Chapter Kinematics in One Dimensin Kinematics deals with the cncepts that are needed t describe mtin. Dynamics deals with the effect that frces have n mtin. Tgether, kinematics and dynamics frm the branch f physics knwn as Mechanics..1 Displacement x r = initial psitin r r r Δx = x x = displacement x r = final psitin 1
.1 Displacement x r =.0 m Δx r = 5.0 m x r = 7.0 m r r r Δx = x x = 7.0 m.0 m = 5.0 m.1 Displacement x r =.0 m Δx r = 5.0 m x r = 7.0 m r r r Δx = x x =.0 m 7.0 m = 5.0 m.1 Displacement x r =.0 m x r = 5.0 m Δx r = 7.0 m r r r Δx = x x (.0) m 7.0 m = 5.0 m =
. Speed and Velcity Average speed is the distance traveled divided by the time required t cver the distance. Average speed = Distance Elapsed time SI units fr speed: meters per secnd (m/s). Speed and Velcity Example 1 Distance Run by a Jgger Hw far des a jgger run in 1.5 hurs (5400 s) if his average speed is. m/s? Distance Average speed = Elapsed time ( )( Elapsed time) ( ) ( 5400 s) =1000 m Distance = Average speed =. m s. Speed and Velcity Average velcity is the displacement divided by the elapsed time. Displacement Average velcity = Elapsed time r v = r r x x = r Δx Δt 3
. Speed and Velcity Example The Wrld s Fastest Jet-Engine Car Andy Green in the car ThrustSSC set a wrld recrd f 341.1 m/s in 1997. T establish such a recrd, the driver makes tw runs thrugh the curse, ne in each directin, t nullify wind effects. Frm the data, determine the average velcity fr each run.. Speed and Velcity r r Δx + 1609 m v = = = + 339.5m s Δ t 4.740 s r r Δx 1609 m v = = = 34.7m s Δ t 4.695 s. Speed and Velcity The instantaneus velcity indicates hw fast the car mves and the directin f mtin at each instant f time. r r Δx v = lim Δt 0 Δt 4
.3 Acceleratin The ntin f acceleratin emerges when a change in velcity is cmbined with the time during which the change ccurs..3 Acceleratin DEFINITION OF AVERAGE ACCELERATION r r r v v a = r Δv = Δt.3 Acceleratin Example 3 Acceleratin and Increasing Velcity Determine the average acceleratin f the plane. v r = 0m s v r = 60 km h t = 0 s t = 9 s r r r v v a = 60 km h 0km h = 9 s 0 s km h = + 9.0 s 5
.3 Acceleratin.3 Acceleratin Example 3 Acceleratin and Decreasing Velcity r r r v v 13m s 8m s a = = = 5.0m 1 s 9 s s.3 Acceleratin 6
.4 Equatins f Kinematics fr Cnstant Acceleratin r r r x x v = r r r v v a = It is custmary t dispense with the use f bldface symbls verdrawn with arrws fr the displacement, velcity, and acceleratin vectrs. We will, hwever, cntinue t cnvey the directins with a plus r minus sign. x x v = v v a =.4 Equatins f Kinematics fr Cnstant Acceleratin Let the bject be at the rigin when the clck starts. x = 0 = 0 t x x v = x v = t x vt = 1 ( v + v)t =.4 Equatins f Kinematics fr Cnstant Acceleratin v v a = v v a = t at = v v v = v + at 7
.4 Equatins f Kinematics fr Cnstant Acceleratin Five kinematic variables: 1. displacement, x. acceleratin (cnstant), a 3. final velcity (at time t), v 4. initial velcity, v 5. elapsed time, t.4 Equatins f Kinematics fr Cnstant Acceleratin v = v + at x 1 ( v + v) t = ( v + v at)t = 1 + x = v t 1 + at.4 Equatins f Kinematics fr Cnstant Acceleratin x = v t + at = 1 1 ( 6.0 m s)( 8.0 s) + (.0m s )( 8.0 s) = + 110 m 8
.4 Equatins f Kinematics fr Cnstant Acceleratin Example 6 Catapulting a Jet Find its displacement. v = 0 m s x =?? a = +31m s v = +6m s.4 Equatins f Kinematics fr Cnstant Acceleratin v v a = t v v t = a x ( ) ( ) ( v v v v t v v ) 1 + = + = 1 a v v x = a.4 Equatins f Kinematics fr Cnstant Acceleratin v v x = a = ( 6 m s) ( 0m s) ( ) 31m s = + 6 m 9
