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Motion

Chapter 2. Motion. Physics: The Most Fundamental Physical Science. Physics is concerned with the basic principles that describe how the universe works. Physics deals with matter, motion, force, and energy. Intro. Physics – Areas of Study. Classical mechanics Waves and sounds

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Motion

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  1. Chapter 2 Motion

  2. Physics:The Most Fundamental Physical Science • Physics is concerned with the basic principles that describe how the universe works. • Physics deals with matter, motion, force, and energy. Intro

  3. Physics – Areas of Study • Classical mechanics • Waves and sounds • Thermodynamics • Electromagnetism • Optics • Quantum mechanics • Atomic and nuclear physics • Relativity Intro

  4. Motion • Motion is everywhere – walking, driving, flying, etc. • This chapter focuses on definition/discussion of: speed, velocity, and acceleration. • There are two basic kinds of motion: • Straight line • Circular Intro

  5. Defining Motion • Position – the location of an object • A reference point and a measurement scale must be given in order to define the position of an object • Motion – an object is undergoing a continuous change in position • Description of Motion – the time rate of change of position • A combination of length and time describes motion Section 2.1

  6. Speed and Velocity • In Physical Science ’speed’ and ’velocity’ have different (distinct) meanings. • Speed is a scalar quantity, with only magnitude • A car going 80 km/h • Velocity is a vector, and has both magnitude & direction • A car going 80 km/h north Section 2.2

  7. distance traveled time to travel distance • Average Speed = • v = d/t or v = Dd/Dt Speed • ( where D means ’change in’) • Over the entire time interval, speed is an average • Distance – the actual path length traveled • Instantaneous Speed – the speed of an object at an instant of time (Dt is very small) • Glance at a speedometer Section 2.2

  8. Instantaneous speed Section 2.2

  9. Velocity is similar to speed except a direction is involved. Displacement – straight line distance between the initial and final position with direction toward the final position Instantaneous velocity – similar to instantaneous speed except it has direction displacement total travel time • Average velocity = Velocity Section 2.2

  10. Displacement and Distance Displacement is a vector quantity between two points.Distance is the actual path traveled. Section 2.2

  11. Describe the speed of the car above. GIVEN: d = 80 m & t = 4.0 s SOLVE: d/t = 80 m/4.0 s = 20 m/s = average speed Velocity would be described as 20 m/s in the direction of motion (east?) Constant Velocity - Example • EQUATION: v = d/t Section 2.2

  12. Constant Velocity – Confidence Exercise • How far would the car above travel in 10s? • EQUATION: v = d/t • REARRANGE EQUATION: vt = d • SOLVE: (20 m/s) (10 s) = 200 m Section 2.2

  13. GIVEN: Speed of light = 3.00 x 108 m/s = (v) Distance to earth = 1.50 x 108 km = (d) EQUATION: v = d/t  rearrange to solve for t  t = d/v t = 0.500 x 103 s = 5.00 x 102 s = 500 seconds 1.50 x 1011 m 3.00 x 108 m/s • SOLVE: t = d/v = Example: How long does it take sunlight to reach Earth? Section 2.2

  14. Sunlight Section 2.2

  15. Acceleration • Changes in velocity occur in three ways: • Increase in magnitude (speed up) • Decrease in magnitude (slow down) • Change direction of velocity vector (turn) • When any of these changes occur, the object is accelerating. • Faster the change  Greater the acceleration • Acceleration – the time rate of change of velocity Section 2.3

  16. change in velocity time for change to occur • Avg. acceleration = Dv t vf – vo t (vf = final & vo = original) • a = = Acceleration • A measure of the change in velocity during a given time period • Units of acceleration = (m/s)/s = m/s2 • In this course we will limit ourselves to situations with constant acceleration. Section 2.3

  17. Constant Acceleration of 9.8 m/s2 • As the velocity increases, the distance traveled by the falling object increases each second. Section 2.3

  18. vf = 90 km/h = 25 m/s vf –vo t 25 m/s – 0 7.0 s • a = = = 3.57 m/s2 Finding Acceleration - Example • A race car starting from rest accelerates uniformly along a straight track, reaching a speed of 90 km/h in 7.0 s. Find acceleration. • GIVEN: vo = 0, vf = 90 km/h, t = 7.0s • WANTED: a in m/s2 Section 2.3

  19. Constant Acceleration = Gravity = 9.8 m/s2 • Special case associated with falling objects • Vector towards the center of the earth • Denoted by “g” • g = 9.80 m/s2 Section 2.3

  20. If frictional effects (air resistance) are neglected, every freely falling object on earth accelerates at the same rate, regardless of mass. Galileo is credited with this idea/experiment. Astronaut David Scott demonstrated the principle on the moon, simultaneously dropping a feather and a hammer. Each fell at the same acceleration, due to no atmosphere & no air resistance. Frictional Effects Section 2.3

  21. Figure 2 • Galileo is also credited with using the Leaning Tower of Pisa as an experiment site. Section 2.3

  22. What about the distance a dropped object will travel? • d = ½ gt2 • This equation will compute the distance (d) an object drops due to gravity (neglecting air resistance) in a given time (t). Section 2.3

  23. Solving for distance dropped - Example • A ball is dropped from a tall building. How far does the ball drop in 1.50 s? • GIVEN: g = 9.80 m/s2, t = 1.5 s • SOLVE: d = ½ gt2 = ½ (9.80 m/s2) (1.5 s)2 = ½ (9.80 m/s2) (2.25 s2) = ?? m Section 2.3

  24. Solving for distance dropped - Example • A ball is dropped from a tall building. How far does the ball drop in 1.50 s? • GIVEN: g = 9.80 m/s2, t = 1.5 s • SOLVE: d = ½ gt2 = ½ (9.80 m/s2) (1.5 s)2 = ½ (9.80 m/s2) (2.25 s2) = 11.0 m Section 2.3

  25. d = ½at2(distance traveled, starting from rest) d = ½gt2 (distance traveled, dropped object) g = 9.80 m/s2 = 32 ft/s2 (acceleration, gravity) • v = d/t (average speed) vf –vo t Dv t • a = = (constant acceleration) Important Equations – Chapter 2 Review

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