1. Lecture #2 Chapter 2 Motion along straight line PHY 2048 Fall 2007

2. Chapter 1 - Introduction I.International System of Units II.Conversion of units III.Dimensional Analysis

3. I.International System of Units QUANTITYUNIT NAMEUNIT SYMBOL Lengthmeterm Timeseconds Masskilogramkg Speedm/s Accelerationm/s 2 ForceNewtonN=kgm/s 2 PressurePascalPa = N/m 2 EnergyJouleJ = Nm=kgm 2 /s 2 PowerWattW = J/s TemperatureKelvinK POWERPREFIXABBREVIATION 10 15 petaP 10 12 teraT 10 9 gigaG 10 6 megaM 10 3 kilok 10 2 hectoh 10 1 dekada 10 -1 deciD 10 -2 centic 10 -3 millim 10 -6 microμ 10 -9 nanon 10 -12 picop 10 -15 femtof

4. Example: 316 feet/h  m/s Conversion of units II. Conversion of units Chain-link conversion method: The original data are multiplied successively by conversion factors written as unity. The units are manipulated like algebraic quantities until only the desired units remain. III. Dimensional Analysis Dimension of a quantity: indicates the type of quantity it is; length [L], mass [M], time [T] Example: x=x 0 +v 0 t+at 2 /2 Dimensional consistency: an equation is dimensionally consistent if each term in it has the same dimensions.

5. Problem solving tactics Explain the problem with your own words. Make a good picture describing the problem. Write down the given data with their units. Convert all data into S.I. system. Identify the unknowns. Find the connections between the unknowns and the data. Write the physical equations that can be applied to the problem. Solve those equations. Check if the values obtained are reasonable  order of magnitude and units.

6. Chapter 2 - Motion along a straight line I.Position and displacement II.Velocity III.Acceleration IV.Motion in one dimension with constant acceleration V.Free fall MECHANICS  Kinematics

7. I.Position and displacement Displacement: Change from position x 1 to x 2  Δx = x 2 -x 1 (2.1) - Vector quantity: Magnitude (absolute value) and direction (sign). - Coordinate ≠ Distance  x ≠ Δx t x Δx = 0 x 1 =x 2 Only the initial and final coordinates influence the displacement  many different motions between x 1 and x 2 give the same displacement. t x Δx >0 x1x1 x2x2 Coordinate system

8. II. Velocity Average velocity: Ratio of the displacement Δx that occurs during a particular time interval Δt to that interval. Motion along x-axis Distance: total length of travel. Displacement ≠ Distance Example: round trip house-work-house  distance traveled = 10 km displacement = 0 - Scalar quantity - Vector quantity  indicates not just how fast an object is moving but also in which direction it is moving. - The slope of a straight line connecting 2 points on an x-versus-t plot is equal to the average velocity during that time interval.

9. Average speed: Total distance covered in a time interval. Instantaneous velocity: How fast a particle is moving at a given instant. Example: A person drives 4 mi at 30mi/h and 4 mi and 50 mi/h  Is the average speed >,<,= 40 mi/h ? <40 mi/h t 1 = 4 mi/(30 mi/h)=0.13 h ; t 2 = 4 mi/(50 mi/h)=0.08 h  t tot = 0.213 h  S avg = 8 mi/0.213h = 37.5mi/h S avg ≠ magnitude V avg S avg always >0 Scalar quantity - Vector quantity

10. When the velocity is constant, the average velocity over any time interval is equal to the instantaneous velocity at any time. Instantaneous speed: Magnitude of the instantaneous velocity. Example: car speedometer. - Scalar quantity Average velocity (or average acceleration) always refers to an specific time interval. Instantaneous velocity (acceleration) refers to an specific instant of time. Slope of the particle’s position-time curve at a given instant of time. V is tangent to x(t) when Δt  0 Instantaneous velocity: Time Position

11. III. Acceleration V Average acceleration: Ratio of a change in velocity Δv to the time interval Δt in which the change occurs. - Vector quantity - The average acceleration in a “v-t” plot is the slope of a straight line connecting points corresponding to two different times. t t t Instantaneous acceleration: Rate of change of velocity with time. - Vector quantity - The instantaneous acceleration is the slope of the tangent line (v-t plot) at a particular time.

12. Example: x(t)=At 2  v(t)=2At  a(t)=2A ; At t=0s, v(0)=0 but a(0)=2A Example: v 1 = -25m/s ; v 2 = 0m/s in 5s  particle slows down, a avg = 5m/s 2 An object can have simultaneously v=0 and a≠0 If particle’s velocity and acceleration have same sign  particle’s speed increases. If the signs are opposite  speed decreases. Positive acceleration does not necessarily imply speeding up, and negative acceleration slowing down. IV. Motion in one dimension with constant acceleration - Average acceleration and instantaneous acceleration are equal.

13. 1D motion with constant acceleration t f – t i = t

14. 1D motion with constant acceleration Using and than substituting into equation for final position yields (1) (2) Equations (1) and (2) are the basic kinematics equations

15. 1D motion with constant acceleration In a similar manner we can rewrite equation for average velocity: and than solve it for x f Rearranging, and assuming

16. 1D motion with constant acceleration These two equations can be combined to yield additional equations. We can eliminate t to obtain Second, we can eliminate the acceleration a to produce an equation in which acceleration does not appear:

17. Equations of motion

19. Kinematics with constant acceleration - Summary

20. Kinematics - Example 1 How long does it take for a train to come to rest if it decelerates at 2.0m/s 2 from an initial velocity of 60 km/h? –Using we rearrange to solve for t: –V f = 0.0 km/h, v i =60 km/h and a= -2.0 m/s 2. –Substituting we have

21. - Equations for motion with constant acceleration: t t t t missing

22. - The motion equations can also be obtained by indefinite integration: V. Free fall Free fall acceleration: (near Earth’s surface) a= -g = -9.8 m/s 2 (in cte acceleration mov. eqs.) Due to gravity  downward on y, directed toward Earth’s center Motion direction along y-axis ( y >0 upwards)

23. Approximations: - Locally, Earth’s surface essentially flat  free fall “a” has same direction at slightly different points. - All objects at the same place have same free fall “a” (neglecting air influence). VI. Graphical integration in motion analysis From a(t) versus t graph  integration = area between acceleration curve and time axis, from t 0 to t 1  v(t) Similarly, from v(t) versus t graph  integration = area under curve from t 0 to t 1  x(t)

24. B1: A rocket is launched vertically from the ground with an initial velocity of 80m/s. It ascends with a constant acceleration of 4 m/s 2 to an altitude of 10 km. Its motors then fail, and the rocket continues upward as a free fall particle and then falls back down. (a) What is the total time elapsed from takeoff until the rocket strikes the ground? (b) What is the maximum altitude reached? (c) What is the velocity just before hitting ground? x y v 0 = 80m/s t 0 =0 a 1 = -g a 0 = 4m/s 2 y 2 =y max y 1 = 10 km v 2 =0, t 2 +v 1, t 1 t1t1 t2t2 t 3 =t 2 t4t4 a 2 = -g 1) Ascent  a 0 = 4m/s 2 2) Ascent  a= -9.8 m/s 2 Total time ascent = t 1 +t 2 = 53.48 s+29.96 s= 83.44 s 3) Descent  a= -9.8 m/s 2 t total =t 1 +2t 2 +t 4 = 53.48 s + 2·29.96 s + 24.22 s=137.62 s h max = y 2  y 2 -10 4 m = v 1 t 2 -4.9t 2 2 = (294 m/s)(29.96s)- (4.9m/s 2 )(29.96s) 2 = 4410 m v3v3