Motion in One Dimension Displacement, Time, Speed, Velocity, Acceleration

1. Motion in One DimensionDisplacement, Time, Speed, Velocity, Acceleration Physics CH. 2

2. One Dimensional Motion • Simplest form of motion • Motion that takes place in only one direction • Commuter train on a track • Forward and backward

3. Frame of Reference • System for specifying the precise location of objects in space and time. • Allows us to take a complex situation and break it down into simpler parts • Complex situation: The commuter train is travelling down its tracks while the Earth is spinning on its axis, moving around the sun, while the sun is moving around the rest of our galaxy which is travelling through space. • Simplified Situation: Choose a train station to be the frame of reference. The station is now the origin for the train, and all motion is measured from this starting point.

4. Displacement ΔX • The change in position of an object • It is the difference between an object’s final position and its initial position. • ΔX = Xf – Xi • Not always equal to the distance travelled • Depends on direction (vector quantity) • Sign indicates direction • Right (East) and Up(North) will be considered positive (+) • Left(West) and Down (South) will be considered negative (-)

5. Vector vs. Scalar • A vector is a quantity that involves both magnitude and direction. • 55 miles north (+ 55 miles) • 3 km south (-3 km) • A scalar is a quantity that does not involve direction. • 63 miles • 18 cm long

6. Average Velocity vavg • Vector Quantity—magnitude and direction • The displacement (Δx) divide by the time (Δt) interval during which the displacement occurred • v avg = Δ x = xf - xi Δ t tf – ti • Velocity is not the same as speed • Speed is a scalar quantity • Speed = distance time The time interval is always positive! The sign of velocity (+ or -) depends on the displacement.

7. Practice Problem 1 • During a race on level ground, Andrea runs with an average velocity of 6.02 m/s to the East. What is Andreas displacement after 137 s?

8. Practice Problem 2 • Scott drives his car with an average velocity of 48.0 km/h to the east. How long will it take him to drive 144 km on a straight highway?

9. Velocity Graph • Position vs. Time • Position is on the y axis • independent variable • Time is on the x axis • dependent variable What does the slope of this line represent?

10. Find the average velocity between … • t = 0 and t=20 • t = 20 and t=40 • t = 0 and t=50

11. What kind of motion does this graph represent?

12. Compare 6 graphs

13. Acceleration • Rate of change of velocity with respect to time. • aavg= Δv = vf- vi Δ t tf – ti • Units

14. Acceleration Vector Quantity

15. Direction of Acceleration Δv = vf - v i • When Δ v is positive, acceleration is positive • When Δ v is zero, acceleration is zero • When Δ v is negative, acceleration is negative

16. Practice Problem 3 • A shuttle bus comes to a sudden stop to avoid hitting a dog. It accelerates uniformly at -4.1 m/s2 as it slows from 9.0 m/s to 0.0 m/s. How long does it take to come to a stop?

17. http://www.physicsclassroom.com/mmedia/index.cfm#kinema

19. Constant Positive Velocity

20. Constant Negative velocity

21. Positive Velocity, Positive Acceleration

22. Positive Velocity, Negative Acceleration

23. Negative Velocity, Negative Acceleration

24. Negative Velocity, Positive Acceleration

25. Instantaneous Velocity • If the position time graph is a straight line (constant velocity) , the instantaneous velocity is the same as the average velocity

26. Instantaneous Velocity • If the position time graph is a curve (not constant velocity), then the instantaneous velocity is the slope of the line tangent to the curve at a given point in time.

27. Instantaneous Acceleration • If the velocity time graph is a curve (not constant acceleration), then the instantaneous acceleration is the slope of the line tangent to the curve at a given point in time.

28. Displacement-Time Graph Wrap Up • On a displacement-time graph … • slope equals velocity. • the "y" intercept equals the initial displacement. • straight lines imply constant velocity. • curved lines imply acceleration. • average velocity is the slope of the straight line • instantaneous velocity is the slope of the line tangent to a curve at any point. • positive slope implies motion in the positive direction. • negative slope implies motion in the negative direction. • zero slope implies a state of rest. • The area under the curve is meaningless

29. Velocity-Time Graph Wrap Up • slope equals acceleration. • the"y" intercept equals the initial velocity. • straight lines imply uniform acceleration. • curved lines imply non-uniform acceleration. • an object undergoing constant acceleration traces a straight line. • average acceleration is the slope of the straight line • instantaneous acceleration is the slope of the line tangent to a curve at any point. • positive slope implies an increase in velocity in the positive direction. • negative slope implies an increase in velocity in the negative direction. • zero slope implies motion with constant velocity. • the area under the curve equals the displacement.

