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Chapter 2 and Chapter 3 Study Guide

by: Nina Nguyen

Chapter 2 and Chapter 3 Study Guide PHYS 1443 - 001

Marketplace > University of Texas at Arlington > Physics > PHYS 1443 - 001 > Chapter 2 and Chapter 3 Study Guide
Nina Nguyen
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Good evening, I have made a study guide for the conceptual questions for Chapter 2 and Chapter 3. I hope this helps you on your exam!
Study Guide
Physics 1443 General Technical Physics Midterm 1 Exam 1
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This 7 page Study Guide was uploaded by Nina Nguyen on Tuesday February 23, 2016. The Study Guide belongs to PHYS 1443 - 001 at University of Texas at Arlington taught by in Spring 2016. Since its upload, it has received 174 views. For similar materials see GENERAL TECHNICAL PHYSICS I in Physics at University of Texas at Arlington.

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Date Created: 02/23/16
Physics 1443­ 012 Chapter Notes Chapter 2: Describing Motion: Kinematics in One Direction  Kinematics is the study of how objects move  Dynamics is the study of forces and why objects move.   Translational Motion is the study of rotating motion.  The position of any particle, or object, can be given by its x and y coordinate.  What is the difference between distance and displacement?   Displacement describes how far away the object is from its starting point.  o Displacement is a vector (aka it is a quantity that has magnitude and direction) o To find displacement:  ∆ x=x −x2 1 ∆ x x x where   means the displacement,  2  is the final position, and  1 is the initial position.   Total Distance is how far an object traveled regardless of direction.  Example:  If Professor Mayfield walked 10 meters to the right, then 10 meters to the left,  what is Professor Mayfield’s displacement? Total distance? In order to find displacement:  ∆ x=x −x 2 1 x First, assume that Professor Mayfield’s initial position ( 2 ) was at 0 meters. You know that he moved 10 meters to the right (in the positive direction), but then he moved 10 meters back (in the negative direction). Therefore, his final position ( x2 ) is right where he started: at 0 meters.   x2=0 x1=0 Therefore,   meters and   meters ∆ x=0meters−0 meters=0 meters And  To find distance:  Simply, just add how far Professor Mayfield walked.  Totaldistance=10meters+ −10meters =20meters Essentially, all distances become a positive number because we don’t care whether he walked in a positive or negative direction.  2­2 Average Velocity  Average Speed: the total distance an object travels divided by the time it takes to travel  that distance  averagespeed= distancetraveled timeelapsed  Average Velocity: an object’s displacement divided by the time it takes to travel that  displacement  o Velocity has a magnitude and includes the direction as well!  averagevelocity= displacement = final position−initial position timeelapsed timeelapsed  Velocity and speed are not the same thing! Also, the average speed is not always equal to  the magnitude of the average velocity! Practice Problem x1=50.0 m 1.   During a 5.00­s interval, a runner’s position changes from   to x2=30.5m . What is this runner’s average velocity? Assume that the runner is moving  along the x­axis of a coordinate system.  Solution: ­6.50 m/s 2. If a student is traveling on a hover board, and her average velocity is 18 km/h, how far  can she travel in 2.5 hours?  Solution: 45 km 2­3 Instantaneous Velocity  Instantaneous Velocity: the average velocity of an object divided by an infinitely small  time interval.  ∆x dx v= lim ∆t = dt ∆t→ 0 Also, the instantaneous velocity equals the slope of the tangent line to the curve. You can determine the instantaneous velocity by taking the derivative of the position function **So… what’s the difference between instantaneous velocity and average velocity? Think of it like this. You woke up late one day, and you’re speeding to class. You live  50.0 miles away, and your class starts in 45.0 minutes. If you arrive just in time, your  average velocity will be about 1.11 miles per minute (50 miles/ 45 minutes). However,  you don’t have to be driving 1.11 miles/min the entire time. Maybe halfway, you saw  construction and slowed down. Your instantaneous velocity would be the velocity you  drove at that instant moment.  Practice Problems m 1. Given: Suppose a jet plane’s position function is  x=At +B , where  A=2.10 2   s B=2.80m. and   What is the displacement of the plane from t1=3.00sandt =5200s ? Solution: 33.6 m 2. Using the information in Problem 1, what is plane’s average velocity?  Solution: 16.8 m/s 3. Using the information in Problem 1, what is the magnitude of the instantaneous velocity  t=5.00 s at  ? Solution: 21.0 m/s 2­4 Acceleration  Acceleration: how rapidly the velocity is changing averageacceleration a ( ) change∈velocity = v 2v 1= ∆ v timeelapsed t2−t1 ∆ t Practice Problem 1. If a car accelerates from rest to 90 km/h in 5.0s, what is the magnitude of the car’s  average acceleration?  