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

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

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 petaP teraT 10 9 gigaG 10 6 megaM 10 3 kilok 10 2 hectoh 10 1 dekada deciD centic millim microμ nanon picop femtof

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.

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.

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

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

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.

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 = h  S avg = 8 mi/0.213h = 37.5mi/h S avg ≠ magnitude V avg S avg always >0 Scalar quantity - Vector quantity

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

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.

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.

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

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

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

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:

Equations of motion

Kinematics with constant acceleration - Summary

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

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

PROBLEMS - Chapter A red car and a green car move toward each other in adjacent lanes and parallel to The x-axis. At time t=0, the red car is at x=0 and the green car at x=220 m. If the red car has a constant velocity of 20km/h, the cars pass each other at x=44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x=76.6m. What are (a) the initial velocity, and (b) the acceleration of the green car? x d=220 m X r1 =44.5 m X r2 =76.6m O v r1 =20km/h v r2 =40km/h X g =220m a g = 2.1 m/s 2 v 0g = m/s

63: At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.2 m/s 2. At the same instant, a truck, traveling with constant speed of 9.5 m/s, overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be traveling at that instant? x (m) t = 0s x=0 Car Truck x=d ? a c = 2.2 m/s 2, v c0 =0 m/s v t = 9.5 m/s Truck Car 97: A proton moves along the x-axis according to the equation: x = 50t+10t 2, where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3s of its motion. (b) Instantaneous velocity of the proton at t = 3s. (c) Instantaneous acceleration of the proton at t = 3s.

(d) Graph x versus t and indicate how the answer to (a) (average velocity) can be obtained from the plot. (e) Indicate the answer to (b) (instantaneous velocity) on the graph. (f) Plot v versus t and indicate on it the answer to (c). x = 50t + 10t 2 v = t 16. An electron moving along the x-axis has a position given by: x = 16t·exp(-t) m, where t is in seconds. How far is the electron from the origin when it momentarily stops? x(t) when v(t)=0??

When a high speed passenger train traveling at 161 km/h rounds a bend, the engineer is 37. When a high speed passenger train traveling at 161 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered into the track from a siding and shocked to see that a locomotive has improperly entered into the track from a siding and is a distance D= 676 m ahead. The locomotive is moving at 29 km/h. The engineer of the is a distance D= 676 m ahead. The locomotive is moving at 29 km/h. The engineer of the high speed train immediately applies the brakes. (a) What must be the magnitude of the high speed train immediately applies the brakes. (a) What must be the magnitude of the resultant deceleration if a collision is to be avoided? (b) Assume that the engineer is at x=0 resultant deceleration if a collision is to be avoided? (b) Assume that the engineer is at x=0 when at t=0 he first spots the locomotive. Sketch x(t) curves representing the locomotive when at t=0 he first spots the locomotive. Sketch x(t) curves representing the locomotive and high speed train for the situation in which a collision is just avoided and is not quite and high speed train for the situation in which a collision is just avoided and is not quite avoided. avoided. x (m) t = 0s t > 0s x=D x=0 Train Locomotive x=D+d L v T =161km/h = m/s = v T0  1D movement with a<0 v L =29 km/h = 8.05 m/s is constant Locomotive Train dLdL

37. Locomotive Train Collision can not be avoided Collision can be avoided - Collision can be avoided: Slope of x(t) vs. t locomotive at t = s (the point were both Lines meet) = v instantaneous locom > Slope of x(t) vs. t train - Collision cannot be avoided: Slope of x(t) vs. t locomotive at t = s < Slope of x(t) vs. t train

- 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)

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)

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 = s s= s 3) Descent  a= -9.8 m/s 2 t total =t 1 +2t 2 +t 4 = s + 2·29.96 s s= s h max = y 2  y m = v 1 t t 2 2 = (294 m/s)(29.96s)- (4.9m/s 2 )(29.96s) 2 = 4410 m v3v3

Announcements - 08/27/04: Teaching assistant: Mads Demenikov Office: 502 Office hours: Wed 8-11 am, 1-3 pm -My webpage: Note: I will post this week’s lectures + solved problems tonight. - Webassign- First assignment available on-line 08/27/04  Due date: Monday, September 6 th, 11:30 pm Chapter 2 -Webassign problems: 5,13,17,24,25,38,44,106 (7th ed.book)