VIEWS: 8 PAGES: 22 POSTED ON: 3/12/2012
Motion in 1-D KINEMATICS The Major Player Isaac Newton 1642-1727 Develops Calculus to explain the theory of Mechanics F ma Kinematics Dynamics We begin here in 1-D Note that vx0 is the same throughout the motion of the object vx 0 v0 cos v y 0 v0 sin KINEMATICS to describe “motion” Kinematics-describes motion Dynamics-concerned with the cause of motion Measurable quantities needed to describe motion are Displacement: Dx=xf-x0 measured in units of distance (e.g., meters) Time or time interval, Dt=tf-t0 measured in sec Dx = xf-xi in fact this is a Vector Two possible directions +/- Coordinate space and time x(t) vs t for armadillo Put them together!!!! Get a rate of change Average Velocity Slope of x versus t graph xt1 xt0 x1 x0 vavg t1 t0 t1 t0 Freedom to choose “origin of time” If initial time is chosen to be t=0 xt1 x0 x1 x0 vavg t1 t1 So let t1=t !!! xt x0 vavg t Sample problem 2-1 “Calculus 101” the derivative graphically “Average velocity at an instant” let the time interval be come very very small Dt dt Instantaneous velocity, v(t) slope of the displacement curve at a point Tangent at P is derivative of curve evaluated at t1 Dxt dxt vt lim t 0 Dt dt Sample problem 2-2 Graphical representation of motion x(t)-position v(t)-velocity a(t)-acceleration Relationship btwn x and v: Calc. 001 Area under velocity plot is the displacement! Dx = vav Dt =Area ??!!! Will Generalize………. t x x0 vt dt 0 Average Acceleration acceleration rate of change of velocity units are distance/time2 average acceleration is the average slope on a velocity vs. time graph vt1 v0 Dvt average acceleration is the rate of change of the aave slope on a displacement t1 0 Dt vs. time graph Instantaneous Acceleration If the time interval between velocity measurements becomes vanishingly small, the acceleration is instantaneous Instantaneous acceleration is the slope of the velocity curve at a point Dvt dvt at lim t 0 Dt dt Example 2-4 x(t)=4 - 27t + t3 a) v(t) ? b) Is there a time, t when v(t) = 0 ? c) Plot results ! Example, constant acceleration: Example, constant acceleration “g” MOTION IN ONE DIMENSION : CONSTANTLY ACCELERATED MOTION Equations based on definition v ( t ) + v0 vavg = (1) 2 vt v0 aave vt v0 a t (2) t xt x0 xt x0 vavgt vavg (3) t MOTION IN ONE DIMENSION (3): CONSTANTLY ACCELERATED MOTION x x0 v0t 1 at 2 2 (4) 2a x x0 v2 v0 2 (5) See blackboard for derivation !!! MOTION IN ONE DIMENSION Special considerations Positive and negative values have meaning (use a coordinate axis or number line) Problems can have as complex a variability as necessary a dv implies v = ò a ( t ) dt and dt x = ò v ( t ) dt dx v implies dt Calculus 102 – Equations of Motion constant Acceleration dv ( t ) a (t) = (Eq. 1) Instantaneous acceleration dt dv ( t ) = a dt (Eq. 1b), re-write (Eq. 1) in differential form v( t ) v ( t ) - v ( 0) = ò ( ) dv v0 Integrate LHS of Eq. 1b ò a ( t ' ) dt ' = a ò dt ' = a t t t v ( t ) - v ( 0) = Integrate RHS of Eq. 1b 0 0 v ( t ) = v ( 0 ) + a t º v0 + a t (Eq. 1c) dx ( t ) v (t) = (Eq. 2) dt dx ( t ) = v ( t ) dt (Eq. 2b), re-write (Eq. 2) in differential form x(t ) x ( t ) - x ( 0) = ò ( ) dx x 0 ò v (t ) t x ( t ) - x ( 0) = 0 ' dt ' = (see black board) 1 x ( t ) - x ( 0 ) = v0 t + at 2 2 1 x ( t ) = x0 + v0 t + at 2 (Eq. 3) 2 Work this out MOTION IN ONE DIMENSION (4): CONSTANTLY ACCELERATED MOTION SPECIAL PROBLEM CLASS: FREE FALL The acceleration of gravity is a constant when near the surface of the earth g = 9.8 m/s2 toward the center of the earth Exercise What is the instantaneous velocity of an object at t = 3 s, if its displacement is described by x(t) = 7.8+9.2t – 2.1 t3 ? What is velocity at t=3.5 s ? What is the acceleration at t=3.0 s? EXAMPLE A ball is initially thrown upward with a velocity of 20.0 m/s at the edge of a 50.0 m tall building. a. How long does the ball take to hit the ground? b. What is the ball’s velocity when it strikes the ground?
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