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MECHANICS

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Motion Along a Straight Line Ps 41

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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

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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

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

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Vector Components Vector Components

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Vector Addition using Components

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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.

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Example: Young and Freedman Problem 1.31/1.38

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DON’T FORGET DIRECTION

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

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Example

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Dot Product Or scalar product Using components

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

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

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Cross Product by Components

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Determinant form

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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

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Motion Along a Straight Line Ps 41

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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

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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

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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.

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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)

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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

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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

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Solution a) b) c)

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

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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.

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Motion at Constant Acceleration Assume that acceleration is constant. Generally

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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

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Motion at Constant Acceleration

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Seat Work #1 Using Derive Hint: Eliminate time

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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.

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Chapter 2 Problem 26 Ignore negative

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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

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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

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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?

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Giancoli 2-42 We can’t ignore negative

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Giancoli 2-42

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Why were there 2 answers to b?

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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.

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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

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

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Chapter 2 Problem 14 What was the cars average velocity?

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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.

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Chapter 2 Problem 19 First find v

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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?

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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)

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Chapter 2 Problem 31

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

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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?

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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!!!

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Chapter 2 Problem 44

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