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Kinematics

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Kinematics. Describing Motion. Reference Frames. Measurements of position, distance or speed must be with respect to a frame of reference. What is the speed of a person with respect to the ground if she walks toward the back of the train at 5km/h while the train moves forward at 40 km/h?.

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Kinematics

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1. Kinematics Describing Motion

2. Reference Frames • Measurements of position, distance or speed must be with respect to a frame of reference. What is the speed of a person with respect to the ground if she walks toward the back of the train at 5km/h while the train moves forward at 40 km/h? Coordinate axes are used to represent the frame of reference

3. Displacement • Defined as change in position What is the displacement of a person who walks 100 m East then 60 m West? Answer: 40 m East • Displacement has both magnitude and direction – it is a vector

4. Representing Displacement • Let x1 be position of object at time t1 • Let x2 be position at time t2 • Then displacement Dx = x2 – x1 • D (Greek letter delta) means change If a person starts at x1 = 40m and walks to the left until reaching x2 = 10m what is the displacement? Answer: Dx = x2 – x1 = 10m – 40m = -30m

5. Average Speed and Velocity • Velocity is speed and direction • Average speed = distance traveled ÷ time elapsed • Average velocity = displacement ÷ time elapsed • Not always equal Find average speed and velocity for a trip 60m North followed by 40m South in 10 seconds Answers: 10 m/s; 2 m/s N

6. Example • During a four second interval a runner’s position changes from x1 = 50m to x2 = 10m. What was the average velocity? • vav= Dx/Dt = (x2 – x1 )/4s = -40m/4s = -10 m/s • What is the average speed? (always positive!) Challenge: convert this velocity to kilometers per hour -10 m/s x 1 km/1000m x 3600 sec/ 1h = -36 km/h Pro tip: use conversion tables on inside front cover of your text

7. How Far? • How far could a runner traveling at an average speed of 36 km/h go in 20. minutes? • 20 min = 1/3 hour • Dx = vavDt= 36 km/h x 1/3h = 12 km = 1.2 x 104 m

8. Three Ways • D = s x T • S = D/T • T = D/s

9. Instantaneous Velocity • Velocity at a particular instant of time • Defined as average velocity over an infinitesimally short time interval • v = lim(as Dt --> 0) Dx/Dt • Finite because both numerator and denominator approach zero; limit approaches a definite value

10. Acceleration • Acceleration is how fast velocity changes • Average acceleration = change of velocity ÷ time elapsed • a av = (v2 – v1)/(t2 –t1) = Dv/Dt

11. Example • Find average acceleration of a car that accelerates along straight road from rest to 80 km/h in 5 seconds • aav = (80 km/h – 0 km/h)/5s = 16 km/h/s

12. Convert to m/s/s • 80 km/h(1000m/1km)(1h/3600s) = 22.2 m/s • a av = (22.2 m/s – 0.0 m/s)/5.0 s = 4.4 m/s/s or 4. 4 m/s2 • Pronounced“meters per second squared” • Challenge: Can an object with zero velocity have non zero acceleration?

13. Object Slowing Down • Called deceleration • Find average acceleration of a car moving to the right(+x direction) 15.0 m/s when driver brakes to 5.0 m/s in 5.0 s? • aav =Dv/Dt = (5.0 m/s – 15.0 m/s) ÷ 5.0 s = • -2.0 m/s2 (acceleration negative) • Would acceleration still be negative if car was moving to the left? • NO! Its acceleration vector would then point to the right and be positive.

14. Constant(Uniform) Acceleration • Let t1 = 0, t2 = t = elapsed time • Re-name x1 = x0 ; v1 = v0 • Average velocity vav = (x – x0)/t • aav = (v – v0)/t • From these it is possible to derive (next slide)

15. For Uniform Acceleration, a • v = v0 +at (a) • x = x0 +v0t + 1/2at2 (b) • v2 = v02+ 2a(x-x0) (c) • vav = (v0+ v)/2 (d) • Each equation can be solved for any of the variables • Problems can be solved more than one way

16. v = v0 +at (a) • Describes change in velocity under uniform acceleration. • tells how fast a particle will be going at time t if at time zero its velocity was v0

17. x = x0 +v0t + 1/2at2 (b) • Describes change in position under uniform acceleration • Sometimes called "equation of motion." • tells where a particle will be at time t, if at time zero it was atx0 moving with velocity v0

18. v2 = v02+ 2a(x-x0) (c) • Velocity equation • Tells what velocity will be after a particle with initial velocity v0 accelerates a distance x - x0 with uniform acceleration a • Take square root to find v • Simplifies to v2 = 2ad when v0 = 0 and d = displacement

19. vav = (v0+ v)/2 (d) • Average (mean) velocity of particle for a trip under uniform acceleration • Provides shortcut way to solve certain problems

20. Special Case of Free Fall* • v = v0 –gt a = -g • y = y0 +v0t - 1/2gt2 • v2 = v02 - 2g(y-y0) • With y up positive, g = 9.80 m/s/s *Free Fall is motion under the influence of gravity alone This form of equations assumes down is positive

