither speeds up nor slows down. The rolling ball maintains a constant speed; this happens because friction overcomes the ball’s inertia and brings it to a stop. The horizontal component of gravity is zero. 2. What is the acceleration of a car that moves at a steady velocity of 100km/hr for 100 seconds? Explain your answer, and state why this question is an exercise in careful reading as well as in Physics. The acceleration is 0. There is no change of velocity in those 100 seconds time interval. The word you have to pay attention to is “steady” that is why is can exercise in careful reading and in Physics because you would try to do the math but at the end the answer would be wrong because the car never accelerated. * Problems 1,5,8 and 10 1. Find the net force produced by a 30-N force and a 20-N force in each of the following cases: a. Both forces act in the same direction. 30+20= 50N b. The two forces act in different directions. 30-20=10N 2. A vehicle changes its velocity from 100 km/h to a dead stop in 10 s. Show that the acceleration in stopping is -10 km/h x s. VF-Vi= 0-100 VT/t = (0-100)/10s = -10 km/h x s 3. A ball is thrown straight up with enough speed so that it is in the air for several seconds. c. What is the velocity of the ball when it reaches its highest point? Velocity is 0 at highest point. d. What is its velocity...
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...EXPERIMENT #6: UNIFORMLY ACCELERATED MOTION: FREE FALL INTRODUCTION: A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. It is a motion of a body where its weight is the only force acting upon it. Our objective in this experiment is to measure the acceleration of a falling object assuming that only force acting on the object is the gravitational force. THEORY: As an object falls freely, it accelerates due to the applied net force. One equation of motion for a body starting from rest and undergoing constant acceleration can be expressed as: d = ½ at2 where d is the distance the object travelled from its starting point, a is the acceleration of the object, and t is the time elapsed since the motion began. PROCEDURE: Materials needed are: * Free fall apparatus * Small and Large metal ball * Smart timer * Clamp rod * Meter stick So first we measured the height of the release mechanism into 50 cm with the use of meter stick. In trial 1, we used small metal ball. Mounted the free fall adapter’s release mechanism horizontally in the clamp. We place the free fall adapter’s timing pad on the table directly below the release mechanism. We placed the metal ball in the release mechanism and pressed the spring loaded rod inward to hold the ball in the mechanism, and tighten the thumbscrew to hold the rod in place. Loosen...
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...Abstract Objective The objective of this experiment was to familiarize ourselves with constant acceleration equations, and to better understand the horizontal and vertical components of a projectile in free fall. Procedure In this experiment we used a PASCO projectile launcher, a photogate and the PASCO equipment to calculate the initial velocity of a ball that we launched off of a table. With this measurement we were able to use constant acceleration equations to calculate the range and the time of flight of the ball. We then used a “time of flight accessory” to obtain a measured value of time of flight. We compared our measured findings to our calculated findings to see the difference. Finally we created four graphs that illustrated horizontal and vertical velocity, as well as horizontal and vertical position all compared to time. Data Our ball was in flight for about .48 s and had an initial velocity of 3.14 m/s. In our vertical velocity verse time graph, I calculated the slope of the line to be 9.8 m/s^2. I am very happy with this result because it is very close to the real acceleration of gravity. Finally, our measured value of time of flight and calculated value deviated by .006s. This was also really cool to see. Sources of Error We did not take air resistance into consideration in this lab. As the ball is flying through the air, it is being slowed down by the air drag. This could have made our measured value of time of flight slower than it...
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...UNIVERSITI TUNKU ABDUL RAHMAN |Centre |: Centre for Foundation Studies (CFS) |Unit Code |: FHSC1014 | |Course |: Foundation in Science |Unit Title |: Mechanics | |Year/ Trimester |: Year 1 / Trimester 1 |Lecturer | | |Session |: 201605 | | | Additional Tutorial 2: Vector and translational kinematics. 1. Find the x and y-components of: (a) a displacement of 200 km, at 30.0o. (b) a velocity of 40.0 km/h, at 120o; and (c) a force of 50.0 N at 330o. [(a) 173 km, 100 km, (b) -20.0 km/h, +34.6 km/h, (c) 43.3 N, -25.0 N] 2. Three forces are applied to an object, as indicated in the drawing. Force [pic] has a magnitude of 21.0 Newton (21.0 N) and is directed 30.0° to the left of the + y axis. Force [pic] has a magnitude of 15.0 N and points along the + x axis. What must be the magnitude and direction (specified by the angle ( in the drawing) of the third force [pic] such that the vector sum of the three forces is 0 N? [18.7 N, 76o] 3. A 200 N block rests on a 30o inclined plane as shown in figure below. If the weight of the block acts vertically downward, what are the components of the...
