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Kinematics and One Dimensional Motion Kinematics Vocabulary Position Kinematics Vocabulary • Kinema means movement Kinematics and One Dimensional Motion • Mathematical description of motion –Position 8.01 – Time Interval – Displacement W02D1 – Velocity; absolute value: speed – Acceleration – Averages of the later two quantities. Coordinate System in One Position Dimension • A vector that points from origin to body. Used to describe the position of a point in space • Position is a function of time • In one dimension: A coordinate system consists of: 1. An origin at a particular point in space 2. A set of coordinate axes with scales and labels 3. Choice of positive direction for each axis: unit vectors G 4. Choice of type: Cartesian or Polar or Spherical xi()txt= ()ˆ Example: Cartesian One-Dimensional Coordinate System 1 Concept Question: Displacement Vector Displacement An object goes from one point in space to another. After Change in position the object arrives at its destination, the magnitude of vector of the object its displacement is: during the time 1) either greater than or equal to interval Δt = t − t 2 1 2) always greater than 3) always equal to 4) either smaller than or equal to G 5) always smaller than Δ≡riix()txt − ()ˆˆ ≡Δ xt () ()21 6) either smaller or larger than the distance it traveled. Average Velocity Instantaneous Velocity G G The average velocity, v () t , is the displacement Δ r divided by the time interval Δt • For each time interval Δ t , we calculate the x- component of the average velocity. As Δ→ t 0 , we G generate a sequence of the x-component of the G ΔΔr x ˆˆ vii()tvt≡= =x () average velocities. The limiting value of this ΔΔtt sequence is defined to be the x-component of the instantaneous velocity at the time t . The x-component of the average velocity is given by Δx xt()+Δt − xt() dx Δx v (t) ≡ lim v = lim = lim ≡ x Δt→0 x Δt→0 Δt→0 vx (t) = Δt Δt dt Δt 2 Instantaneous Velocity Table Problem: Hedge Fund x-component of the velocity is equal to the slope of the Ride Home tangent line of the graph of x-component of position vs. A hedge fund manager usually takes the train home time at time t and is met at the train station exactly at 6:30 by his chauffer who drives him 6 miles to his estate. One day he leaves work early, arriving at the train station at 6:00. Finding his cell phone discharged, he jogs towards his home at 6 mph. After 24 minutes, he meets his chauffer who turns around and takes him home. a How much earlier than usual does he arrive home? b What else can you determine from this information? Stuck? position Graphical Representation • Represent the Problem in New Way Estate –Graphical Å12 minÆ L – Pictures with descriptions i m ly o y l – Pure verbal u a a d u s s in o – Equations e T U • Could You Solve it If? Å6ÆÅ6Æ – The problem were simplified? er nag – You knew some other fact/relationship Ma Station – Solve any part of problem, even simple time one 0 24 30 3 CQ: Representing Motion - 1d CQ: Representing Motion - 1d Which of these is a possible graph of the position, Which of these is a possible graph of the x- x(t)? component of velocity? Average Acceleration Instantaneous Acceleration Change in instantaneous velocity divided by For each time interval Δt, we calculate the x- the time interval Δt = t − t component of the average acceleration. As Δ t → 0 , 2 1 we generate a sequence of x-component of average G vv− accelerations. The limiting value of this sequence is G Δv ΔΔvvxxˆˆˆˆ()xx,2 ,1 ai≡= = iii = =ax defined to be the x-component of the instantaneous ΔΔΔΔtt t t acceleration at the time t. G Δvdvvt+Δ t − vt The x-component of the average acceleration ˆˆxx ˆxx( ) ( ) ˆˆ aiii(tat )=≡xx ( ) lim a = lim = lim ii ≡ Δ→tt00 Δ→ΔΔttdt Δ→ t 0 Δv a = x dv x Δt a (t) = x x dt 4 Concept Question: One- Instantaneous Acceleration Dimensional Kinematics The x-component of acceleration is equal to the slope of the tangent line of the graph of the x-component of the The graph shows the position as a velocity vs. time at time t function of time for two trains running on parallel tracks. For times greater than t =0, which is true: 1. At time tB, both trains have the same velocity. 2. Both trains speed up all the time. 3. Both trains have the same velocity at some time before tB, . 