Physics 207: Lecture 3, Pg 1 Physics 207, Lecture 3 l Today (Finish Ch. 2 & start Ch. 3)  Examine systems with non-zero acceleration (often constant)  Solve 1D problems with zero and constant acceleration (including free-fall and motion on an incline)  Use Cartesian and polar coordinate systems  Perform vector algebra

Physics 207: Lecture 3, Pg 2 “2D” Position, Displacement time (sec) position -2,2 -1,2 0,2 1,2 2,2 3,2 (x,y meters) origin position vectors x y

Physics 207: Lecture 3, Pg 3 “2D” Position, Displacement time (sec) position -2,2 -1,2 0,2 1,2 2,2 3,2 (x,y meters) origin position vectors x y displacement vectors

Physics 207: Lecture 3, Pg 4 Position, Displacement, Velocity time (sec) displacement vectors velocity vectors 1  2 2  3 3  4 4  5 5  6 Velocity always has same magnitude & length  CONSTANT x y

Physics 207: Lecture 3, Pg 5 Acceleration l The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval in which that change occurs. l Bold fonts are vectors l The average acceleration is a vector quantity directed along ∆v

Physics 207: Lecture 3, Pg 6 Acceleration l Particle motion often involves non-zero acceleration  The magnitude of the velocity vector may change  The direction of the velocity vector may change (true even if the magnitude remains constant)  Both may change simultaneously l E.g., a “particle” with smoothly decreasing speed v0v0 v1v1 v1v1 v1v1 v3v3 v5v5 v2v2 v4v4 v0v0

Physics 207: Lecture 3, Pg 7 Average Acceleration l Question: A sprinter is running around a track. 10 seconds after he starts he is running north at 10 m/s. At 20 seconds he is running at 10 m/s eastwards. l What was his average acceleration in this time (10 to 20 secs)?

Physics 207: Lecture 3, Pg 8 Average Acceleration l Question: A sprinter is running around a track. 10 seconds after he starts he is running north at 10 m/s. At 20 seconds he is running at 10 m/s eastwards. l What was his average acceleration in this time (10 to 20 secs)? 14 m/s /10 s= 1.4 m/s 2 to the SE N

Physics 207: Lecture 3, Pg 9 Instantaneous Acceleration l Average acceleration l The instantaneous acceleration is the limit of the average acceleration as ∆v/∆t approaches zero

Physics 207: Lecture 3, Pg 10 Position, velocity & acceleration for motion along a line If the position x is known as a function of time, then we can find both the instantaneous velocity v x and instantaneous acceleration a x as a function of time! t vxvx t x axax t

Physics 207: Lecture 3, Pg 11 Position, velocity & acceleration for motion along a line If the position x is known as a function of time, then we can find both the instantaneous velocity v x and instantaneous acceleration a x as a function of time! t vxvx t x axax t

Physics 207: Lecture 3, Pg 12 Going the other way…. l Particle motion with constant acceleration  The magnitude of the velocity vector changes a titi tt 0 t tftf  v = area under curve = a  t

Physics 207: Lecture 3, Pg 13 Going the other way…. l Particle motion with constant acceleration  The magnitude of the velocity vector changes l A particle with smoothly increasing speed: v1v1v1v1 v0v0v0v0 v3v3v3v3 v5v5v5v5 v2v2v2v2 v4v4v4v4 a v t 0 v f = v i + a  t = v i + a (t f - t i )  v = area under curve = a  t a titi tt 0 t tftf

Physics 207: Lecture 3, Pg 14 So if constant acceleration we can integrate twice t axax x t xixi vxvx t vivi

Physics 207: Lecture 3, Pg 15 Two other relationships l If constant acceleration then we also get: Slope of x(t) curve

Physics 207: Lecture 3, Pg 16 Example problem l A particle moves to the right first for 2 seconds at 1 m/s and then 4 seconds at 2 m/s. l What was the average velocity? l Two legs with constant velocity but …. Slope of x(t) curve vxvx t

Physics 207: Lecture 3, Pg 17 Example problem l A particle moves to the right first for 2 seconds at 1 m/s and then 4 seconds at 2 m/s. l What was the average velocity? l Two legs with constant velocity but …. l We must find the displacement (x 2 –x 0 ) l And x 1 = x 0 + v 0 (t 1 -t 0 ) x 2 = x 1 + v 1 (t 2 -t 1 ) l Displacement is (x 2 - x 1 ) + (x 1 – x 0 ) = v 1 (t 2 -t 1 ) + v 0 (t 1 -t 0 ) l x 2 –x 0 = 1 m/s (2 s) + 2 m/s (4 s) = 10 m in 6 seconds or 5/3 m/s Slope of x(t) curve vxvx t

Physics 207: Lecture 3, Pg 18 A particle starting at rest & moving along a line with constant acceleration has a displacement whose magnitude is proportional to t 2 1. This can be tested 2. This is a potentially useful result Displacement with constant acceleration

Physics 207: Lecture 3, Pg 19 Free Fall l When any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity. l This acceleration on the Earth’s surface, caused by gravity, is typically written as “little” g l Any object, be it a baseball or an elephant, experiences the same acceleration (g) when it is dropped, thrown, spit, or hurled, i.e. g is a constant.

