 Download Presentation Mechanics Motion in One Dimension Vectors and Two Dimensional Motion

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Mechanics Motion in One Dimension Vectors and Two Dimensional Motion. Chapter 1 (p.2-22) Chapter 2 (p.23-50) Chapter 3 (p.51-78). Basic Concepts. Problem solving strategy: 1.Read problem 2.Draw diagram 3.Identify data 4.Choose equation(s) 5.Solve 6.Evauate and check answer SI Units: I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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1. MechanicsMotion in One DimensionVectors and Two Dimensional Motion Chapter 1 (p.2-22) Chapter 2 (p.23-50) Chapter 3 (p.51-78)

2. Basic Concepts Problem solving strategy: 1.Read problem 2.Draw diagram 3.Identify data 4.Choose equation(s) 5.Solve 6.Evauate and check answer SI Units: Length=meter Mass=Kilogram Time=Second *** Remember significant digits ***

3. Key Equations Displacement: an objects change in position given by final minus initial coordinates. Average Velocity V = x/t Instantaneous Velocity V = lim x/t t→0

4. Acceleration • Average Acceleration: a = v/t • Instantaneous Acceleration: of an object at a certain time equals the slope of a velocity-time graph at that instant. • One dimension motion with constant acceleration: • V = vo+at • V = (vo+v)/2 • X = vot+.5at2 • V2 = vo2+2ax

5. Projectile Motion Problem Solving Strategy • Select a coordinate system • Resolve the initial velocity vector into x and y components • Treat the horizontal motion and the vertical motion independently • Analyze the horizontal motion of the projectile with constant velocity • Analyze vertical motion of the projectile with constant acceleration

6. Equations for Projectile Motion x = vxot vy = vyo-gt y = vyot-.5gt2 vy2 = vyo2-2gy g = 9.8 m/s2 (on earth…but you can do projectile motion on other planets if you know the local value of g. Also, some electric field problems can be solved in the same manner as projectile problems!)

7. Relative Velocity Different displacements and velocities may be observed for objects in motion depending on the frame of reference. For example: A mouse is running at 1.2 m/s backwards on a canoe that is moving forward at 3.4 m/s on a calm pond. Vmc= -1.2 m/s (velocity of mouse relative to canoe) Vcp= +3.4 m/s (velocity of canoe relative to pond) Vmp=vmc+vcp=+2.2 m/s (velocity of mouse relative to the pond)

8. A physical quantity that requires both direction and magnitude Vector

9. Has only magnitude and no direction Scalar quantity

10. Vector Scalar Scalar Scalar Vector Vector Scalar Vector Scalar Scalar Vector Vector Scalar Scalar Vector Scalar Determine which of the following quantities are vectors, and which are scalars. Momentum Speed kinetic energy Work Acceleration Force Power Electric field electric potential electric charge Torque magnetic field Temperature Time Displacement density

11. _______ is described as the limit of the average velocity as the time interval becomes infinitesimally short. Instantaneous Velocity

12. The average velocity on a position-time graph is the _____ of the straight line connecting the final and initial points. slope

13. The ______ is zero in a projectile problem when the object is traveling through the air and is acted on only by gravity. A) Vertical Velocity C) Horizontal Velocity B) Horizontal Acceleration D) Speed B) Horizontal Acceleration

14. The ______ is zero in a projectile motion problem when the object reaches the highest point of its trajectory. A) Speed C) Acceleration B) Horizontal Velocity D) Vertical velocity D) Vertical Velocity

15. The 3 equations that describe the motion of an object moving with constant acceleration along the x-axis are: V = vo+at DX = vot+.5at2 V2 = vo2+2aDx Also… Dx= ½ (v+vo)t

16. A race car starting from rest accelerates at a rate 5 m/s2. What is the velocity of the car after it has traveled 100 ft? V2 = vo2 + 2ax V2 = (0)2+2(5m/s2)(30.5m) = √(305m2/s2) = ± 17.5m/s

17. A golf ball is released from rest at the top of a really tall building. Neglecting air resistance, calculate the position and velocity of the ball after 2 seconds. V = at = (-9.8m/s2)(2s) = -19.6m/s Y = .5at2 = .5(-9.8m/s2)(2s)2 = -19.6m

18. A passenger at the rear of a train, traveling at 15m/s relative to the earth, throws a ball with a speed of 15m/s in the opposite direction of the trains motion. What is the balls velocity relative to the earth? vte = 15m/s vbt = -15m/s vbe = vbt + vte = 0

19. Adding vectors: 1) Select a coordinate system 2) Draw a sketch and label vectors 3) Find x and y components of all 4) Find the resultant components in both directions 5) Use ______ to find the magnitude of the resultant vector Pythagorean theorem

20. The _____ acceleration of an object at a certain time equals the slope of a velocity-time graph at that instant. instantaneous

21. The average velocity of an object moving along a line can be either positive or negative depending on the sign of the displacement. True or False True

22. The slope of a line tangent to the position-time curve at a point is equal to the ________ of the object at that time. Instantaneous velocity

23. The displacement of an object during a time interval can be found by determining the _________ on a velocity-time graph. Area under the curve

24. What does the area under the curve of an acceleration time graph tell you about the motion of an object? How much its velocity changed during that time interval.