4. Somewhere on the graph, both trains have the same acceleration. Summary: Constant Example: Free Fall Acceleration • Choose coordinate system with x-axis • Acceleration a = constant x vertical, origin at ground, and positive unit vector pointing upward. a =−g =−(9.8 m ⋅s-2 ) • Acceleration: x • Velocity vx (t) = vx,0 + axt vx (t) = vx,0 − gt • Velocity 1 2 1 2 • Position x()txvtat=+0,0xx + x(t) = x + v t − gt 2 • Position: 0 x,0 2 5 Concept Question: One- Concept Question: One Dimensional Kinematics Dimensional Kinematics You are throwing a ball straight up in the air. At the highest point, the ball’s A person standing at the edge of a cliff throws one ball straight up and another ball straight 1) velocity and acceleration are zero down, each at the same initial speed. Neglecting air resistance, which ball hits the 2) velocity is nonzero but its acceleration ground below the cliff with the greater speed: is zero 1. ball initially thrown upward; 3) acceleration is nonzero, but its velocity is zero 2. ball initially thrown downward; 4) velocity and acceleration are both 3. neither; they both hit at the same speed. nonzero. Worked Example: Runner A runner accelerates from rest with a Problem Solving Strategies: constant x-component of acceleration One-Dimensional Kinematics 2.0 m s-2 for 2.0 s and then travels at a constant velocity for an additional 6.0 s. How far did the runner travel? 6 I. Understand – get a conceptual I. Understand – get a grasp of the problem conceptual grasp of the Question 1: How many objects are involved in the problem problem? • stage 1: constant acceleration Question 2: How many different stages of motion occur? • stage 2: constant velocity. Question 3: For each object, how many independent directions are needed to describe the motion of that object? Tools: Coordinate system Question 4: What choice of coordinate system best suits Kinematic equations the problem? Question 5: What information can you infer from the problem? II. Devise a Plan II. Devise a Plan • Sketch the problem Stage 1: constant acceleration • Choose a coordinate system Initial conditions: x = 0 v = 0 0 x,0 • Write down the complete set of equations for the position and 1 velocity functions; identify any specified quantities; clean up the Kinematic Equations: 2 x()tat= x vtxx()= at equations. 2 Final Conditions: end acceleration at tt= a • Finding the “missing links”: count the number of independent 1 equations and the number of unknowns. position: x ≡==xt() t at2 aaxa2 • You can solve a system of n independent equations if you have velocity vx,a ≡ vx (t = ta ) = axta exactly n unknowns. • Look for constraint conditions 7 III. Devise a Plan III. Solve Stage 2: constant velocity, time interval [ttab, ] • Design a strategy for solving a system of equations. • runs at a constant velocity for the time tt− ba • Check your algebra and dimensions. • final position xbbaxaba≡==+xt() t x v, ( t − t ) • Substitute in numbers. • Check your results and units. III. Solve III. Solve • three independent equations 1 x = at2 • Solve for distance the runner has axa2 traveled vatx,axa= 1122 x =+xvtt( −) xb≡==xt() t b at xa + at xa() t b −= t a att xab − at xa baxaba, 22 • Six symbols: x , x , v , a , t , t b a x,a x a b • Three given quantities specified in problem a , t , t x a b 8 IV. Review IV. Review Numerical results: • Check your results, do they make sense (think!) Runner accelerated for ta = 2.0s -2 • Check limits of an algebraic expression (be creative) Initial acceleration: ax =⋅2.0m s Runs at a constant velocity for • Think about how to extend model to cover more ttba−=6.0s general cases (thinking outside the box) Total time of running ttba= +=+=6.0s 2.0s 6.0s 8.0s • Solved problems act as models for thinking about Total distance running new problems. (Mechanics provides a foundation of 112-2 -212 of solved problems.) xattattotal=− x a b x a =()2.0 m ⋅ s()() 2.0s 8.0s −() 2.0 m ⋅ s() 2.0s =× 2.8 10 m 22 -2 -1 Final velocity vatxa, = x a =⋅(2.0m s)( 2.0s) =⋅ 4.0m s Summary: Time Dependent Table Problem: One Acceleration Dimensional Kinematics • Acceleration is a non- Bus and car constant function of time At the instant a traffic light turns green, a car starts from atx () rest with a given constant acceleration, 5.0 × 10 − 1 m ⋅ s -2 . Just as the light turns green, a bus, traveling with a given • Change in velocity constant speed, 1.6 × 10 1 m ⋅ s -1 , passes the car. The car tt′= vt()−= v atdt (′ ) ′ speeds up and passes the bus some time later. How far xx,0 ∫ x down the road has the car traveled, when the car passes t′=0 the bus? • Change in position tt′= x()tx−= vtdt (′ ) ′ 0 ∫ x t′=0 9 Example: Non-constant Example: Non-constant acceleration acceleration • Consider an object released at time t = 0 with an initial x- Velocity: v x tt′= component of velocity x ,0 , located at position 0 , and vt()=+ v atdt (′′ ) xx,0 ∫ x accelerating according to t′=0 tt′= ′ ′ 223tt′= 11tt= tt= 2 =+vbbtbtdtvbtbtbt() ++′′ =+ ′ + ′ + ′ xx,0∫ 0 1 2 ,0 0t′=0 1tt′=00 2 ′= atx ()=++ b01 btbt 2 t′=0 23 1 =+−vbtbt2 x,0 02 1 Position: tt′= xt()=+ x v ( t′′ ) dt = 0 ∫ x t′=0 tt′= 11 111 • Find the velocity and position as a function of time.
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