Physics 207: Lecture 3, Pg 20 Gravity facts: l g does not depend on the nature of the material !  Galileo ( ) figured this out without fancy clocks & rulers! l Feather & penny behave just the same in vacuum l Nominally, g = 9.81 m/s 2  At the equatorg = 9.78 m/s 2  At the North poleg = 9.83 m/s 2

Physics 207: Lecture 3, Pg 21 When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? A. Both v = 0 and a = 0 B. v  0, but a = 0 C. v = 0, but a  0 D. None of the above y Exercise 1 Motion in One Dimension

Physics 207: Lecture 3, Pg 22 When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? A. Both v = 0 and a = 0 B. v  0, but a = 0 C. v = 0, but a  0 D. None of the above y Exercise 1 Motion in One Dimension

Physics 207: Lecture 3, Pg 23 In driving from Madison to Chicago, initially my speed is at a constant 65 mph. After some time, I see an accident ahead of me on I-90 and must stop quickly so I decelerate increasingly “fast” until I stop. The magnitude of my acceleration vs time is given by, A.  B.  C.  a t Home Exercise More complex Position vs. Time Graphs Question: My velocity vs time graph looks like which of the following ? v t v v

Physics 207: Lecture 3, Pg 24 In driving from Madison to Chicago, initially my speed is at a constant 65 mph. After some time, I see an accident ahead of me on I-90 and must stop quickly so I decelerate increasingly fast until I stop. The magnitude of my acceleration vs time is given by, A.  B.  C.  a t Home Exercise More complex Position vs. Time Graphs Question: My velocity vs time graph looks most like which of the following ? v t v v

Physics 207: Lecture 3, Pg 25 Exercise 2 1D Freefall A. v A < v B B. v A = v B C. v A > v B l Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v 0, Alice straight down and Bill straight up.. ( l The speed of the balls when they hit the ground are v A and v B respectively. (Neglect air resistance.) v0v0v0v0 v0v0v0v0 BillAlice H vAvAvAvA vBvBvBvB

Physics 207: Lecture 3, Pg 26 Exercise 2 1D Freefall A. v A < v B B. v A = v B C. v A > v B l Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v 0, Alice straight down and Bill straight up. The speed of the balls when they hit the ground are v A and v B respectively. v0v0 v0v0 BillAlice H vAvA vBvB

Physics 207: Lecture 3, Pg 27 Exercise 2 1D Freefall : Graphical solution l Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v 0, Alice straight down and Bill straight up. vxvx t cliff back at cliff turnaround point ground v0v0 -v 0 v ground  v= -g  t identical displacements (one + and one -)

Physics 207: Lecture 3, Pg 28 The graph at right shows the y velocity versus time graph for a ball. Gravity is acting downward in the -y direction and the x-axis is along the horizontal. Which explanation best fits the motion of the ball as shown by the velocity-time graph below? A. The ball is falling straight down, is caught, and is then thrown straight down with greater velocity. B. The ball is rolling horizontally, stops, and then continues rolling. C. The ball is rising straight up, hits the ceiling, bounces, and then falls straight down. D. The ball is falling straight down, hits the floor, and then bounces straight up. E. The ball is rising straight up, is caught and held for awhile, and then is thrown straight down. Home Exercise,1D Freefall

Physics 207: Lecture 3, Pg 29 Problem Solution Method: Five Steps: 1) Focus the Problem - draw a picture – what are we asking for? 2) Describe the physics -what physics ideas are applicable -what are the relevant variables known and unknown 3) Plan the solution -what are the relevant physics equations 4) Execute the plan -solve in terms of variables -solve in terms of numbers 5) Evaluate the answer -are the dimensions and units correct? -do the numbers make sense?

Physics 207: Lecture 3, Pg 30 A science project l You drop a bus off the Willis Tower (442 m above the side walk). It so happens that Superman flies by at the same instant you release the car. Superman is flying down at 35 m/s. l How fast is the bus going when it catches up to Superman?

Physics 207: Lecture 3, Pg 31 A “science” project l You drop a bus off the Willis Tower (442 m above the side walk). It so happens that Superman flies by at the same instant you release the car. Superman is flying down at 35 m/s. l How fast is the bus going when it catches up to Superman? l Draw a picture  y t yiyi 0

Physics 207: Lecture 3, Pg 32 A “science” project l Draw a picture  l Curves intersect at two points y t yiyi 0

Physics 207: Lecture 3, Pg 33 Home exercise of a 1D motion problem l A cart is initially traveling East at a constant speed of 20 m/s. When it is halfway (in distance!) to its destination its speed suddenly increases and thereafter remains constant. All told the cart spends a total of 10 s in transit with an average speed of 25 m/s. l What is the speed of the cart during the 2 nd half of the trip? l This is a variation on the in class example but much more difficult. And

