Kinematics - (describing how things move)

1. Kinematics - (describing how things move) Change in position (How far you travel in a given direction) How far you travel How fast you travel (in a given direction) How fast you travel Rate of change of velocity

2. Describing Motion There are lots of different ways to describe motion…. Words Sketches Time elapsed photographs Physical Expressions (Equations) Graphical Representation

3. Kinematics Equations that Make Sense! Average speed: sav = d / change in t SI unit: m/s sav = d / ∆t = d / tf - ti Average velocity:vav = ∆d / ∆t SI unit: m/s vav = (df - di) / t df = di + vavt Average acceleration:aav = ∆v / ∆t SI unit: m/s/s = m/s2 aav = (vf - vi) / t vf = vi+ aavt Note: if the time intervals are very small we call these quantities instantaneous

4. Using Split Times! 3.1 2.1 1.7 1.4 1.2(5) Determine the average velocity for each distance interval Determine average velocity of the object over the time recorded = 25m - 0m/ 14.5 s vav = df - di / t = 1.7 m / s Determine the average acceleration over the time recorded a = vf - vi / t = (1.25 m/s - 3.1m/s) / 14.5 s = - 0.13 m / s2 Note: the a is negative because the change in v is negative!!

5. Reference Frames All measurements are made relative to a frame of reference.. Sitting at your desk right now you don’t seem to be moving but you are infact revolving at 30 km/s (67,000 mi/h) around the sun The earth also has a rotational speed about its axis. People on earth experience different tangential speeds depending on their latitude. The space shuttle at Kennedy Space Center is already traveling at 410 m/s (917 mi/h) before it even gets off the ground. A geosynchronous satellite that is in a stationary orbit over the earth is traveling at over 3080 m/s (6890 mi/h) Someone moving south at 5 km/h on a train which itself is moving south at 80 km/h will be moving at 85 km/h relative to the earth. If the person had been moving north on the train their velocity would have been 75 km/h relative to the earth.

6. Solving Kinematics Problems • Assign a coordinate system – Define which directions are positive and negative. • Write down your known variables and show unknowns with a question mark. • Write down the kinematics expression that will allow you to solve for one variable. All the others in your expression should be known. Rearrange if necessary. • Substitute numbers and units into your physical expression. • Solve the equation for your unknown and include the correct units. • Check your answer. Does the magnitude of your answer make sense? • Do the units come out right? • Can you use another expression to check your answer?

7. Example 1 Distance Run by a Jogger How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m/s? t = 5400s sav = 2.22 m/s d = ? sav = d / t So…… d = savt = (2.22 m/s)(5400s) d = 12000 m Check: Units come out right (m) when multiplied

8. Example 2 The World’s Fastest Jet-Engine Car Andy Green in the car ThrustSSC set a world record of 341.1 m/s in 1997. To establish such a record, the driver makes two runs through the course, one in each direction,to nullify wind effects. From the data, determine the averagevelocity for each run. a) t = 4.740 s x = +1609m vav = ? vav = x/ t = (+1609m)/ (4.740 s) vav = + 339.5 m/s b) t = 4.695 s x = -1609m vav = ? vav = x/ t = (-1609m)/ (4.695 s) vav = - 342.7 m/s

9. Example 3Acceleration and Increasing Velocity Determine the average acceleration of the plane. vi = 0 km/h vf = 260 km/h ti = 0s tf = 29s aav = (vf - vi) / (tf – ti) aav = (+260 km/h – 0 km/h) / (29s – 0s) aav = + 9.0 km/h /s

10. Graphical Representation of Motion Kinematics Relationships Through Graphing: 1. The slope of a d-t graph at any time tells you the av. velocity of the object. 2. The slope of a v-t graph at any time tells you the av. acceleration of the object. The area under a v-t graph tells you the displacement of the object during that time. 4. The area under a a-t graph tells you the change in velocity of the object during that time

11. Constant Motion On the d-t graph at any point in time … vav = ∆d / ∆t On the v-t graph at any point in time… aav = vf - vi / t On the a-t graph the area between the line and the x-axis is…. aav = (10 - 10)m/s / (5 - 0)s vav = (50 - 0)m / (5 - 0)s Area of rectangle = b x h aav = 0 m/s2 Area = 5s x 0 m/s2 = 0 m/s vav = 10 m/s Looking at the area between the line and the x-axis…. The area thus represents…. The slope is constant on this graph so the velocity is constant ∆v = aav∆t Area of rectangle = b x h Area = 5s x 10 m/s = 50 m Change in velocity Which is of course displacement