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...Steering Behaviors For Autonomous Characters Craig W. Reynolds Sony Computer Entertainment America 919 East Hillsdale Boulevard Foster City, California 94404 craig_reynolds@playstation.sony.com http://www.red.com/cwr/ cwr@red.com Keywords: Animation Techniques, Virtual/Interactive Environments, Games, Simulation, behavioral animation, autonomous agent, situated, embodied, reactive, vehicle, steering, path planning, path following, pursuit, evasion, obstacle avoidance, collision avoidance, flocking, group behavior, navigation, artificial life, improvisation. Abstract This paper presents solutions for one requirement of autonomous characters in animation and games: the ability to navigate around their world in a life-like and improvisational manner. These “steering behaviors” are largely independent of the particulars of the character’s means of locomotion. Combinations of steering behaviors can be used to achieve higher level goals This paper divides motion behavior into three levels. It will focus on the (For example: get from here to there while avoiding obstacles, follow this corridor, join that group of characters...) middle level of steering behaviors, briefly describe the lower level of locomotion, and touch lightly on the higher level of goal setting and strategy. Introduction Autonomous characters are a type of autonomous agent intended for use in computer animation and interactive media such as games and virtual reality. These agents represent a This stands...
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...Acceleration, in physics, is the rate at which the velocity of an object changes over time. An object's acceleration is the net result of any and all forces acting on the object, as described by Newton's Second Law. [1] The SI unit for acceleration is the metre per second squared (m/s2). Accelerations are vector quantities (they have magnitude and direction) and add according to the parallelogram law.[2][3] As a vector, the calculated net force is equal to the product of the object's mass (a scalar quantity) and the acceleration. For example, when a car starts from a standstill (zero relative velocity) and travels in a straight line at increasing speeds, it is accelerating in the direction of travel. If the car changes direction there is an acceleration toward the new direction. When accelerating forward, passengers in the car experience a force pushing them back into their seats. They experience sideways forces when changing direction. If the speed of the car decreases, this is acceleration in the opposite direction, sometimes called deceleration.[4]Mathematically, there is no separate formula for deceleration, as both are changes in velocity. In everyday use and in kinematics, the speed of an object is the magnitude of its velocity (the rate of change of its position); it is thus a scalar quantity.[1] The average speed of an object in an interval of time is the distance travelled by the object divided by theduration of the interval;[2] the instantaneous speed is the limit of...
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...Discussion #3 For this experiment we measured gravitational acceleration and velocity of a cart getting pushed up a ramp. First we had to make a prediction of how a velocity and acceleration graph would look like with a cart going up the ramp. After that we actually started to do the experiment. We then went to the computer which would help us graph our measurements of each time we did the experiment. It measured velocity, acceleration, and position of the cart each time. We did the experiment about a couple times until we got a good looking graph, then we recorded it on our lab reports and used it for the rest of our remaining results. Before using that, we took a measurement of the angle of the ramp which turned out to be 4.04 degrees. After that, we then took the graphs we did that were on the computer and we used different tools to find out the acceleration and slope of each specific time in the reading the lab report told us to do. From there after we were done, we then waited till the whole class was done and we all wrote down what our readings were for each measurement. Our measurements were; 4.04 degrees for angle A, .63 kg for the mass of the cart, .582 with an uncertainty of .009 for our acceleration from the average slope, .58 with an uncertainty of .009 for th average acceleration from STATS , 1.13 for mass of the cart with added mass, .569 with an uncertainty of .032 for acceleration from average slope with the doubled mass, and finally 8.199 with an uncertainty...
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