3But First a Review Significant Figures Non-zero digits are always significant.Any zeros between two significant digits are significant.A final zero or trailing zeros in the decimal portion are significant.Ex has 4 significant figuresEx. 2,500 has 2 significant figuresEx x 103 has 4 significant figuresMultiplication/Division – Determined by the LEAST number of significant figuresAddition/Subtraction – Determined by LEAST number of decimal of places in the decimal portion
4VectorsVectors are physical quantities with both magnitude and direction and cannot be represented by just a single numberDisplacement vs. DistanceVelocity vs. SpeedRepresented by AThe magnitude of A is represented by |A| or AP2AP1P2B=-AP1
5Vector Addition Tip to tail method or Parallelogram method Vector addition is commutative(a)(b)
24Displacement Is a vector quantity, usually denoted by x. Change in the position of a point. (we can approximate objects to be a particle)Remember, since it’s a vector, it’s important to note both magnitude and direction.Define positive displacement to be a movement along the positive x-axis
25Average VelocityP1At time t1 the car is at point P1 and at time t2 the car is at point P2We can define P1 and P2 to have coordinates x1 and x2 respectivelyΔx=x2-x1Average velocityP2
26VelocityVelocity is the change in displacement per unit time in a specific direction.It is a vector quantity, usually denoted by vHas SI unit of m/sAverage velocity can be useful but it does not paint the complete picture.The winner of a race has the highest average velocity but is not necessarily the fastest.
27Instantaneous Velocity Velocity at a specific instant of timeDefine instant as an extremely short amount of time such that it has no duration at all.Instantaneous Velocitytop speed of km/h(Sport version. Picture only shows regular version)
28x-t Graph Average Velocity Instantaneous Velocity Average velocity is the slope of the line between two pointsInstantaneous VelocityInstantaneous velocity is the slope of the tangent line at a specific pointxxx2x1oot1t2tt1t
29Sample ProblemA Bugatti Veyron is at rest 20.0m from an observer. At t=0 it begins zooming down the track in a straight line. The displacement from the observer varies according to the equationa) Find the average velocityfrom t=0s to t=10sb) Find the average velocityfrom t=5s to t=10sc) Find the instantaneous velocity at t=10s
39Falling ObjectsMost common example of constant acceleration is free fall.Freely falling bodies are objects moving under the influence of gravity alone. (Ignore air resistance)Attracts everything to it at a constant rate.Note: because it attracts objects downwards acceleration due to gravity isGalileo Galilei formulated the laws of motion for free fall
40Freely FallingA freely falling body is any body that is being influenced by gravity alone, regardless of initial motion.Objects thrown upward or downward or simply released are all freely falling
41Example Giancoli 2-42A stone is thrown vertically upward with a speed of m/s. (a) How fast is it moving when it reaches a height of 11.0m? (b) How long will it take to reach this height? (c) Why are there two answers for b?
45SummaryThese 4 equations will allow you to solve any problem dealing with motion in one direction as long as acceleration is CONSTANT!184.108.40.206.
46Problems from the Book (Giancoli 6th ed) 14- Calculate the average speed and average velocity of a complete round-trip, in which the outgoing 250 km is covered at 95km/hr, followed by a 1 hour lunch break and the return 250km is covered at 55km/hr.Start95 kph1 hour breakEnd55 kph
47Chapter 2 Problem 14Average speed = change in distance / change in timeFor first legFor returnTotal timeAverage speed
48Chapter 2 Problem 14What was the cars average velocity?
49Chapter 2 Problem 19A sports car moving at constant speed travels 110m in 5.0s. If it then brakes and comes to a stop in 4.0 s, what is its acceleration in m/s2? Express the answer in terms of g’s where g=9.80 m/s2.
51Chapter 2 Problem 31A runner hopes to complete a 10,000m run in less than min. After exactly 27.0 min, there are still 1100m to go. The runner must then accelerate at 0.20m/s2 for how many seconds in order to achieve the desired time?
52Chapter 2 Problem 31 Find average v at 27 min The runner will then accelerate for t seconds covering some distance d,and will then cover the remaining distance in (180-t). (a is now zero)