Physics 101 Lecture 3 Motion in 1D Dr. Ali ÖVGÜN EMU Physics Department
Motion along a straight line q Motion q Position and displacement q Average velocity and average speed q Instantaneous velocity and speed q Acceleration q Constant acceleration: A special case q Free fall acceleration
Motion Everything moves! Motion is one of the main topics in Physics I Simplification: Consider a moving object as a particle, i.e. it moves like a particle a point object LAX Newark
4 Basic Quantities in Kinematics
One Dimensional Position r q Motion can be defined as the change of position over time. q How can we represent position along a straight line? q Position definition: n n n Defines a starting point: origin (r = 0), r relative to origin Direction: positive (right or up), negative (left or down) It depends on time: t = 0 (start clock), r(t=0) does not have to be zero. q Position has units of [Length]: meters. r = + 2.5 m i r = - 3 m i For motion along a straight line, the direction is represented simply by + and signs. + sign: Right or Up. - sign: Left or Down.
Displacement q Displacement is a change of position in time.!!! q Displacement: Δr = rf ( t f ) ri ( ti ) n f stands for final and i stands for initial. q It is a vector quantity. q It has both magnitude and direction: + or - sign q It has units of [length]: meters. r 1 (t 1 ) = + 2.5 m i r 2 (t 2 ) = - 2.0 m i Δr = -2.0 m - 2.5 m = -4.5 m i r 1 (t 1 ) = - 3.0 m i r 2 (t 2 ) = + 1.0 m i Δr = +1.0 m + 3.0 m = +4.0 m i
Distance and Position-time graph q Displacement in space n n From A to B: Δr = r B r A = 52 m 30 m i = 22 m i From A to C: Δr = r c r A = 38 m 30 m = 8 m i q Distance is the length of a path followed by a particle n n from A to B: d = r B r A = 52 m 30 m = 22 m from A to C: d = r B r A + r C r B = 22 m + 38 m 52 m = 36 m q Displacement is not Distance.
Velocity q Velocity is the rate of change of position. q Velocity is a vector quantity. q Velocity has both magnitude and direction. displacement q Velocity has a unit of [length/time]: meter/second. distance q We will be concerned with three quantities, defined as: n Average velocity n Average speed v s avg avg Δx = = Δt x x Δt total distance = Δt f i n Instantaneous velocity Δx dx v = lim = Δ t 0 Δt dt displacement
Average Velocity q q q q Average velocity Δx x f xi vavg = = Δt Δt is the slope of the line segment between end points on a graph. Dimensions: length/time (L/T) [m/s]. SI unit: m/s. It is a vector (i.e. is signed), and displacement direction sets its sign.
Average Speed q Average speed q Dimension: length/time, [m/s]. q Scalar: No direction involved. q Not necessarily close to V avg : n n s avg = total distance Δt S avg = (6m + 6m)/(3s+3s) = 2 m/s V avg = (0 m)/(3s+3s) = 0 m/s
Instantaneous Velocity q Instantaneous means at some given instant. The instantaneous velocity indicates what is happening at every point of time. q Limiting process: n Chords approach the tangent as Δt => 0 n Slope measure rate of change of position q Instantaneous velocity: q It is a vector quantity. Δx dx v = lim = Δ t 0 Δt dt q Dimension: length/time (L/T), [m/s]. q It is the slope of the tangent line to x(t). q Instantaneous velocity v(t) is a function of time.
Uniform Velocity q Uniform velocity is the special case of constant velocity q In this case, instantaneous velocities are always the same, all the instantaneous velocities will also equal the average velocity q Begin with Δx x f xi then x = x + v Δ x v x = = f i x t Δt Δt v x(t) v x Note: we are plotting velocity vs. time v(t) x f x i 0 t 0 t i t f t
Average Acceleration q Changing velocity (non-uniform) means an acceleration is present. q Acceleration is the rate of change of velocity. q Acceleration is a vector quantity. q Acceleration has both magnitude and direction. q Acceleration has a dimensions of length/time 2 : [m/s 2 ]. q Definition: n n Average acceleration Instantaneous acceleration a avg = Δv Δt = v t f f v t i i a = lim Δt 0 Δv Δt = dv dt = d dt dx dt = 2 d v dt 2
Average Acceleration q Average acceleration q Velocity as a function of time q It is tempting to call a negative acceleration a deceleration, but note: q n a avg v f = ( t) Δv Δt = = v i v t f f + v t a i i avg Δt Note: we are plotting velocity vs. time When the sign of the velocity and the acceleration are the same (either positive or negative), then the speed is increasing n When the sign of the velocity and the acceleration are in the opposite directions, the speed is decreasing Average acceleration is the slope of the line connecting the initial and final velocities on a velocity-time graph
Instantaneous and Uniform Acceleration q The limit of the average acceleration as the time interval goes to zero lim Δt 0 q When the instantaneous accelerations are always the same, the acceleration will be uniform. The instantaneous acceleration will be equal to the average acceleration q Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph a = Δv = Δt dv dt = d dt dx dt 2 d x = 2 dt
Special Case: Motion with Uniform Acceleration (our typical case) q Acceleration is a constant q Kinematic Equations (which we will derive in a moment)! r =! v! r + v t + 1 at 0 0 2! v! = 0 +! at! 2! v 2 0!!! 2 = v + 2aΔx
Problem-Solving Hints q q Read the problem Draw a diagram n Choose a coordinate system, label initial and final points, indicate a positive direction for velocities and accelerations q q q q Label all quantities, be sure all the units are consistent n Convert if necessary Choose the appropriate kinematic equation Solve for the unknowns n You may have to solve two equations for two unknowns Check your results v 2 2 0 = v + 2aΔx v = v 0 + 1 2 at Δx = v0t + at 2
Example q An airplane has a lift-off speed of 30 m/s after a take-off run of 300 m, what minimum constant acceleration? 2 v = v 0 + at Δx = v t + 1 at 0 q What is the corresponding take-off time? 2 v 2 2 0 = v + 2aΔx Δx = v0t + at v = v + at 1 2 0 2 v 2 2 = v + 2aΔx 0
September 8, 2008
September 8, 2008
Free Fall Acceleration y q Earth gravity provides a constant acceleration. Most important case of constant acceleration. q Free-fall acceleration is independent of mass. q Magnitude: a = g = 9.8 m/s 2 q Direction: always downward, so a g is negative if we define up as positive, a = -g = -9.8 m/s 2 q Try to pick origin so that x i = 0
Free Fall for Rookie q A stone is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The stone just misses the edge of the roof on the its way down. Determine q (a) the time needed for the stone to reach its maximum height. q (b) the maximum height. q (c) the time needed for the stone to return to the height from which it was thrown and the velocity of the stone at that instant. q (d) the time needed for the stone to reach the ground q (e) the velocity and position of the stone at t = 5.00s
Example Example
Summary q This is the simplest type of motion q It lays the groundwork for more complex motion q Kinematic variables in one dimension n Position r(t) m L n Velocity v(t) m/s L/T n Acceleration a(t) m/s 2 L/T 2 n All depend on time n All are vectors: magnitude and direction vector: q Equations for motion with constant acceleration: missing quantities!!! n v = v 0 + at r n!!!! 1 2 r = r0 + v0t + 2 at v " 2! 2 n v = v +!!! a( x x ) t 0 2 0
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September 8, 2008
September 8, 2008
September 8, 2008
24.5 m/s and 0.1 m/s September 8, 2008