Slide 1 / 246 New Jersey Center for Teaching and Learning Progressive Science Initiative This material is made freely available at www.njctl.org 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. Click to go to website: www.njctl.org Slide 2 / 246 Kinematics in Two Dimensions www.njctl.org Slide 3 / 246 Table of Contents: Kinematics in 2D Click on the topic to go to that section Kinematics in One Dimension (Review) Adding Vectors in Two Dimensions Basic Vector Operations Vector Components Projectile Motion General Problems
Slide 4 / 246 Key Terms and Equations Kinematics Equations: 2-Dimensional Equations v = v 0 + a t v x = v cos( θ) v y = v sin( θ) v 2 = v 02 + 2 a Δx x = x 0 + v 0 t + 1/2 a t 2 v = √(v x2 + v y2 ) θ = tan -1 (v y / v x ) Right Triangle Equations: a 2 + b 2 = c 2 SOH CAH TOA sin(θ) = Opposite/Hypotenuse cos(θ) = Adjacent/Hypotenuse tan(θ) = Opposite/Adjacent Slide 5 / 246 Kinematics in One Dimension Return to Table of Contents Slide 6 / 246 Review of 1-D Kinematics · Kinematics is the description of how objects move with respect to a defined reference frame. · Displacement is the change in position of an object. · 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.
Slide 7 / 246 Review of 1-D Kinematics · 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 Slide 8 / 246 Starting from rest, you 1 accelerate at 4.0 m/s 2 for 6.0s. What is your final velocity? v = v o + at v = 0 + 4(6) v = 24m/s http:/ / njc.tl/ 3i Slide 9 / 246 You have an initial velocity of 2 -3.0 m/s. You then experience an acceleration of 2.5 m/s 2 for 9.0s; what is your final velocity? v = v o + at v = -3 + 2.5(9) v = 19.5m/s http:/ / njc.tl/ 3j
Slide 10 / 246 How much time does it take to 3 come to rest if your initial velocity is 5.0 m/s and your acceleration is -2.0 m/s 2 ? v = v o + at 0 = 5 + -2t t = 2.5s http:/ / njc.tl/ 3k Slide 11 / 246 4 An object moves at a constant speed of 6 m/s. This means that the object: A Increases its speed by 6 m/s every second B Decreases its speed by 6 m/s every second Doesn’t move C Has a positive acceleration D Moves 6 meters every second E http:/ / njc.tl/ 3l Slide 12 / 246 5 A snapshot of three racing cars is shown on the 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 has the lowest average speed? A Car I Car II B Car III C All three cars have the same average speed D E More information is required http:/ / njc.tl/ 3m
Slide 13 / 246 Motion at Constant Acceleration In physics there is another approach in addition to algebraic which is called graphical analysis. The following formula v = v 0 + at can be interpreted by the graph. We just need to recall our memory from math classes where we already saw a similar formula y = mx + b. Slide 14 / 246 Motion at Constant Acceleration From these two formulas we can find some analogies: Velocity v ⇒ y (depended variable of x), v 0 ⇒ b (intersection with vertical axis), t ⇒ x (independent variable), a ⇒ m ( slope of the graph- the ratio between rise and run Δ y/ Δx). y = m x + b v = a t + v 0 (or v = v 0 + a t) Slide 15 / 246 Motion at Constant 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.
Slide 16 / 246 6 The velocity as a function of time is presented by the graph. What is the acceleration? 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 7 The velocity as a function of time is presented by the graph. Find the acceleration. a = slope = Δ v/ Δ t = (0 m/s - 25 m/s)/10 s = -2.5 m/s 2 http:/ / njc.tl/ 3o Slide 18 / 246 Motion at Constant 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.