.4 Equatins f Kinematics fr Cnstant Acceleratin Equatins f Kinematics fr Cnstant Acceleratin v = v + at 1 x ( v + v)t = v = v + ax x = v t + 1 at.5 Applicatins f the Equatins f Kinematics Reasning Strategy 1. Make a drawing.. Decide which directins are t be called psitive (+) and negative (-). 3. Write dwn the values that are given fr any f the five kinematic variables. 4. Verify that the infrmatin cntains values fr at least three f the five kinematic variables. Select the apprpriate equatin. 5. When the mtin is divided int segments, remember that the final velcity f ne segment is the initial velcity fr the next. 6. Keep in mind that there may be tw pssible answers t a kinematics prblem..5 Applicatins f the Equatins f Kinematics Example 8 An Accelerating Spacecraft A spacecraft is traveling with a velcity f +350 m/s. Suddenly the retrrckets are fired, and the spacecraft begins t slw dwn with an acceleratin whse magnitude is 10.0 m/s. What is the velcity f the spacecraft when the displacement f the craft is +15 km, relative t the pint where the retrrckets began firing? x a v v t +15000 m -10.0 m/s? +350 m/s 10
.5 Applicatins f the Equatins f Kinematics.5 Applicatins f the Equatins f Kinematics x a v v t +15000 m -10.0 m/s? +350 m/s v = v + ax v = v + ax v = ± = ± 500m s ( 350 m s) + ( 10.0m s )( 15000 m).6 Freely Falling Bdies In the absence f air resistance, it is fund that all bdies at the same lcatin abve the Earth fall vertically with the same acceleratin. If the distance f the fall is small cmpared t the radius f the Earth, then the acceleratin remains essentially cnstant thrughut the descent. This idealized mtin is called free-fall and the acceleratin f a freely falling bdy is called the acceleratin due t gravity. g = 9.80 m s r 3.ft s 11
.6 Freely Falling Bdies g = 9.80 m s.6 Freely Falling Bdies Example 10 A Falling Stne A stne is drpped frm the tp f a tall building. After 3.00s f free fall, what is the displacement y f the stne?.6 Freely Falling Bdies y a v v t? -9.80 m/s 0 m/s 3.00 s 1
.6 Freely Falling Bdies y a v v t? -9.80 m/s 0 m/s 3.00 s y = v t + at = 1 1 ( 0m s)( 3.00 s) + ( 9.80 m s )( 3.00 s) = 44.1m.6 Freely Falling Bdies Example 1 Hw High Des it G? The referee tsses the cin up with an initial speed f 5.00m/s. In the absence if air resistance, hw high des the cin g abve its pint f release?.6 Freely Falling Bdies y a v v t? -9.80 m/s 0 m/s +5.00 m/s 13
.6 Freely Falling Bdies y a v v t? -9.80 m/s 0 m/s +5.00 m/s v = v + ay v v y = a = v v y = a ( 0 m s) ( 5.00 m s) ( 9.80m s ) = 1.8 m.6 Freely Falling Bdies Cnceptual Example 14 Acceleratin Versus Velcity There are three parts t the mtin f the cin. On the way up, the cin has a vectr velcity that is directed upward and has decreasing magnitude. At the tp f its path, the cin mmentarily has zer velcity. On the way dwn, the cin has dwnward-pinting velcity with an increasing magnitude. In the absence f air resistance, des the acceleratin f the cin, like the velcity, change frm ne part t anther?.6 Freely Falling Bdies Cnceptual Example 15 Taking Advantage f Symmetry Des the pellet in part b strike the grund beneath the cliff with a smaller, greater, r the same speed as the pellet in part a? 14
.7 Graphical Analysis f Velcity and Acceleratin Δx + 8 m Slpe = = = + 4 m s Δ t s.7 Graphical Analysis f Velcity and Acceleratin.7 Graphical Analysis f Velcity and Acceleratin 15
.7 Graphical Analysis f Velcity and Acceleratin Δv + 1 m s Slpe = = = + 6m Δ t s s 16