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MECHANICS Motion Along a Straight Line Ps 41 But 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. 0.002500 has 4 significant figures Ex. 2,500 has 2 significant figures Ex. 2.500 x 103 has 4 significant figures Multiplication/Division – Determined by the LEAST number of significant figures Addition/Subtraction – Determined by LEAST number of decimal of places in the decimal portion Vectors Vectors are physical quantities with both magnitude and direction and cannot be represented by just a single number Displacement vs. Distance Velocity vs. Speed Represented by A The magnitude of A is represented by |A| or A P2 P1 A P2 P1 B=-A Vector Addition Tip to tail method or Parallelogram method Vector addition is commutative F1 F2 F2 F1 (a) (b) Vector Components Vector Components A Ax Ay Ax A cos Ay A sin Vector Addition using Components Example: Young and Freedman Problem 1.31/1.38 A postal employee drives a delivery truck along the route shown. Determine the magnitude and direction of the resultant displacement. Example: Young and Freedman Problem 1.31/1.38 D D1 D2 D3 D1 2.6km D2 4.0km D3 3.1km D 9.7 km Example: Young and Freedman Problem 1.31/1.38 D Dx D y D1 D2 D3 Dx D1x D2 x D3 x D y D1 y D2 y D3 y D1x 0km D1 y 2.6km D2 x 4.0km D2 y 0km D3 x 3.1 cos(45) 2.192km D3 y 3.1sin(45) 2.192km Dx 0 4 2.192 6.192km Dy 2.6 0 2.192 4.792km D D x2 Dy2 7.829 7.8km Example: Young and Freedman Problem 1.31/1.38 DON’T FORGET DIRECTION Dx 6.192km Dy 4.792km 4.792 tan 37.7 6.192 1 D 7.8km _ 38 NofE Unit Vectors Unit vectors are unitless vectors with a magnitude of 1. Primarily used to point a direction. Represented by iˆ A Ax iˆ Ay ˆj Az kˆ Note scalar times vector Example G 2iˆ 4 ˆj G Dot Product Or scalar product Using components A B A B cos iˆ iˆ ˆj ˆj kˆ kˆ 1 iˆ ˆj ˆj kˆ iˆ kˆ 0 A B ( Axiˆ Ay ˆj Az kˆ) ( Bxiˆ By ˆj Bz kˆ) Ax Bx Ay By Az Bz Example What is the angle between the vectors G 2iˆ 4 ˆj H 3iˆ ˆj H Compute up to 3 sig figs. Solution G H G H cos 20 10 cos Gx H x G y H y (2)(3) (4)(1) 2 2 cos ( ) 81.9 200 1 G Cross Product Or vector product C A B C A B sin Direction is dictated by the right hand rule Anti-commutative A B B A Cross Product by Components iˆ iˆ ˆj ˆj kˆ kˆ 0 iˆ ˆj ˆj iˆ kˆ ˆj kˆ kˆ ˆj iˆ kˆ iˆ iˆ kˆ ˆj A B ( Axiˆ Ay ˆj Az kˆ) ( Bxiˆ By ˆj Bz kˆ) ( Ay Bz Az By )iˆ ( Az Bx Ax Bz ) ˆj ( Ax By Ay Bx )kˆ Determinant form iˆ A B Ax Bx ˆj ˆ k Ay By Az Bz iˆ A B Ax Bx ˆj ˆ k Ay By Az Bz iˆ A B Ax Bx ˆj ˆ k Ay By Az Bz iˆ A B Ax Bx ˆj ˆ k Ay By Az Bz Example Vector A has a magnitude to 5 and lies in the direction of the x-axis. Vector B has a magnitude of 2 and lies along the xyplane at a 30o angle with the x-axis. Find AxB. Solution Let C A B C A B sin (5)(2) / 2 5 Motion Along a Straight Line Ps 41 Displacement 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 Average Velocity P1 At time t1 the car is at point P1 and at time t2 the car is at point P2 We can define P1 and P2 to have coordinates x1 and x2 respectively Δx=x2-x1 Average velocity vave x2 x1 x t2 t1 t P2 Velocity Velocity is the change in displacement per unit time in a specific direction. It is a vector quantity, usually denoted by v Has SI unit of m/s Average 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. Instantaneous Velocity Velocity at a specific instant of time Define instant as an extremely short amount of time such that it has no duration at all. Instantaneous Velocity x dx v lim t 0 t dt top speed of 431.072 km/h (Sport version. Picture only shows regular version) x-t Graph Average Velocity x Instantaneous Velocity x x2 x1 o t1 t2 o t Average velocity is the slope of the line between two points t1 t Instantaneous velocity is the slope of the tangent line at a specific point Sample Problem A 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 equation x 20.0m (1.39m s2 )t 2 a) Find the average velocity from t=0s to t=10s b) Find the average velocity from t=5s to t=10s c) Find the instantaneous velocity at t=10s Solution x0 20m x10 20m (1.39m s2 )(10)2 159m a) vave 159 20 139 13.9 m s 10 0 10 2 x5 20m (1.39m s2 )(5) 54.75m b)vave 159 54.75 104 .25 20.9 m s 10 5 5 m )t 2 ] d [ 20 m ( 1 . 