Velocity. UF PHY2053, Lecture 3: Motion in One Dimension

Transcription

1 Velocity 1

2 Average Velocity vs Average Speed Average Velocity is the change in position during a time interval vav,x Δx Δt x f - xi tf - ti Average Speed is the distance traveled during a time interval (hint: car model ) distance traveled average speed total time 2 =

3 Graphical Representation, Constant velocity Rise of the graph, Δx Run of the graph, Δt 3

4 Graphical Representation, Constant velocity v av,x = Δx Δt 4

5 Average Velocity, Non Constant The motion is nonconstant velocity The average velocity is the slope of the blue line joining two points 5

6 Clicker Question #1 (5 min) A car is driving from city A to city B. It drives the first half of the distance at a constant velocity of 15 m/s, and then drives the second half of the distance at a constant velocity of 30 m/s. What is the average velocity of the car for this trip? 6

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8 Instantaneous Velocity The limit of the average velocity as the time interval becomes infinitesimally short, or as the time interval approaches zero vx lim Δx Δt 0 Δt = lim tf ti x f - xi tf - ti The instantaneous velocity indicates what is happening at every point of time Instantaneous speed can be defined by analogy. It is the average speed as the total time becomes infinitesimally short. (= abs. value of instantaneous velocity) 8

9 Instantaneous Velocity Δx Δt 9

10 Instantaneous Velocity Δx Δt 10

11 Instantaneous Velocity Δx Δt 11

12 Instantaneous Velocity Δx Δt 12

13 The Inverse Problem: From Velocity to Position 13

14 Computing positions from velocity distributions [m/s] v x t [s] v x = 5 m/s for 2 seconds (6s 4s) Change in position: x = 2s 5m/s = 10 m Notice: if v x were negative, this would indicate movement in the negative direction 14

15 A more complex distribution.. [m/s] v x t [s] Nontrivial shape How would we compute the change in position? Start from a rough approximation.. 15

16 A more complex distribution.. [m/s] v x t [s] Rough approximation: assume that the velocity is constant in increments of one second. From t=2s to t=3s, velocity ~ 4 m/s, distance traveled ~ 4 m. 16

17 A more complex distribution.. [m/s] v x t [s] Rough approximation: assume that the velocity is constant in increments of one second. From t=2s to t=3s, velocity ~ 4 m/s, distance traveled ~ 4 m. Do this for the entire time range of motion; sometimes undershoot, sometimes overshoot 17

18 A more complex distribution.. [m/s] v x t [s] Improve approximation: assume that the velocity is constant in increments of one half second. Better description of change of position, more precise calculation 18

19 A more complex distribution.. [m/s] v x t [s] Improve approximation: assume that the velocity is constant in increments of one half second. Better description of change of position, more precise calculation Computation with increments of a quarter second would be better; Can continue to infinity.. 19

20 A more complex distribution.. [m/s] v x Δx t [s] Can continue to infinity.. At which point we find that the distance traveled equals the area under the graph of velocity vs time 20

21 Clicker Problem #3 [2 min] [m/s] v x t [s] For the velocity distribution on the left, what is the average velocity from t=2s to t=8 s? 21

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23 Acceleration 23

24 Average and Instantaneous Acceleration the rate of change of velocity has physical relevance By analogy with velocity, define average acceleration: aav,x Δv x Δt and instantaneous acceleration: = vx,f - vx,i tf - ti ax lim Δv x Δt 0 Δt = lim tf ti vxf - vxi tf - ti 24

25 Graphical Representation [m/s] v x B A t [s] By analogy with velocity in the graph of position vs time: Average acceleration between A and B is the slope of the blue line connecting them Instantaneous acceleration is the slope of the tangent at the point of interest (A) 25

26 Graphical Representation [m/s] v x By analogy with velocity in the graph of position Graphs of Velocity B vs Time vs will time: be particularly useful and interesting: Average acceleration From the area under the between graph obtain A and Δx B is the From the slope of the chord slope obtain of the avg blue accel. line A From the slope of the tangent, connecting inst. accel. them t [s] Instantaneous acceleration is the slope of the tangent at the point of interest (A) 26

27 Experiment: Motion Sensor Cart 27

28 Motion along a line with constant acceleration 29

29 Graphical representation Special case: slope of the velocity vs time graph is constant! Take advantage of this and derive some practical formulas: First, the obvious one: 30

30 Position vs Time for Const Accel. Reminder: Change in position is the area under the velocity graph Triangular area = ½(Δv x Δt) = ½(a x Δt 2 ) Rectangular area = v ix Δt Total area = Rectangle + Triangle 31

31 Connecting Change in Position, Initial and Final Velocities Recall: so, Δt=Δv x/ a x Triangular area = ½(Δv x Δt) = ½(a x Δt 2 ) Rectangular area = v ix Δt Total area = Rectangle + Triangle 32

32 Example Problem: A Mustang GT500 goes from 0 to 60 mph in 4.1 seconds. What is the average acceleration over this period? The stopping distance from 60 mph has been measured to be 120 ft. Assuming constant deceleration, find its magnitude. [all accelerations should be computed in m/s 2 ] 33

33 Free Fall 34

34 Free Fall Special case of motion with constant acceleration, acceleration of Earth s gravitational field g= 9.80 m/s 2 Important: as much as possible try to keep a fixed convention about what is the positive y direction Recommend: positive y is up. 35

35 Free Fall Height after 1, 2, 3.. s Acceleration of Earth s gravitational field g= 9.80 m/s 2 Recall: Let us consider v i = 0, y i =0: start from standing still at origin t 1s 2s 3s 4s y ½gs 2 2gs 2 4.5gs 2 8gs 2 y/y(1s)

36 Experiment: 1:4:9:16 Free Fall demonstration 37

37 Supporting Materials 38

38 Demonstrations Cart Motion Sensor 1:4:9:16 Free Fall demonstration 39