PROCESSING THE DATA (PART A) 1. Describe the difference between the two lines on your graph made in Step 6. Explain why the lines are different. Referring to graph on the right the difference between the two lines is that one line is at a faster speed than the other in the same amount of time. While one is steeper the other one is not as steep. 2. How would the graph change if you walked toward the Motion Detector rather than away from it? Test your answer using the Motion Detector. Since
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Calculate the velocity of the object after 3 seconds and before it hits the ground. What can be the height it is thrown? 6. Calculate the velocity of the car which has initial velocity 24m/s and acceleration 3m/s² after 15 second. 7. The car which is initially at rest has an acceleration 7m/s² and travels 20 seconds. Find the distance it covers during this period. 8. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the
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the Student Formula Competition Acceleration Event Chantharasenawong C*. and Promoppatum P. Department of Mechanical Engineering‚ King Mongkut’s University of Technology Thonburi *corresponding author: chawin.cha@kmutt.ac.th ABSTRACT This article aims to quantitatively investigate the advantages gained when racecars are deliberately positioned far from the starting line in the Acceleration Event of the TSAE Student Formula competition. A racecar acceleration model‚ verified with an actual
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Law applied to the smaller mass‚ M1‚ implies (see Figure 2b): FNET(on M1) = M1a T– M1g = M1a (2) The tension T can be eliminated from equations (1) and (2) obtain: M2g – M1g = M2a + M1a (3) The magnitude of the acceleration‚ a‚ of the system is then: (4) The numerator‚ (M2 – M1)g‚ is the net force causing the system to accelerate. The denominator‚ (M2 + M1)‚ is the total mass being accelerated. Equation (4) can be written: (5) Where FNET
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is moving with a uniform speed c) When the body is moving with a non-uniform speed. 4. The brakes applied to a car produced an acceleration of 6 ms-2 in the opposite direction of the motion. If the car takes 2s to stop after the application of brakes‚ calculate the distance it travels during this time. 5. What is positive acceleration and negative acceleration? 6. Why is the motion of Satellites around their planets considered an accelerated motion? 7. Study the time (t) versus distance
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Acceleration Velocity Displacement Distance Time Definition 1. Acceleration is the rate of change of velocity with time. Velocity is a vector physical quantity; both magnitude and direction are required to define it. the length of an imaginary straight path‚ typically distinct from the path actually travelled by P. Distance is a numerical description of how far apart objects are. In physics or everyday usage‚ distance may refer to a physical length‚ or an estimation Time in physics is
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Determine the acceleration in a quick sprint. Question What would the participant’s acceleration be if he/she sprints forward in a positive direction? Hypothesis/Prediction When a person sprints forward‚ it means he/she speeds up. Consequently‚ the acceleration should be positive. When the velocity accelerates at a constant rate‚ the acceleration should remain constant. Therefore‚ if the participant is moving toward a positive direction and the speed increases‚ then the acceleration should be positive
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Chapter 4 Problems 1‚ 2‚ 3 = straightforward‚ intermediate‚ challenging Section 4.1 The Position‚ Velocity‚ and Acceleration Vectors 1. A motorist drives south at 20.0 m/s for 3.00 min‚ then turns west and travels at 25.0 m/s for 2.00 min‚ and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip‚ find (a) the total vector displacement‚ (b) the average speed‚ and (c) the average velocity. Let the positive x axis point east. 2. A golf ball is hit off a tee at the
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Flywheels Laboratory Experiment 4 Aziz Darwish H00124728 14th November‚ 2012 Mechanical Engineering B51PX Praxis Mounif Abdallah Contents Page number Abstract/Introduction 1 Aim/Objective 1 Theory 1-2 Apparatus (Equipment) 3 Procedure 3 Calculations 3-4 Results
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1 1 2 3 C Motion I 7 (a) From 1 January 2009 to 10 January 2009‚ the watch runs slower than the actual time by 9 minutes. Therefore‚ when the actual time is 2:00 pm on 10 January 2009‚ the time shown on the watch should be 1:51 pm on 10 January 2009. Practice 1.1 (p. 6) D (a) Possible percentage error 10 −6 = × 100% 24 × 3600 = 1.16 × 10 % 1 (b) = 1 000 000 days 10 −6 –9 It would take 1 000 000 days to be in error by 1 s. (b) Percentage error 9 = × 100% 9 × 24 × 60 = 6.94 × 10–2%
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