Key Terms and Equations Click on the topic to go to that section - PDF document

Slide 1 / 246 Slide 2 / 246 New Jersey Center for Teaching and Learning Progressive Science Initiative Kinematics in Two This material is made freely available at www.njctl.org Dimensions and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. www.njctl.org Click to go to website: www.njctl.org Slide 3 / 246 Slide 4 / 246 Table of Contents: Kinematics in 2D Key Terms and Equations Click on the topic to go to that section Kinematics in One Dimension (Review) Kinematics Equations: 2-Dimensional Equations Adding Vectors in Two Dimensions v = v 0 + a t v x = v cos( θ) v y = v sin( θ) v 2 = v 02 + 2 a Δx Basic Vector Operations x = x 0 + v 0 t + 1/2 a t 2 v = √(v x2 + v y2 ) θ = tan -1 (v y / v x ) Vector Components Right Triangle Equations: Projectile Motion a 2 + b 2 = c 2 General Problems SOH CAH TOA sin(θ) = Opposite/Hypotenuse cos(θ) = Adjacent/Hypotenuse tan(θ) = Opposite/Adjacent Slide 5 / 246 Slide 6 / 246 Review of 1-D Kinematics · Kinematics is the description of how objects move with respect to a defined reference frame. Kinematics in One · Displacement is the change in position of an object. Dimension · Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time. · Instantaneous velocity is the limit as the time becomes infinitesimally short. · Average acceleration is the change in velocity divided by the time. Return to Table of Contents

Slide 7 / 246 Slide 8 / 246 Starting from rest, you 1 Review of 1-D Kinematics accelerate at 4.0 m/s 2 for 6.0s. What is your final velocity? · Instantaneous acceleration is the limit as the time interval becomes infinitesimally small. · There are four equations of motion for constant acceleration, each requires a different set of quantities. v = v o + at x = x o + v o t + ½ at 2 v 2 = v o2 + 2 a(x - x o ) v = v + v o 2 v = v o + at v = 0 + 4(6) v = 24m/s http:/ / njc.tl/ 3i Slide 9 / 246 Slide 10 / 246 You have an initial velocity of How much time does it take to 2 3 -3.0 m/s. You then experience come to rest if your initial an acceleration of 2.5 m/s 2 for velocity is 5.0 m/s and your 9.0s; what is your final acceleration is -2.0 m/s 2 ? velocity? v = v o + at v = v o + at 0 = 5 + -2t v = -3 + 2.5(9) t = 2.5s v = 19.5m/s http:/ / njc.tl/ 3j http:/ / njc.tl/ 3k Slide 11 / 246 Slide 12 / 246 4 An object moves at a constant speed of 6 m/s. 5 A snapshot of three racing cars is shown on the This means that the object: diagram. All three cars start the race at the same time, at the same place and move along a straight track. As they approach the finish line, which car A Increases its speed by 6 m/s every second has the lowest average speed? Decreases its speed by 6 m/s every second B A Car I Doesn’t move C Has a positive acceleration Car II D B Moves 6 meters every second Car III E C All three cars have the same average speed D More information is required E http:/ / njc.tl/ 3l http:/ / njc.tl/ 3m

Slide 13 / 246 Slide 14 / 246 Motion at Constant Acceleration Motion at Constant Acceleration In physics there is another approach in addition to algebraic which is called graphical analysis. From these two formulas we can find some analogies: Velocity v ⇒ y (depended variable of x), The following formula v = v 0 + at can be interpreted by the graph. v 0 ⇒ b (intersection with vertical axis), t ⇒ x (independent variable), We just need to recall our memory from math classes a ⇒ m ( slope of the graph- the ratio between rise and run where we already saw a similar formula y = mx + b. Δ y/ Δx). y = m x + b v = a t + v 0 (or v = v 0 + a t) Slide 15 / 246 Slide 16 / 246 6 The velocity as a function of time is presented by the graph. Motion at Constant Acceleration What is the acceleration? The formula a = Δ v/ Δ t shows that the value of acceleration is the same as the slope on a graph of velocity versus time. a = slope = Δ v/ Δ t = (10 m/s -2 m/s)/40 s = 0.2 m/s 2 http:/ / njc.tl/ 3n Slide 17 / 246 Slide 18 / 246 7 The velocity as a function of time is presented by the graph. Motion at Constant Acceleration Find the acceleration. The acceleration graph as a function of time can be used to find the velocity of a moving object. When the acceleration is constant it can be shown on the graph as a straight horizontal line. a = slope = Δ v/ Δ t = (0 m/s - 25 m/s)/10 s = -2.5 m/s 2 http:/ / njc.tl/ 3o