12. Changing Motion On the d-t graph at any point in time … vav = ∆d / ∆t On the v-t graph at any point in time… aav = vf - vi / t On the a-t graph the area between the line and the x-axis is…. aav = (20 - 0)m/s / (5 - 0)s The slope is constantly increasing on this graph so the velocity is increasing at a constant rate Area of rectangle = b x h aav = 4 m/s2 Area = 5s x 4 m/s2 = 20 m/s Looking at the area between the line and the x-axis…. The area thus represents…. The slope of a tangent line drawn at a point on the curve will tell you the instantaneous velocity at this position Area of triangle = 1/2 (b x h) Change in velocity Area = 1/2 (5s x 20 m/s) = 50 m Which is of course displacement

13. To determine the velocity at any point in time you need to find the slope of the distance-time graph. This means that you need to find the slope of the tangent line drawn at the point of interest. By selecting two points spaced evenly on either side of the point of interest, a line can be drawn between them that has the same slope as the tangent. (shown below). Slope between 1s and 3s shows the velocity at 2s The velocity at 2s is p/t = (18m - 2m)/ (3s - 1s) = 8m/s The acceleration is given by the slope of the velocity-time graph. Therefore: a = v / t = 20m/s / 5s = 4m/s2

14. Example Problem • A student is late for the school bus. She runs east down the road at 3 m/s for 30s, then thinks that she has dropped her calculator so stops for 10s to check. She jogs back west at 2 m/s for 10s, stops for 5 s then accelerates uniformly from rest to 4 m/s east over a 10 second period. • Sketch the velocity-time graph of the student’s motion • Determine the total distance and displacement of the student during this time • Determine the student’s average velocity during this time

15. Velocity-Time Graph of the Student’s Motion

16. Total distance traveled by the student is…. dtotal = d1 + d2 + d3 + d4 + d5 dtotal = s1t1 + s2  t2+ s3 t3 + s4 t4 + s5 t5 dtotal = (3m/s)(30s) + (0m/s)(10s) + (2m/s)(10s) + ………(0m/s)(5s) + (1/2(4m/s)(10s) dtotal = 130 m Total displacement by the student is…. + east - west dtotal =  d1 +  d2 +  d3 +  d4 +  d5 dtotal = v1t1 + v2 t2+ v3 t3 + v4 t4 + v5 t5 dtotal = (3m/s)(30s) + (0m/s)(10s) + (-2m/s)(10s) + ………(0m/s)(5s) + (1/2(4m/s)(10s) dtotal = + 90 m (East)

17. Average velocity of the student is….. vav = dtotal / ttotal = + 90m East / 65s vav = 1.4 m/s East + 90 m + 20 m - 20 m

18. More Kinematics Equations that Make Sense! df = di + vavt but vav = (vi + vf) / 2 df = di + (vi + vf) /2t but vf = vi + aavt df = di + (vi + (vi + aav t) /2t df = di + vi t + 1/2 aavt2 or… d = vi t + 1/2 aavt2 df = di + (vi + vf) /2t but t = (vf - vi ) / aav df = di + (vi + vf) /2(vf - vi) /aav d = (vi + vf) /2(vf - vi) /aav So… d = (vf2 - vi2 ) /2aav and… vf2 = vi2 + 2aav d

19. Free Fall Free fall describes the motion of an object which is only under the influence of gravity. I.e. a ball thrown upwards or dropped An object in free fall experiences a constant uniform acceleration of 9.8 m/s2 in the downwards direction Kinematics equations can be used for solving free fall problems by replacing aav in the expressions with g where g is 9.8 m/s2 downwards vf = vi + g t dv = vi t + 1/2 g t2 vf2 = vi2 + 2gdv Air resistance limits the time of free fall. Eventually a falling object will reach a constant velocity downwards known as its terminal velocity - vi is negative + vi is positive g is positive g is negative + - If you define up as the positive direction, g must be negative because the velocity gets less positive over time If you define up as the negative direction, g must be positive because the velocity gets less negative over time

20. Example 1 A Falling Stone A stone is dropped from the top of a tall building. After 3.00s of free fall, what is the displacement y of the stone? Freely Falling Bodies

21. Freely Falling Bodies ? - 9.8 m/s2 0 m/s 3.00 s

22. Freely Falling Bodies dv = vi t + ½ g t2 = (0 m/s)(3.00s) + ½ (-9.8 m/s2)(3.00s)2 = - 44.1 m

23. Example 2 How High Does it Go? The referee tosses the coin up with an initial speed of 5.00m/s. In the absence if air resistance, how high does the coin go above its point of release? Freely Falling Bodies

24. Freely Falling Bodies ? - 9.80 m/s2 0 m/s + 5.00 m/s

25. Freely Falling Bodies vf2 = vi2 + 2 g dv = 1.28 m