Kinematics in Two permission of the owners. NJCTL maintains its - PDF document

Slide 1 / 125 Slide 2 / 125 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 Kinematics in Two permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to Dimensions 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 / 125 Slide 4 / 125 Table of Contents Click on the topic to go to that section Kinematics in One Dimension (Review) · Kinematics in One Adding Vectors in Two Dimensions · Dimension Basic Vector Operations · Vector Components · Projectile Motion · Return to Table of Contents Slide 5 / 125 Slide 6 / 125 Review of 1-D Kinematics Review of 1-D Kinematics · Instantaneous acceleration is the limit as the time interval · Kinematics is the description of how objects move with becomes infinitesimally small. respect to a defined reference frame. · There are four equations of motion for constant acceleration, · Displacement is the change in position of an object. each requires a different set of quantities. · Average speed is the distance traveled divided by the time v = v o + at it took; average velocity is the displacement divided by the time. x = x o + v o t + ½ at 2 · Instantaneous velocity is the limit as the time becomes v 2 = v o2 + 2 a(x - x o ) infinitesimally short. v = v + v o · Average acceleration is the change in velocity divided by 2 the time.

Slide 7 / 125 Slide 8 / 125 1 A snapshot of three racing cars is shown on the 2 A car and a delivery truck both start from rest and diagram. All three cars start the race at the same accelerate at the same rate. However, the car time, at the same place and move along a straight accelerates for twice the amount of time as the track. As they approach the finish line, which car truck. What is the final speed of the car compared has the lowest average speed? to the truck? Car I Half as much A A B Twice as much Car II B Answer Answer C Four times as much Car III C D One quarter as much All three cars have the same average speed D http:/ / njc.tl/ 3m Slide 9 / 125 Slide 10 / 125 3 A car and a delivery truck both start from rest and 4 A modern car can develop an acceleration four accelerate at the same rate. However, the car times greater than an antique car like “Lanchester accelerates for twice the amount of time as the 1800”. If they accelerate over the same distance, truck. What is the traveled distance of the car what would be the velocity of the modern car compared to the truck? compared to the antique car? Half as much Half as much A A The same The same B B Answer Answer Twice as much Twice as much C C Four times as much Four times as much D D Slide 11 / 125 Slide 12 / 125 Motion at Constant Acceleration Graphing Motion at Constant Acceleration Below we can find the geometric explanation to the acceleration a =Δv/Δt. In physics there is another approach in addition to algebraic which is called graphical analysis. The formula v = v 0 + at can be If slope is equal to: m = Δy/Δx interpreted by the graph. We just need to recall our memory from Then consider a graph with velocity on the y-axis and time on the math classes where we already saw a similar formula y = mx + b. x-axis. What is the slope for the graph on the right? From these two formulas we can make some analogies: v # y (dependent 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 Δy/Δx).

Slide 13 / 125 Slide 14 / 125 Motion at Constant Acceleration 5 The velocity as a function of time is presented by the graph. What is the acceleration? The graph on the right has a slope of Δv/Δt, which is equal to acceleration. Therefore, the slope of a velocity vs. time graph is equal to acceleration. (slope) (slope of velocity vs. time) y =Δy/Δx a =Δv/Δt Answer http:/ / njc.tl/ 3n Slide 15 / 125 Slide 16 / 125 6 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. Answer http:/ / njc.tl/ 3o Slide 17 / 125 Slide 18 / 125 Motion at Constant Acceleration 7 Which of the following statements is true? 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 A The object slows down is equivalent to the Answer B The object moves with a constant velocity shadowed area (4x12 = 48). C The object stays at rest The change in velocity during first 12 seconds D The object is in free fall is 48 m/s. http:/ / njc.tl/ 3q

Slide 19 / 125 Slide 20 / 125 Analyzing Position vs Time Graphs 8 The following graph shows acceleration as a function of time of a moving object. What is the change in velocity during first 10 seconds? Recall earlier in this unit that slope was used to describe motion. x (m) The slope in a position vs. time graph is Δx/Δt, which is equal to Δx velocity. Δt v = Δx/Δt Answer Therefore, slope is equal to velocity on a position vs. time t (s) graph. http:/ / njc.tl/ 3r http:/ / njc.tl/ 2t Slide 21 / 125 Slide 22 / 125 Analyzing Position vs Time Graphs 9 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 positive slope is a positive velocity, a negative slope is a negative velocity, and a slope of zero means zero velocity. 10 m negative slope zero slope A positive slope v > 0 v < 0 v = 0 B 24 m x x x (m) (m) (m) 36 m C 48 m D Answer t (s) t (s) t (s) A positive velocity means moving in the positive direction, a negative velocity means moving in the negative direction, and zero velocity means not moving at all. http:/ / njc.tl/ 2t http:/ / njc.tl/ 3s Slide 23 / 125 Slide 24 / 125 11 What is the velocity of the object? 10 Which of the following is true? 2 m/s A 4 m/s B 6 m/s C 8 m/s D The object increases its velocity A B The object decreases its velocity Answer Answer The object’s velocity stays unchanged C The object stays at rest D http:/ / njc.tl/ 3t http:/ / njc.tl/ 3v

Slide 25 / 125 Slide 26 / 125 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 What happens when it We call this acceleration the "acceleration due to (negative acceleration) goes down? gravity" and it is denoted by g. g = -9.8 m/s 2 It slows down. What happens when it g = 9.8 m/s 2 (negative acceleration) 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 27 / 125 Slide 28 / 125 It stops momentarily. On the way up: On the way down: v = 0 v 0 g = -9.8 m/s 2 v 1 t = 0 s v a a t = 3 s It speeds up. v 2 a t = 2 s It slows down. (negative acceleration) v 1 (negative acceleration) g = -9.8 m/s 2 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 v 0 An object is thrown upward It returns with its v 1 with initial velocity, v original velocity. o t = 3 s t = 0 s v Slide 29 / 125 Slide 30 / 125 12 A ball is thrown straight up from point A it reaches a maximum For any object thrown straight up into the air, what height at point B and falls back to point C. Which of the does the velocity vs time graph look like? following is true about the direction of the ball’s velocity and acceleration between A and B? An object is thrown upward with initial velocity, v o v (m/s) It stops momentarily. v = 0 D A g = -9.8 m/s 2 B E t (s) Answer C It returns with its original velocity but in the opposite direction. http:/ / njc.tl/ 42