Displacement, Velocity and Acceleration

1. Chapter 2 Displacement, Velocity and Acceleration

2. Solve the following Average Acceleration Problem Find the acceleration of an amusement park ride that falls from rest to a speed of 28 m/s in 3.0 s. Answer 9.3 m/s2

3. Motion • The simplest form is one dimension like a train moving back and forth along one track • Depends on a frame of reference = a coordinate system for specifying the precise location of objects in space (against which you measure changes in position) • Why use a frame of reference? Think of all motion around the train – Earth, Solar System, Galaxy, etc.

4. Displacement • The change in position of an object – the length of the straight line drawn from its initial position to its final position • ∆x = xf - xi • Change in position = final position – initial position • Not always equal to the distance traveled and can equal zero if initial and final position is the same • Can be negative or positive

7. Average velocity is displacement divided by the time interval in m/s Velocity • Average velocity is displacement divided by the time interval in m/s

9. Velocity • Can be positive or negative if the displacement is negative. • Time interval is always positive • Velocity is not the same as speed – describes motion with both a direction and a numerical value  a magnitude • Speed has no direction, only a magnitude

11. Velocity can be interpreted graphically • Instantaneous velocity may not be the same as average velocity  the velocity of an object at some instant or specific point – when position versus time graph is not a straight line

12. Acceleration • Measures the rate of change in velocity • The magnitude of acceleration is calculated by dividing the total change in an object’s velocity by the time interval which the change occurs

14. Aavg = ∆v / ∆t (vf – vi / tf – ti) • Average acceleration = change in velocity / time required for change • Units are written as meters per second per second or m/s2 • Acceleration has direction and magnitude

15. Solve the following Average Acceleration problem • Find the acceleration of an amusement park ride that falls from rest to a speed of 28 m/s in 3.0 seconds. • Answer 9.3 m/s2

16. Motion with Constant Acceleration • The following image was taken 10 times in 1 second and shows that the ball travels a greater distance each 0.10 s as its velocity increases

18. Because acceleration is constant, the velocity increases by exactly the same during each time interval

19. Displacement with constant uniform acceleration • Displacement depends on acceleration, initial velocity and time • ∆x = ½(vi + vf) ∆t • Displacement = ½(initial velocity + final velocity)(time interval)

20. Practice Problem-Displacement with Uniform Acceleration • A bicyclist accelerates from 5.0 m/s to a velocity of 16 m/s in 8 s. Assuming uniform acceleration, what distance does the bicyclist travel during this time interval? • Answer 84 meters

21. Velocity w/ constant uniform acceleration • vf = vi + a∆t • final velocity = initial velocity + (acceleration x time interval)

22. Displacement w/ constant uniform accleration • ∆x = vi∆t + ½a(∆t)2 • displacement = (initial velocity x time interval) + ½ acceleration x (time interval)2

23. Final Velocity after any displacement • vf2 = vi2 + 2a∆x • (final velocity)2 = (initial velocity)2 + 2(acceleration)(displacement) • Take the square root of the right side to find final velocity • Square root may be positive or negative – the right value is obtained by reasoning based on the direction of motion

24. Falling Objects • Aug. 2 1971 – astronaut David Scott demonstrated a feather and hammer hitting the ground at the same time on the moon • When there is no air resistance, all objects fall with the same acceleration regardless of their mass – freefall

25. Free Fall • In the absence of air resistance, it is found that all bodies at the same location above the earth fall vertically with the same acceleration.( decreases with increasing altitude and varies slightly with latitude) • Furthermore, if the distance of the fall is small compared to the radius of the earth, the acceleration remains essentially constant. • Since the acceleration is constant in free-fall, the equations of kinematics can be used.

27. The time between the images is the same but the displacement at each time interval increases which means the velocity was not constant – apple and feather are accelerating

28. Freefall acceleration is denoted with the symbol g. On Earth, the magnitude of g is approximately 9.81 m/s2 or 32 ft/s2 • The acceleration of objects in freefall near the surface of Earth is a = g = -9.81 m/s2 • All kinematics equations can be used for free fall just replace x with y

29. What goes up must come down! • We know that when you throw a ball up, it will continue to move upward for some time, stop at the peak and then change direction and begin to fall – objects thrown into the air have a downward acceleration as soon as they are released • Why?

31. Although the velocity of the ball is zero at its peak of upward motion, the acceleration is equal to -9.81 m/s2 • Freely falling objects always have the same downward acceleration

32. The downward acceleration is the same when the object is moving up, at rest at the top of its path, and when moving down • What changes are the position and the magnitude and direction of velocity

34. When an object is thrown up in the air, it has positive velocity and negative acceleration = object is slowing down • When the object begins moving down, it has negative velocity and acceleration • Negative velocity and a negative acceleration according to table 2-3 indicates the object is speeding up – freefall acceleration

35. Free Fall Equation