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Unformatted text preview: Motion in 1D KINEMATICS The Major Player Isaac Newton 16421727 Develops Calculus to explain the theory of Mechanics F = ma
Kinematics Dynamics We begin here in 1D 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" Kinematicsdescribes motion
Dynamicsconcerned with the cause of motion Measurable quantities needed to describe motion are Displacement: x=xfx0 measured in units of distance (e.g., meters) Time or time interval, t=t t f 0 measured in sec x = xfxi 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
vavg x( t1 )  x( t0 ) x1  x0 = t1  t0 t1  t0 Freedom to choose "origin of time" If initial time is chosen to be t=0 vavg = x( t1 )  x( 0) x1  x0 t1 t1 So let t1=t !!! vavg x( t )  x0 = t Sample problem 21 "Average velocity at an instant" let the time interval be come very very small "Calculus 101" the derivative graphically t dt
Tangent at P is derivative of curve evaluated at t1 Instantaneous velocity, v(t) slope of the displacement curve at a point x( t ) dx( t ) v( t ) = lim t dt t 0 Sample problem 22 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! x = vav t =Area ??!!! Will Generalize.......... x  x0 = v( t ) dt 0 t Average Acceleration acceleration rate of change of velocity units are distance/time2 average acceleration is the average slope on a velocity vs. time graph average acceleration is the rate of change of the slope on a displacement vs. time graph aave v( t1 )  v( 0) v( t ) = t1  0 t 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 v( t ) dv( t ) a ( t ) = lim t dt t 0 Example 24 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
vavg = v ( t ) + v0 2 (1) (2) (3) aave =
vavg v( t )  v( 0 ) t v( t ) = v( 0 ) + a t
x( t ) = x0 + vavg t x ( t )  x0 = t MOTION IN ONE DIMENSION (3):
CONSTANTLY ACCELERATED MOTION x = x0 + v0t + 1 at 2 2
2 2a ( x  x0 ) = v 2  v0 (4) (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 dv a= dt dx v= dt implies v = a (t ) d t and implies x = v t d t () Calculus 102 Equations of Motion constant Acceleration
a ( t) = dv ( t ) dt dv ( t ) = a dt v ( t )  v ( 0) = v ( t )  v ( 0) = (Eq. 1) Instantaneous acceleration (Eq. 1b), rewrite (Eq. 1) in differential form ( ) dv
v0 t 0 v( t ) Integrate LHS of Eq. 1b
t 0 v ( t ) = v ( 0) + a t v0 + a t v ( t) = dx ( t ) dt dx ( t ) = v ( t ) dt x ( t )  x ( 0) = x ( t )  x ( 0) =
x( t ) x 0 t a ( t ' ) dt ' = a dt ' = a t (Eq. 1c) Integrate RHS of Eq. 1b (Eq. 2) (Eq. 2b), rewrite (Eq. 2) in differential form ( ) dx v( t )
' 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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 Fall '10
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 Physics, mechanics

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