21. Concept Check (1) • The velocity and acceleration of an object • (a) must be in the same direction • (b) must be in opposite directions • (c) can be in the same or opposite directions • (d) must be in the same direction or zero (c) Can be in the same or opposite directions. Example: a rock thrown upward

22. Concept Check(2) • At the top of its path the velocity and acceleration of a brick thrown upward are (a) both non zero (b) both zero (c) velocity is zero acceleration is non zero (d) acceleration is zero, velocity is non zero (c) is correct; v = 0, a = 9.80 m/s/s downward

23. Problem Solving Tips • Read and re-read the problem • Make a diagram with all given info • Ask yourself “what is problem asking?” • Ask which physics principles apply • Look for most applicable equations • Be sure problem lies within their range of validity

24. Special Tricks • Break problem up into parts, like up and down part of path of an object thrown up • Use symmetry • Take situation to an extreme and look for a constraint • Choose reference frame that makes problem easiest

25. Problem solving… • Do algebraic calculations • Be aware you may have to solve equations simultaneously • Do arithmetic at end • Check units and significant figures • Ask, “is answer reasonable?”

26. Your Choice • You may choose up positive or negative; same with left or right • You may put the zero of coordinates anywhere you choose • Generally make choices that minimize number of negative quantities

27. Examples • What time is required for a car to travel 30.0 m while accelerating from rest at a uniform 2.00 m/s2 • x = ½ a t2 • t2 = 2x/a • t = (2x/a)1/2 • t = (2(30m)/2.00 m/s2) = 5.48s a = 2.00m/s2 x0 = 0 v0 = 0 x = 30m

28. Distance to Brake • A car traveling 28 m/s brakes at -6.0 m/s2. What distance is required to stop. • Method 1. Use v2 = v02+ 2a(x-x0) Solve for x = x0 + (v2 – v02) ÷ 2a x = (0 – 28 m/s2)÷ 2(-6.0 m/s2) = 65m • Method 2. Use vav = (v0+ v)/2 = 14 m/s v = v0 +at ; t = (v – v0)/a = -28m/s ÷ -6.0 m/s2 = 4.67 s x = vavt = 14 m/s (4.67s) = 65 m

29. Moral of Story • There is always more than one way to solve a (kinematics) problem! • You do not have to use the teacher’s way or the textbook way so long as the method and answer are correct. • Warning: Sometimes it is possible to get a correct answer using an incorrect method – no credit

30. Free Fall • Which falls faster, an elephant or a mouse?

31. Galileo’s Experiment • He asked, which would reach the ground first, a marble or a cannonball? Courtesy Dan Heller Photography

32. Galileo’s Discovery • All objects accelerate to earth equally (regardless of mass) • Air resistance must be neglected for this to be true • Acceleration due to gravity a = g = 9.80 m/s2

33. Velocity Reached • In free fall an object’s speed increases by about 10 m/s in each second. • v = gt ~ 10t • g ~ 10 m/s2

34. Distance Fallen From Rest • Distance fallen in t seconds d = 1/2gt2

35. Example: Ball Thrown Upward • Person throws ball upward with v = 15.0 m/s. (a) How high does it go? (b) How long is it in the air? • Choose y positive up, negative down • Choose y = 0 at throw height • Use v2 = v02 +2ay • y = (v2 – v02)/2a = (0 – (15.0m/s)2)/2(-9.80m/s2) = 11.5 m

36. Time In The Air • Use y = v0t + ½ a t2 • 0 = (15.0 m/s)t + ½(-9.80 m/s2)t2 • Factor: (15.0 m/s – 4.90 m/s2t)t = 0 • Two solutions: t =0 corresponds to instant ball thrown • T = 15.0 m/s ÷ 4.90 m/s2 = 3.06 s corresponds to instant ball returns to ground

37. Another Way? • Can you think of another way? • Find time to rise from v = v0 +at t = (v – v0)/a = (15m/s – 0)/9.80 m/s = 1.53s Then use average velocity 7.5 m/s x 1.53s = 11.5 m to find height. Total time in air is double 1.53s by symmetry. Gravity gives back speed on the way down at the same rate it took it away on the way up.

38. Puzzlers • What is the velocity of a ball thrown upward at its maximum height? • What is its acceleration at that height? • Are velocity and acceleration always in the same direction

39. Useful Form of Kinematics Equations for Free Fall • v = v0 - gt (a) • y = y0 +v0t - 1/2gt2 (b) • v2 = v02 - 2g(y-y0) (c) • vav = (v0+ v)/2 (d) • Here assuming up is positive • g = 9.80 m/s2

40. Graphing Motion • Position Graph (x or y vs. t) • Slope is velocity • If curved slope defined as slope of tangent to the curve at that point • Velocity Graph • Slope is acceleration • Area under graph is displacement • Acceleration Graph • Constant and non zero for uniform acceleration

41. Concept Check (3) • Which of these could be the velocity graph of a rock thrown upward, then falling downward? (Assume up is positive) (a) (b) (c) (d) (e)