Slide 19 / 246 Slide 20 / 246 Motion at Constant Acceleration 8 What is the velocity of the object at 5 s? In order to find the change in velocity for a certain limit of A 1 m/s time we need to calculate the area under the acceleration versus time graph. 2 m/s B 3 m/s C 4 m/s D The change in velocity during first 12 seconds E 5 m/s is equivalent to the shadowed area (4 χ 12 = 48). The change in velocity during first 12 seconds is 48 m/s. http:/ / njc.tl/ 3p Slide 21 / 246 Slide 22 / 246 10 The following graph shows acceleration as a function of 9 Which of the time of a moving object. What is the change in velocity following during first 10 seconds? statements is true? Δ v = area = (3 χ 10) = 30 m/s A The object speeds up The object slows down B C The object moves with a constant velocity The object stays at rest D The object is in free fall E http:/ / njc.tl/ 3q http:/ / njc.tl/ 3r Slide 23 / 246 Slide 24 / 246 11 The graph represents the relationship between 12 Which of the velocity and time for an object moving in a following is true? straight line. What is the traveled distance of the object at 9 s? A 10 m 24 m B A The object increases its velocity 36 m C B The object decreases its velocity D 48 m The object’s velocity stays unchanged C 56 m E The object stays at rest D More information is required E http:/ / njc.tl/ 3s http:/ / njc.tl/ 3t

Slide 25 / 246 Slide 26 / 246 13 What is the initial position of the object? 14 What is the velocity of the object? A 2 m 2 m/s A 4 m 4 m/s B B 6 m 6 m/s C C 8 m 8 m/s D D E 10 m E 10 m/s http:/ / njc.tl/ 3u http:/ / njc.tl/ 3v Slide 27 / 246 Slide 28 / 246 15 16 What is the initial position of the object? The graph represents the position as a function of time of a moving object. What is the velocity of the object? 2 m A 4 m B 5 m/s A 6 m C -5 m/s B 8 m D 10 m/s C 10 m E -10 m/s D 0 m/s E http:/ / njc.tl/ 3u http:/ / njc.tl/ 3w Slide 29 / 246 Slide 30 / 246 Free Fall It stops momentarily. What happens at the v = 0 top? All unsupported objects fall towards the earth with the g = -9.8 m/s 2 same acceleration. It speeds up We call this acceleration the "acceleration due to What happens when it (negative acceleration) goes down? gravity" and it is denoted by g. g = -9.8 m/s 2 It slows down. What happens when it (negative acceleration) g = 9.8 m/s 2 goes up? g = -9.8 m/s 2 Keep in mind, ALL objects accelerate towards the earth at the same rate. An object is thrown upward It returns with its What happens when it with initial velocity, v lands? original velocity. o g is a constant!

Slide 31 / 246 Slide 32 / 246 It stops momentarily. On the way up: On the way down: v = 0 v 0 g = -9.8 m/s 2 t = 0 s a v 1 a v t = 3 s It speeds up. v 2 a (negative acceleration) t = 2 s It slows down. v 1 g = -9.8 m/s 2 (negative acceleration) t = 1 s v 2 g = -9.8 m/s 2 a a v 1 v 2 t = 1 s v 2 t = 2 s a v a An object is thrown upward It returns with its v 0 v 1 original velocity. with initial velocity, v o t = 3 s t = 0 s v Slide 33 / 246 Slide 34 / 246 For any object thrown straight up into the 17 An object moves with a constant acceleration of 5 air, what does the velocity vs time graph m/s 2 . Which of the following statements is true? look like? An object is thrown upward with initial velocity, v o A The object’s velocity stays the same v The object moves 5 m each second (m/s) B It stops momentarily. v = 0 The object’s acceleration increases by 5 m/s2 C g = -9.8 m/s 2 each second The object’s acceleration decreases by 5 m/s2 D t (s) each second the object’s velocity increases by 5 m/s each E second It returns with its original velocity but in the opposite direction. E http:/ / njc.tl/ 3x Slide 35 / 246 Slide 36 / 246 18 A truck travels east with an increasing velocity. 19 A car and a delivery truck both start from rest and Which of the following is the correct direction of accelerate at the same rate. However, the car the car’s acceleration? accelerates for twice the amount of time as the truck. What is the final speed of the car compared to the truck? A B A Half as much The same B C Twice as much C D Four times as much D E One quarter as much E A http:/ / njc.tl/ 3y