Physics 207: Lecture 3, Pg 34 Home exercise of a 1D motion problem l A cart is initially traveling East at a constant speed of 20 m/s. When it is halfway (in distance!) to its destination its speed suddenly increases and thereafter remains constant. All told the cart spends a total of 10 s in transit with an average speed of 25 m/s. l What is the speed of the cart during the 2 nd half of the trip? l Dynamical relationships (only if constant acceleration): And

Physics 207: Lecture 3, Pg 35 Drawing the picture l Plus the average velocity l Knowns:  x 0 = 0 m  t 0 = 0 s  v 0 = 20 m/s  t 2 = 10 s  v avg = 25 m/s  relationship between x 1 and x 2 l Four unknowns x 1 v 1 t 1 & x 2 and must find v 1 in terms of knowns t2t2 t0t0 t1t1 v 1 ( > v 0 ) a 1 =0 m/s 2 v 0 a 0 =0 m/s 2 x0x0 x1x1 x2x2

Physics 207: Lecture 3, Pg 36 Using l Four unknowns l Four relationships l Focus on t2t2 t0t0 t1t1 v 1 ( > v 0 ) a 1 =0 m/s 2 v 0 a 0 =0 m/s 2 x0x0 x1x1 x2x2

Physics 207: Lecture 3, Pg 37 Using t2t2 t0t0 t1t1 v 1 ( > v 0 ) a 1 =0 m/s 2 v 0 a 0 =0 m/s 2 x0x0 x1x1 x2x2

Physics 207: Lecture 3, Pg 38 Using l Eliminate unknowns first x 1 next t 1 then x 2 t2t2 t0t0 t1t1 v 1 ( > v 0 ) a 1 =0 m/s 2 v 0 a 0 =0 m/s 2 x0x0 x1x1 x2x & Mult. by 2/ t 2 1

Physics 207: Lecture 3, Pg 39 Fini l Plus the average velocity l Given:  v 0 = 20 m/s  t 2 = 10 s  v avg = 25 m/s t2t2 t0t0 t1t1 v 1 ( > v 0 ) a 1 =0 m/s 2 v 0 a 0 =0 m/s 2 x0x0 x1x1 x2x2

Physics 207: Lecture 3, Pg 40 Tips: l Read !  Before you start work on a problem, read the problem statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning of all the terms used in stating the problem. l Watch your units (dimensional analysis) !  Always check the units of your answer, and carry the units along with your numbers during the calculation. l Ask questions !

Physics 207: Lecture 3, Pg 41 Context Rich Problem: For discussion l On a bright sunny day you are walking around the campus watching one of the many construction sites. To lift a bunch of bricks from a central area, they have brought in a helicopter. As the pilot is hovering he/she accidentally releases the bricks when they are 1000 m above the ground. A worker, directly below, stands still for 10 seconds before walking away. (Let g = 10 m/s 2 ) There is no wind or other effects. Does the worker live? (Criteria for living…..the worker moves before the bricks strike the ground)

Physics 207: Lecture 3, Pg 42 Problem: 1. We need to find the time it takes for the brick to hit the ground. 2. If t > 10 sec. then the worker is assured survival m

Physics 207: Lecture 3, Pg 43 Problem #1 (At home) You are writing a short adventure story for your English class. In your story, two submarines on a secret mission need to arrive at a place in the middle of the Atlantic ocean at the same time. They start out at the same time from positions equally distant from the rendezvous point. They travel at different velocities but both go in a straight line. The first submarine travels at an average velocity of 20 km/hr for the first 500 km, 40 km/hr for the next 500 km, 30 km/hr for the next 500 km and 50 km/hr for the final 500 km. In the plot, the second submarine is required to travel at a constant velocity, which you wish to explicitly mention in the story. What is that velocity?  a. Draw a diagram that shows the path of both submarines, include all of the segments of the trip for both boats.  b. What exactly do you need to calculate to be able to write the story?  c. Which kinematics equations will be useful?  d. Solve the problem in terms of symbols.  e. Does you answer have the correct dimensions (what are they)?  f. Solve the problem with numbers.

Physics 207: Lecture 3, Pg 44 Problem #2 (At home) As you are driving to school one day, you pass a construction site for a new building and stop to watch for a few minutes. A crane is lifting a batch of bricks on a pallet to an upper floor of the building. Suddenly a brick falls off the rising pallet. You clock the time it takes for the brick to hit the ground at 2.5 seconds. The crane, fortunately, has height markings and you see the brick fell off the pallet at a height of 22 meters above the ground. A falling brick can be dangerous, and you wonder how fast the brick was going when it hit the ground. Since you are taking physics, you quickly calculate the answer.  a. Draw a picture illustrating the fall of the brick, the length it falls, and the direction of its acceleration.  b. What is the problem asking you to find?  c. What kinematics equations will be useful?  d. Solve the problem in terms of symbols.  e. Does you answer have the correct dimensions?  f. Solve the problem with numbers.

Physics 207: Lecture 3, Pg 45 See you Wednesday Assignment: l For Wednesday, Read through Chapter 4.4