Slide 19 / 246 Motion at Constant Acceleration In order to find the change in velocity for a certain limit of time we need to calculate the area under the acceleration versus time graph. The change in velocity during first 12 seconds is equivalent to the shadowed area (4 χ 12 = 48). The change in velocity during first 12 seconds is 48 m/s. Slide 20 / 246 8 What is the velocity of the object at 5 s? A 1 m/s 2 m/s B 3 m/s C 4 m/s D 5 m/s E http:/ / njc.tl/ 3p Slide 21 / 246 9 Which of the following statements is true? A The object speeds up The object slows down B The object moves with a constant velocity C D The object stays at rest The object is in free fall E http:/ / njc.tl/ 3q
Slide 22 / 246 10 The following graph shows acceleration as a function of time of a moving object. What is the change in velocity during first 10 seconds? Δ v = area = (3 χ 10) = 30 m/s http:/ / njc.tl/ 3r Slide 23 / 246 11 The graph represents the relationship between velocity and time for an object moving in a straight line. What is the traveled distance of the object at 9 s? A 10 m 24 m B 36 m C 48 m D E 56 m http:/ / njc.tl/ 3s Slide 24 / 246 12 Which of the following is true? A The object increases its velocity The object decreases its velocity B The object’s velocity stays unchanged C The object stays at rest D More information is required E http:/ / njc.tl/ 3t
Slide 25 / 246 13 What is the initial position of the object? A 2 m B 4 m 6 m C 8 m D 10 m E http:/ / njc.tl/ 3u Slide 26 / 246 14 What is the velocity of the object? 2 m/s A 4 m/s B 6 m/s C 8 m/s D 10 m/s E http:/ / njc.tl/ 3v Slide 27 / 246 15 What is the initial position of the object? 2 m A 4 m B 6 m C 8 m D 10 m E http:/ / njc.tl/ 3u
Slide 28 / 246 16 The graph represents the position as a function of time of a moving object. What is the velocity of the object? 5 m/s A -5 m/s B 10 m/s C -10 m/s D 0 m/s E http:/ / njc.tl/ 3w Slide 29 / 246 Free Fall All unsupported objects fall towards the earth with the same acceleration. We call this acceleration the "acceleration due to gravity" and it is denoted by g. g = 9.8 m/s 2 Keep in mind, ALL objects accelerate towards the earth at the same rate. g is a constant! Slide 30 / 246 It stops momentarily. What happens at the v = 0 top? g = -9.8 m/s 2 It speeds up What happens when it (negative acceleration) goes down? g = -9.8 m/s 2 It slows down. What happens when it (negative acceleration) goes up? g = -9.8 m/s 2 An object is thrown upward It returns with its What happens when it with initial velocity, v original velocity. lands? o
Slide 31 / 246 It stops momentarily. v = 0 g = -9.8 m/s 2 It speeds up. It slows down. (negative acceleration) (negative acceleration) g = -9.8 m/s 2 g = -9.8 m/s 2 An object is thrown upward It returns with its with initial velocity, v original velocity. o Slide 32 / 246 On the way up: On the way down: v 0 t = 0 s a v 1 a v t = 3 s v 2 a t = 2 s v 1 t = 1 s v 2 a a v 1 v 2 t = 1 s v 2 t = 2 s a v a v 0 v 1 t = 3 s t = 0 s v Slide 33 / 246 For any object thrown straight up into the air, what does the velocity vs time graph look like? An object is thrown upward with initial velocity, v o v (m/s) It stops momentarily. v = 0 g = -9.8 m/s 2 t (s) It returns with its original velocity but in the opposite direction.
Slide 34 / 246 17 An object moves with a constant acceleration of 5 m/s 2 . Which of the following statements is true? A The object’s velocity stays the same The object moves 5 m each second B The object’s acceleration increases by 5 m/s2 C each second The object’s acceleration decreases by 5 m/s2 D each second the object’s velocity increases by 5 m/s each E second E http:/ / njc.tl/ 3x Slide 35 / 246 18 A truck travels east with an increasing velocity. Which of the following is the correct direction of the car’s acceleration? A B C D E A Slide 36 / 246 19 A car and a delivery truck both start from rest and accelerate at the same rate. However, the car accelerates for twice the amount of time as the truck. What is the final speed of the car compared to the truck? A Half as much The same B Twice as much C Four times as much D E One quarter as much http:/ / njc.tl/ 3y
Recommend
More recommend
