PES 1110 Fall 2013, Spendier Lecture 3/Page 1. Review Last time we talked about the first two physical quantities of kinematics: a = = units : v= =

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1 PES Fall 3, Spendier Lecture 3/Page Review Last time we talked about the first two physical quantities of kinematics: Position, x - Where an object is located, how far and what direction from origin - displacement: x = x x - distance: d = x + x Velocity, v - How fast an object is going and direction of motion. Slope of the positionversus-time graph. displacment x x ( t ) x ( t ) m Average velocity: v avg = = = units: elapsed time t t t s Instantaneous velocity: time derivative of position x dx lim t v= = t dt Today: continue chapter - Acceleration - Equations of motion ( d motion constant acceleration) Mathematical means of describing the relationships among these variables Acceleration Acceleration: a is the rate at which the velocity changes What does a physicist mean when they say the word rate? It means divide by time. Velocity is the rate at which position changes: so it is position divided by time. v v -v m / s m a = = units : = avg t t -t s s m/s per every second most descriptive way - how much did the velocity change by A world s land speed record was set by Colonel John P. Stapp when on March 9 th 954, he rode a rocket-propelled sled that moved down a track at km/h. He and his sled were brought to a stop in.4s. Assuming constant acceleration, what acceleration did he experience? t = s, t =.4 s v = km/h, v = m/s need to convert km/h into m/s

2 PES Fall 3, Spendier Lecture 3/Page km m hr v = = 83.3 m / s hr km 36s then a avg, r v v -v m / s 83.3 m / s = = = t t -t.4s s =.4 / m s He experienced a negative acceleration. Negative acceleration indicates acceleration in the opposite direction to motion. This usually indicates breaking or deceleration. We can also express this result in terms of g = m/s (magnitude of the average acceleration due to earth s gravity) The force of gravity that pulls you toward the earth. In fact, acceleration forces are measured in g-forces (more in chapter 5), where g is equal to the force of acceleration due to gravity near the Earth's surface (9.8 m/s, or 3 ft/s). What is his acceleration in g? g.4 m / s.7g = 9.8 m / s The deceleration was.7 g He did not die - however a constant 5 g-s for a minute can be deadly. Some g-forces we experience: Rollercoaster:.7 g a sneeze -.9 g car staring.5 g, (Formula car.7 g) car cornering.9g (Formula car -3g) A rabbit accelerates from rest to.m/s in.5 s. Find the rabbit s average acceleration t = s, t =. s v = m/s, v =. m/s a avg, r v v -v. m / s m / s = = = t t -t.5s s = 4.4 / m s Instantaneous acceleration So to make this instantaneous at one instant of time, we have to find the slope at this time, the slope of what graph? Position vs time: slope gives velocity Velocity vs time: slope gives acceleration

3 PES Fall 3, Spendier Lecture 3/Page 3 Again we are doing this magnifying business The instantaneous velocity (or simply acceleration) a= lim v = dv t t dt When you read the book (section.6) you can see that we can combine what we know about instantaneous velocity dv d d dx d x a= = v= = dt dt dt dt dt The acceleration of a particle at any instant is the second derivative of its poison x(t) with respect to time. %%%%%%%%%%%%% aside What makes a rollercoaster thrilling...is it the fast speed? How fast does it go - 7 mph (i looked it up) - well - I can go 7 miles per hour in my car and it is not fun So what makes a rollercoaster ride exciting? It is acceleration. More precisely it is the change in acceleration. The change in acceleration is caller jerk and defined as: = = = = dt dt dt dt dt dt dt dt 3 da d d dv d d dx d x Jerk = ( a) Jerk is the third derivative of position with respect to time. When rollercoaster are designed engineers look at jerk. What instrument is used to measure jerk a jerk meter. %%%%%%%%%%%%%% end aside Direction (and therefore) sign of acceleration There is one more thing about acceleration a tricky thing what about the plus and minus for acceleration. Velocity is easy what does a positive velocity mean? Going to the right. What does a negative velocity mean? Going to the left. Make my position positive, where should I stand? To the right. The direction (and therefore) sign of acceleration is more complicated than position s or velocity s. You cannot say that a positive acceleration means speeding up and a negative means slowing down. 3 3

4 PES Fall 3, Spendier Lecture 3/Page 4 From our equation: v a avg, r = t we see that acceleration is in the same directions as the change in velocity. Let s say we have this moving object. The object moves to the right and slows down: From m/s to 5m/s in a time t: v = m/s v = 5 m/s Then the average velocity v v -v 5 m / s m / s 5 m / s aavg, r = = = = t t t t This is an example where the object moves into the positive direction but the acceleration is negative, since the object is slowing down. In General: When a and v have the same sign, speed increases (mean absolute value increases). When a and v have the opposite sign, speed decreases (mean absolute value decreases). Keep all these in mind as we get into problem solving Constant Acceleration Acceleration is constant with time. Vertical motion: Near the surface of the earth, gravity can be considered as providing a constant D acceleration acting towards the center of the earth. We will talk about gravity a lot more later in the course but mention it here while we are looking a D acceleration. Real life: What do you notice when you throw up a ball. If we throw a ball straight up, it turns around at some point and comes back done due to gravity motion graphs? Position vs time graph corresponds to a parabola. 4