25. The velocity of a car decreases from 30m/s to 10m/s in a time interval of 2 seconds. What is the acceleration? A = 10 – 30 2.0 A = -10m/s2

26. Which equation represents velocity as a function of displacement? V2 = vo2+2ax

27. Which equation represents displacement as a function of time? X = vot+.5at2

28. A plane traveling at 4m/s at a height of 100m above the ground drops a package. Where does that package fall relative to the point at which it was dropped? Horizontal: x = vxot =(40m/s)t Y = -.5gt2 -100m=-.5(9.8)t2 T=4.51s X=(40)(4.51)=180m

29. A stone is thrown upward from the top of a building at an angle of 30º to the horizontal and with an initial speed of 20m/s. The height of the building is 45m. How long is the stone in flight?Where does it land? At what angle does it strike the ground? vxo=vocosθo=17.3m/s vyo=vosinθo= 10m/s Y=vyot-.5gt2 -45m=(10)t-.5(9.8)t2 T=4.22s X=vxot= 73m Vy=vyo-gt= -31.36 m/s Vx=vox=17.3 m/s Tanf=vy/vx f=61o below horizontal

30. A golf ball is released from rest at the top of a tall building. Neglecting air resistance calculate the position and velocity of the ball after 3 seconds. V=at=(-9.8)t V=(-9.8)(3)= -29.4m/s Y=.5at2=.5(-9.8)t2 Y=.5(-9.8)(3)2= -44.1m

31. A stone is thrown from the top of a building with an initial speed of 20m/s straight upward. The building is 50m high and the stone just misses the edge of the roof on the way down. Determine a) the time for the stone to reach its max height, b) the max height, c)the time needed for the stone to return to the level of the thrower. • 20m/s + (-9.8m/s2)t=0 • t=2.04s • b) ymax=(20m/s)(2.04s) + .5(-9.8m/s2)(2.04)2 • ymax= 20.4m • c) y=vot + .5at2 • 0=20t – 4.90t2 • 0=t(20-4.90t) • t=0 t=4.80s

32. Explain what each of these equations are used for in a projectile motion problem involving a plane dropping a package to the ground. • y= -.5 gt2b) Vy= vyo- gt c)Dx=vxot • This equation can be used to determine the time of flight (initial vertical velocity is zero for a dropped object) • This equation is used to find the vertical component of the velocity just before the package hits the ground. • Since the package is dropped from the plane it has a horizontal velocity and moves horizontally at a constant rate (zero acceleration) the entire time it is in the air.

33. What equation is used to find the maximum height in a projectile motion problem? And in what column, horizontal or vertical, should this equation solved? The max height is found using y= (vosinθo)t - .5gt2 and it should be in the vertical column.

34. A stone is thrown upwards from the top of a building at an angle of 30º to the horizontal and with an initial speed of 20m/s. The height of the building is 45m. • How long is the stone in fight? • What is the speed of the stone just before it strikes the ground? Vxo= vocosθ= +17.3m/s Vyo= vosinθ = +10m/s y= vyot- .5gt2 -45m= (10m/s)t- .5(9.8)t2 T= 4.22s vy= vyo- gt= -31.4m/s v=√(vx2+ vy2)= 35.9m/s see slide #30 to find how to calculate horizontal distance, or angle of final velocity vector.

35. Explain all variables in this equation? Vy2=vyo2-2g Vy2 = final velocity in the y direction Vyo2 = initial velocity in the y direction g = downward acceleration due to gravity (9.8m/s2 on earth )

36. Define instantaneous velocity and give its equation. Instantaneous velocity is defined as the limit of the average velocity as t goes to zero. v= lim r/t t→0

37. Define average velocity and give its equation. Average velocity of an object moving along the x-axis during some time interval is equal to the displacement of the object divided by the time interval over which the displacement occurred. V= x/t = (xf-xi)/(tf-ti)

38. A car traveling at a constant speed of 30m/s (67mi/h) passes a trooper. One second later, the trooper sets off to chase the car with a constant acceleration of 30m/s2. How long does it take the trooper to overtake the car? Car Trooper a=zero a=30m/s2 (!!) xo=30 meters xo=0 meters v=30m/s (constant ) vo=0m/s xc=xo+vt xt= .5at2 Set their positions equal and solve for time: xo + vt = .5at2 t=[v±(v2+2axo)½]/a  t=2.73s