Solution: 5.0 m/s2 The acceleration tells us how quickly the velocity changes, whereas the velocity tells us how quickly the position changes! If an object has a velocity of zero, it does not mean that the acceleration is zero, nor does a zero acceleration mean that the velocity is zero!  Deceleration: the magnitude of the object’s velocity is decreasing, and then the velocity  and acceleration is pointing in opposite directions.  o It does not mean that the acceleration is negative.   Instantaneous Acceleration: the limiting value of the average acceleration as we let the  change in time approach zero. ∆ v dv d dx d x a=lim ∆t = dt= dt dt) = 2 ∆t→ 0 dt One way to find acceleration is to take the second derivative of the position function in respect to time. Practice Problem 1. You see a flying squirrel moving in a straight line. His position is given by the function  m 2 of  x= ( )0 2 t +2.80m. What is his average acceleration during s t1=3.00sand t 25.00 s? Solution: 4.20 m/s 2. Using the information in Problem 1, what is the flying squirrel’s instantaneous  acceleration?  Solution: 4.20 m/s 2­5 Motion at Constant Acceleration Important equations to remember include:  ∆ x x−x 0 ´= ∆ t= t v−v 0 a= t Hint: a commonly asked problem is to determine the velocity of an object after any elapsed time  has passed. Usually, you’ll be given the constant acceleration.  In order to solve this, use the following equation:  v=v +at 0 If you have constant acceleration (must be constant!), you can use: v +v ´= 0 2 Also, learn how to use:  v +v x=x +0 0 t ( )2 or 1 2 x=x +0 t0 at 2 The last equation you need to get familiar with is: 2 2 v =v 02a x−( 0) The last four equations are only applicable if the acceleration is constant! Practice Problem 1. You are designing an airport for small planes. The airplane must reach a speed of at least  s2 27.8 m/s before takeoff. The plane can accelerate at 2.00 m/ . If the runway was 150  m long, can the airplane reach the speed needed to take off?  Solution: No, only reached 24.5 m/s 2. If not, what is the minimum length at which the runway can be?  Solution: 193m 2­7 Freely Falling Objects Galileo states that: at a given location on the Earth and in the absence of air resistance, all  objects fall with the same constant acceleration.  m We call this the acceleration due to gravity, which its magnitude= 9.80  2 s y=v t+ at 2 Important equation:  0 2 Practice Problem 1. You are dropping a ball from a tower that is 70.0 m high. How far will it have fallen after 1.00s? 2.00s? 3.00s? Ignore air resistance.  Solution: 4.90m, 19.6m,44.1m 2. Now, suppose that same ball (from Problem 1) was thrown with a velocity of 3.00 m/s.  What is the ball’s position after 1.00s? 2.00s?  Solution: 7.90m, 25.6m Be cautious: Velocity and acceleration are not always in the same direction. Also, the ball does  not have zero acceleration at the top of the parabola; it has a zero velocity.  Chapter 3: Kinematics in Two or Three Dimensions; Vectors If the vectors are in the same direction, you can simply add or subtract them. You can use the Pythagorean Theorem if the vectors are perpendicular to one another.  Resultant Displacement is the sum of all of the vectors.  **The magnitude of the resultant vector is smaller than the sum of the magnitudes of its  components.  Vectors must be drawn from tail­to­tip! 3­3 Subtraction of Vectors, and Multiplication of a Vector by a  Scalar The negative of a vector has the same magnitude but the opposite sign (+/­), but a vector can  never have a negative magnitude. The sign tells us direction only.  3­4 Adding Vectors by Components A vector can be broken down into its x and y components. The components are drawn at a  perpendicular of the vector. Components are drawn from the tip of a vector.  vy=Vsinθ vx=Vcosθ 2 2 V= v√+x y vy tanθ= v x In order to add vectors using its components, 1. Resolve each vector into its components 2. Find the sum of the x components and the sum of the y components  3­5 Unit Vectors Has a magnitude of exactly one The components have a “hat” on the i, j, k, components  3­6 Vector Kinematics We can use vectors to talk about velocity and acceleration in a two and three dimensional  motion.  The displacement vector is the vector that characterizes the change in position  ∆ r=r´ 2r ´1 The average and instantaneous vectors are determined by using the same methods in Chapter 2.  3­8 Solving Problems Involving Projectile Motions’ Horizontal Motion Vertical Motion v =v v =v x x0 y y0 x=x +0 t x0 1 2 y=y +0 t−y0t 2 2 2 v =v 02a x−( 0) ** I suggest that you do Problem 3­6 on page 65 in our textbook.  Asymmetric projectile motion problems must be broken down into two steps. 


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