5 PES Fall 3, Spendier Lecture 3/Page 5 Demonstration: - Drop a whiteboard eraser and flat piece of paper simultaneously. observation: paper falls more slowly - Drop a whiteboard eraser and a crunched up piece of paper simultaneously. observation: both objects fall at the same rate Inference: After removing effects of air resistance, all objects accelerate downwards at the same rate. Gravitational acceleration: given by symbol g On earth: g = m/s By definition, up is positive and since objects accelerator down, g is negative All calculations in this course will ignore air resistance! So for free fall problems: constant acceleration A simplification in notation: We can always assume that the initial time t starts at zero Then elapsed time: t -t = is t = t hence I do not need a subscript t= t t= t = t Now by making this choice along comes a change in notation call the final position x = x call the initial position x = x The initials will have a subscript of zero to indicate to others and ourselves that we made this choice that the initial time is zero. Better notation: v = v x = x v = v x = x 5

6 PES Fall 3, Spendier Lecture 3/Page 6 For example, with this new notation, the equation for average velocity becomes x x x x x x vavg = = = t t t t Equations of motion (along straight line for constant acceleration) Mathematical means of describing the relationships among position, velocity, and acceleration. When the acceleration is constant, the average acceleration and instantaneous acceleration (see velocity vs time graph) are equal and we can write v v v - v v - v a = aavg = = = t t t t We can now solve for v v(t) = v +at Eq. our first Eq. of motion ==> missing x-x Test for t =, v = v as it should be So in this first equation, we have no information about x or x. How can we relate Eq. to position? Through average velocity: v avg x x ( t) x ( t) x( t) x ( t) t t t t = = = Eq. For a linear velocity function - due to constant acceleration- the average velocity over any time interval is the average at the beginning - v - and the velocity at the end - v -then for interval t = to t the average velocity is v avg = v + v ( ) Eq.3 sub Eq.3 into Eq. and using Eq. x(t) - x (t) = ( v + v ) = ( v + [ v + at ]) = v + at t Hence: x( t) x ( t) v t at = + Eq. 4 nd Eq. of motion ==> missing v our parabola! Test: for t =, x = x as it should be. 6

7 PES Fall 3, Spendier Lecture 3/Page 7 Then we can use some more algebra by combining Eq.4 and Eq. to obtain other equations of motion like: ( ) v = v + a x x Eq. 5 3rd Eq. of motion ==> missing time x( t) x ( t) vt at x( t) x ( t) = ( v + v) t Eq. 7 5th Eq. of motion ==> missing a = Eq. 6 4th Eq. of motion ==> missing v Table - in the book has all these summarized. We have five possible equations (see boxed ones), each involving three of the motion variables. They are valid in cases of constant acceleration only. Book section -8 In case you have already taken calculus - the equations can be derived in a different way. As a physicist it is always useful to be able to derive something in more than one way. Now we need to practice using these equations: A fighter plane must reach a speed of 75 m/s from rest in a distance of just 75m. What acceleration is required to achieve this? picture: knowns: v = m/s, v = 75 m/s, x = m, x = 75 m unknowns: time and acceleration to find a, we can use: v = v + a x x ( ) ( ) ( 75 m / s) -( m / s) ( m ) v - v a= = = 37.5 m / s = 3.8 m / s x-x 75-7

8 PES Fall 3, Spendier Lecture 3/Page 8 A car is traveling on a straight road with a speed of 3.m/s when the driver hits the brakes causing a constant deceleration of.5m/s. How long does it take and how far does the car go while stopping? Step : draw a picture Step : List of known variables The initial position: x = (always determined by your picture) Initial velocity: v = 3 m/s Acceleration: a = -.5 m/s^ (arrows of velocity vectors for same time interval get smaller) final velocity: v = Step 3: List of unknowns Time t how long does it take Final position x how far Step 4: Use appropriate equations of motion a) How long does it take the car to stop? v= v+ at I know everything else but time inside of it v v m / s 3 m / s = t= = s a.5 m / s It takes the car s to stop. b) How far does the car go? x= x+ ( v ) t+ at = + ( 3 m / s)( s) + (.5 m / s )( s) = 36 m+ (.5 m / s )44 s = 36m 8m= 8m So the car travels 8 m until it stops. 8

9 PES Fall 3, Spendier Lecture 3/Page 9 A ball is thrown vertically upwards at a speed of 5. m/s. a) At what time does the ball reach its maximum height? t =.55 s b) What is the ball's maximum height? x = 3.9 m c) How long is the ball in the air for if it is caught at the same height from which it was thrown? t = 5. s d) What is the ball's velocity as it is caught? v = 5 m/s down 9