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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.

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Presentation on theme: "MECHANICS. Motion Along a Straight Line Ps 41 But First a Review  Significant Figures  Non-zero digits are always significant.  Any zeros between."— Presentation transcript:

1 MECHANICS

2 Motion Along a Straight Line Ps 41

3 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 has 4 significant figures  Ex. 2,500 has 2 significant figures  Ex x 10 3 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

4 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 P1P1 P2P2 A P1P1 P2P2 B=-A

5 Vector Addition  Tip to tail method or Parallelogram method  Vector addition is commutative (a)(b)

6 Vector Components  Vector Components

7 Vector Addition using Components

8 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.

9 Example: Young and Freedman Problem 1.31/1.38

10

11  DON’T FORGET DIRECTION

12 Unit Vectors  Unit vectors are unitless vectors with a magnitude of 1.  Primarily used to point a direction.  Represented by  Note scalar times vector

13 Example

14 Dot Product  Or scalar product  Using components

15 Example  What is the angle between the vectors Compute up to 3 sig figs.  Solution

16 Cross Product  Or vector product  Direction is dictated by the right hand rule  Anti-commutative

17 Cross Product by Components

18 Determinant form

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20

21

22 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 xy- plane at a 30 o angle with the x-axis. Find AxB.  Solution Let

23 Motion Along a Straight Line Ps 41

24 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

25 Average Velocity  At time t 1 the car is at point P 1 and at time t 2 the car is at point P 2  We can define P 1 and P 2 to have coordinates x 1 and x 2 respectively Δ x=x 2 -x 1  Average velocity P1P1 P2P2

26 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.

27 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  top speed of km/h (Sport version. Picture only shows regular version)

28 x-t Graph  Average Velocity  Average velocity is the slope of the line between two points  Instantaneous Velocity  Instantaneous velocity is the slope of the tangent line at a specific point x t x t x2x2 x1x1 o t1t1 t2t2 o t1t1

29 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 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

30 Solution  a)  b)  c)

31 Acceleration  Acceleration describes the rate of change of velocity with time.  Average Acceleration  Vector quantity denoted by  Instantaneous acceleration

32 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.

33 Motion at Constant Acceleration  Assume that acceleration is constant.  Generally

34  Feel Free to use v f, v i, v 0 whatever notation you’re more comfortable with  BUT be consistent through out the entire problem

35 Motion at Constant Acceleration

36 Seat Work #1  Using  Derive  Hint: Eliminate time

37 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/s 2, estimate the speed of the car just before braking.

38 Chapter 2 Problem 26  Ignore negative

39 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.  Note: because it attracts objects downwards acceleration due to gravity is  Galileo Galilei formulated the laws of motion for free fall

40 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

41 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?

42 Giancoli 2-42  We can’t ignore negative

43 Giancoli 2-42

44  Why were there 2 answers to b?

45 Summary  These 4 equations will allow you to solve any problem dealing with motion in one direction as long as acceleration is CONSTANT!  1.  2.  3.  4.

46 Problems from the Book (Giancoli 6 th 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. 95 kph 55 kph Start End 1 hour break

47 Chapter 2 Problem 14  Average speed = change in distance / change in time  For first leg  For return  Total time  Average speed

48 Chapter 2 Problem 14  What was the cars average velocity?

49 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/s 2 ? Express the answer in terms of g’s where g=9.80 m/s 2.

50 Chapter 2 Problem 19  First find v

51 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/s 2 for how many seconds in order to achieve the desired time?

52 Chapter 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)

53 Chapter 2 Problem 31

54  Almost there Quadratic Equation  Results in t=357s or t=3.11s  t<180 seconds so t=3.11s  whew

55 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?

56 Chapter 2 Problem 44  Let  t w be the time it takes for the stone to reach the top of the window.  x w be the height above the window stone was dropped  At t=t w +0.28s the stone is now x w -2.2m!!!

57 Chapter 2 Problem 44


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