30. Acceleration-Time Graph Wrap Up • On an acceleration-time graph … • slope is meaningless. • the"y" intercept equals the initial acceleration. • an object undergoing constant acceleration traces a horizontal line. • zero slope implies motion with constant acceleration. • the area under the curve equals the change in velocity.

31. Displacement, CONSTANT Acceleration, Time Equation • v avg = Δx v avg = vi + vf Δt 2 • Δ x = vi + vf Δ t 2 • 2Δx = (vi+ vf) Δt • Δx = ½ (vi + vf) Δ t Applies only with constant acceleration

32. Practice Problem 4 • A racing car reaches a speed of 42 m/s. It un engages its “parachute” and braking system uniformly decelerates until it comes to a stop 5.5 s later. What distance does the race car travel during braking?

33. Initial Velocity, Acceleration, Time Equation • aavg= Δ v = vf - vi Δ t tf – ti • aavg= vf- vi tf– ti • a = vf - vi Δ t • a Δ t = vf– vi • a Δ t + vi= vf • vf = vi + a Δ t

34. Displacement, Velocity, Acceleration Equation • Δ x = ½ (vi + vf) Δt • vf = vi + a Δ t • Δ x = ½ (vi + vi + a Δ t )Δ t • Δ x = ½ (2vi + a Δ t ) Δt • Δ x = ½ (2vi Δt + a Δt2 ) • Δ x = vi Δ t + ½ a Δ t2

35. Practice Problem 5 • A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s2 for 15 s before takeoff. • What is its speed at takeoff? • How long must the runway be for the plane to be able to take off?

36. Initial Velocity, Acceleration, Displacement Equation vf - vi = a 2Δ x (vi + vf) (vf- vi) (vi + vf)= a (2Δx) (vf2- vi2) = 2a Δ x • Δ x = ½ (vi + vf) Δt • 2Δx = (vi + vf) Δ t • 2Δ x = Δ t (vi + vf) vf= vi +a Δ t • vf = vi +a 2Δ x (vi + vf)

37. Practice Problem 6 • A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2 . What is the velocity of the stroller after it has travelled 4.75 m?

38. Falling Objects • The Hammer and the Feather • David Scott • Apollo 15 • 8/2/1971 • http://www.youtube.com/watch?v=KDp1tiUsZw8

39. An initially-stationary object is allowed to fall freely under gravity drops a distance which is proportional to the square of the elapsed time. This image, spanning half a second, was captured with a stroboscopic flash at 20 flashes per second. During the first 1/20th of a second the ball drops one unit of distance (here, a unit is about 12 mm); by 2/20ths it has dropped at total of 4 units; by 3/20ths, 9 units and so on

40. Free Fall Acceleration • If air resistance is disregarded, all objects dropped near the surface of a planet fall with the same constant acceleration • Acceleration due to gravity (g) • g = - 9.81 m/s2 • g = - 32 ft/s2

41. Object Tossed Straight Up • Acceleration is constant during upward and downward motion • Up • Positive velocity • Negative acceleration • Slows down as it rises • At the top of the path, the velocity is zero • Down • Negative velocity • Negative acceleration • Speeds up as it falls

42. Practice Problem • Jason hits a volleyball so that it moves with an initial straight upward velocity of 6.0 m/s. • How high will the ball go? • How long does it take for the ball to come back to Jason?

43. Reaction Time • Get in groups of 2-3 • Have one lab partner hold a meter stick vertically between the thumb and forefinger of a second lab partner. • The meter stick should be held so that the zero mark is between the fingers of the second lab partner with the 1 cm mark above. • The second lab partner should not be touching the meter stick and his/her catching hand must be resting on a table. • Without warning, the first lab partner should drop the meter stick so that it falls between the thumb and forefinger. The second lab partner catches the meter stick as quickly as he/she can. • Calculate the reaction time from free fall acceleration and the distance the meter stick has fallen.