39 s2 c)dv (2.78 m s 2 )(t ) dt dt v10 (2.78m s 2 )(t ) 27.8 m s Acceleration Acceleration describes the rate of change of velocity with time. Average Acceleration aave v2 v1 v t2 t1 t Vector quantity denoted by a Instantaneous acceleration v dv d x a lim 2 t 0 t dt dt 2 WARNING Just because acceleration is positive (negative) does not mean that velocity is also positive (negative). Just because acceleration is zero does not mean velocity is zero and vice versa. Motion at Constant Acceleration Assume that acceleration is constant. v a c t v v0 a t v v0 at Generally ti t0 0 v v0 at Feel Free to use vf, vi, v0 whatever notation you’re more comfortable with BUT be consistent through out the entire problem Motion at Constant Acceleration x x0 x vave t t0 t v v0 v0 at v0 vave 2 2 vave v0 12 at 2 1 x v0t 2 at x x0 v0t 2 at 1 2 Seat Work #1 Using v v0 at x x0 v0t 12 at2 Derive v v 2ax 2 2 0 Hint: Eliminate time Giancoli Chapter 2 Problem 26 In coming to a stop a car leaves skid marks 92 m long on the highway. Assuming a deceleration of 7.00m/s2, estimate the speed of the car just before braking. Chapter 2 Problem 26 x 92m a 7.00 m s 2 v 0ms v0 ? v 2 v02 2ax 0 v02 2(7.00)(92) v02 1288 Ignore negative v0 35.89 m s 36 m s Falling Objects Most 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. g 9.80 m s 2 Note: because it attracts objects downwards acceleration due to gravity is a g 9.80m s 2 Galileo Galilei formulated the laws of motion for free fall Freely Falling A 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 Example Giancoli 2-42 A stone is thrown vertically upward with a speed of 18.0 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? Giancoli 2-42 x 11m a 9.80 m s 2 v0 18.0 m s v? t ? v 2 v02 2ax v 2 (18) 2 2(9.80)(11) v 108.4 v 10.4 m s 2 We can’t ignore negative Giancoli 2-42 x x0 v0t 1 2 at 2 11.0 18.0t 1 2 (9.80)t 2 (4.90)t 18.0t 11 0 2 18 (18) 2 4(4.9)(11) t 2(4.9) t 0.775s t 2.90s b b 4ac 2a 2 Giancoli 2-42 Why were there 2 answers to b? Summary These 4 equations will allow you to solve any problem dealing with motion in one direction as long as acceleration is CONSTANT! v v0 at 1. 2. x x0 v0t 2 at 3. v2 v02 2ax 4. 1 vave v v0 2 2 Problems 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. Start 95 kph 1 hour break End 55 kph Chapter 2 Problem 14 Average speed = change in distance / change in time speed ave For first leg d t 250km 250km t1 2.6316hr 95kph 95 kph t 1 For return 250km 250km t 2 4.5455hr 55kph 55kph t 2 Total time ttotal t1 1 t2 8.1771hr Average speed d 500 speed ave 61.15kph 61kph t 8.1771 Chapter 2 Problem 14 What was the cars average velocity? Chapter 2 Problem 19 A 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. Chapter 2 Problem 19 First find v x 110 m vave 22 m s t 5s v 22 m s aave 5.5 m s 2 t 4s 1g m aave 5.5 s 2 ( ) 0.56g ' s 9.80 m s 2 Chapter 2 Problem 31 A runner hopes to complete a 10,000m run in less than 30.0 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? Chapter 2 Problem 31 Find average v at 27 min vcurrent x 8900 m 1 min 5.49 m sec v0 t 27 min 60 sec The runner will then accelerate for t seconds covering some distance d, v v0 at 5.49 0.20(t ) d v0t 12 at2 5.49(t ) 12 (0.20)t 2 and will then cover the remaining distance in (180-t). (a is now zero) x 1100 d v t 180 t Chapter 2 Problem 31 v(180 t ) 1100 d (v0 at)(180 t ) 1100 d 180v0 180at v0t at 2 1100 d 1100 180v0 180at v0t at d 2 1100 180v0 180at v0t at v0t at 2 1100 180v0 180at at 1 2 1 2 2 2 1100 180(5.49) 180(0.20)t (0.5)(0.20)t 2 0.10t 36t 111.8 0 2 Chapter 2 Problem 31 Almost there Quadratic Equation b b 4ac 2a 2 Results in t=357s or t=3.11s t<180 seconds so t=3.11s whew Chapter 2 Problem 44 A falling stone takes 0.28s to travel past a window 2.2m tall. From what height above the top of the window did the stone fall? Chapter 2 Problem 44 Let x0 0 v0 0 m s tw be the time it takes for the stone to reach the top of the window. xw be the height above the window stone was dropped At t=tw+0.28s the stone is now xw-2.2m!!! Chapter 2 Problem 44 x x0 v0t 1 2 at 2 x 1 2 ( 9 . 80 ) t 2 x 4.90t 2 xw 4.90t w2 xw 2.2 4.90(t w 0.28) 2 xw 2.2 4.90(t w2 0.56t w 0.0784) xw 2.2 4.90t w2 2.744t w 0.38416 2.2 2.744t w 0.38416 t w 0.662s xw 2.1m