Mechanics The \Kinematics Equations" Lana Sheridan De Anza College Oct 1, 2018 Last time • acceleration • graphs of kinematic quantities Overview • finish discussion of graphs • the kinematics equations (constant acceleration) • applying the kinematics equations 36 Chapter 2 Motion in One Dimension In Figure 2.10c, we can tell that the car slows as it moves to the right because its 36 Chapter 2 Motiondisplacement in One Dimension between adjacent images decreases with time. This case suggests the car moves to the right with a negative acceleration. The length of the velocity arrow Chapter 2 Motion in One Dimension 36 decreasesIn Figure in 2.10c, time weand can eventually tell that thereaches car slows zero. as From it moves this todiagram, the right we because see that its the displacementacceleration betweenand velocity adjacent arrows images are notdecreases in the withsame time. direction. This case The suggests car is moving the car withIn moves Figurea positive to the2.10c, velocity, right we with canbut atellwith negative that a negative the acceleration. car acceleration. slows as The it moves (This length totype of the the of right velocitymotion because arrowis exhib its - decreasesdisplacementited by a carin time thatbetween andskids eventually adjacent to a stop images afterreaches its decreases zero.brakes From are with applied.)this time. diagram, This The case wevelocity see suggests that and the accelthe - accelerationcareration moves are to inandthe opposite rightvelocity with directions.arrows a negative are Innot acceleration. terms in the of same our The earlier direction. length force ofThe discussion,the car velocity is moving imaginearrow withdecreasesa force a positive pulling in timevelocity, on and the but eventuallycar with opposite a negative reaches to the acceleration. zero. direction From (This itthis is moving:typediagram, of motionit weslows see isdown. thatexhib the- itedacceleration Each by a car purple that and skids accelerationvelocity to a arrowsstop arrow after are its innot brakesparts in the (b)are same andapplied.) direction.(c) of The Figure velocityThe 2.10car and is moving theaccel same- erationwithlength. a positive are Therefore, in oppositevelocity, these but directions. diagramswith a negative In represent terms acceleration. of our motion earlier (This of forcea particletype discussion, of undermotion constant imagine is exhib accel - - aitederation. force by pullinga This car that important on skids the car to analysis aopposite stop after model to itsthe brakeswill direction be arediscussed itapplied.) is moving: in theThe it next velocityslows section. down. and accel- erationEach arepurple in opposite acceleration directions. arrow Inin termsparts (b)of our and earlier (c) of forceFigure discussion, 2.10 is the imagine same length.aQ force uick Therefore,pulling Quiz 2.5 on Whichthesethe car diagrams one opposite of the represent to following the direction motion statements itof is a moving:particle is true? under it (a)slows constantIf adown. car accelis trav-- eration. Eacheling This purpleeastward, important acceleration its acceleration analysis arrow model mustin willparts be be eastward.(b) discussed and (c) (b) in of Ifthe Figurea nextcar is section.2.10 slowing is the down, same length.its acceleration Therefore, thesemust bediagrams negative. represent (c) A particle motion with of a constantparticle under acceleration constant accelcan - Qeration. uicknever QuizThis stop 2.5important and Which stay stopped. analysisone of the model following will be statements discussed isin true? the next (a) If section. a car is trav- eling eastward, its acceleration must be eastward. (b) If a car is slowing down, Q itsuick acceleration Quiz 2.5 Whichmust be one negative. of the following(c) A particle statements with constant is true? acceleration (a) If a car iscan trav- nevereling eastward,stop and stay its accelerationstopped. must be eastward. (b) If a car is slowing down, x 2.6its acceleration Analysis must Model: be negative. Particle (c) A particle with constant acceleration can never stop and stay stopped. Slope ϭ vxf Under Constant Acceleration x 2.6If the accelerationAnalysis of Model:a particle varies Particle in time, its motion can be complex and difficult Slope ϭ v Kinematics Graphsx xf Underto analyze. Constant A very common Acceleration and simple type of one-dimensional motion, however, is i x that2.6 in Analysiswhich the acceleration Model: is Particle constant. In such a case, the average acceleration Slope ϭ vxi If the acceleration of a particle varies in time, its motion can be complex and difficult Slope ϭ vxf t a over any time interval is numerically equal to the instantaneous acceleration a t toUnder xanalyze.,avg AConstant very common Acceleration and simple type of one-dimensional motion, however, is x xi at any instant within the interval, and the velocity changes at the same rate through- thatIf the in acceleration which the acceleration of a particle isvaries constant. in time, In itssuch motion a case, can the be averagecomplex acceleration and difficult aSlope ϭ vxi out the motion. This situation occurs often enough that we identify it as an analysis t a over any time interval is numerically equal to the instantaneous acceleration a x t tox,avg analyze. A very common and simple type of one-dimensional motion, however, xis i atmodel: any instant the particle within the under interval, constant and theacceleration. velocity changes In the at discussion the same rate that through follows,- we vSlopex ϭ v that in which the acceleration is constant. In such a case, the average acceleration a xi outgenerate the motion. several This equations situation that occurs describe often the enough motion that of we a particleidentify forit as this an model.analysis t ax,avg over any time interval is numerically equal to the instantaneous acceleration ax t If we replace ax,avg by ax in Equation 2.9 and take ti 5 0 and tf to be any later time Slope ϭ ax model:at any instant the particle within under the interval, constant and acceleration. the velocity Inchanges the discussion at the same that rate follows, through we - v ax generatet, we find several that equations that describe the motion of a particle for this model. axt out the motion. This situation occurs often enough that we identify it as an analysis vxf 2 vxi 5 Slope ϭ a model:If we thereplace particle ax,avg underby ax in constant Equation acceleration. 2.9 and take In ti the 0 anddiscussion tf to be thatany laterfollows, time we vxi x vxf ax 5 vx t, we find that v generate several equations that describe thet 2motion0 of a particle for this model. axtxi v 2 v If we replace ax,avg by ax in Equation 2.9xf andxi take ti 5 0 and tf to be any later time v Slope ϭ ax v t or a 5 xi t xf t, we find that x v t 2 0 axixt b vxf 5 vxi 1 axt v (for2 v constant ax) (2.13) t or xf xi vxi t vxf a 5 x t 2 0 vxi This powerful expression enables us to determine an object’s velocity at any time abx vxf 5 vxi 1 axt (for constant ax) (2.13) t t if we know the object’s initial velocity v and its (constant) acceleration a . A t or xi x Thisvelocity–time powerful expressiongraph for this enables constant-acceleration us to determine an motion object’s is shownvelocity in at Figure any time 2.11b. ax b Slope ϭ 0 vxf 5 vxi 1 axt (for constant ax) (2.13) t Theif we graph know is the a straight object’s line, initial the velocityslope of vwhichxi and isits the (constant) acceleration acceleration ax; the (constant) ax. A velocity–timeslope is consistent graph withfor this a 5constant-acceleration dv /dt being a constant. motion Notice is shown that in the Figure slope 2.11b. is posi - ϭ This powerful expression xenablesx us to determine an object’s velocity at any time ax Slope 0 ax Thetive, graph which is indicatesa straight aline, positive the slope acceleration. of which isIf the the acceleration acceleration a xwere; the (constant)negative, the t if we know the object’s initial velocity vxi and its (constant) acceleration ax. A slopeslope is of consistent the line inwith Figure ax 5 2.11bdvx/dt wouldbeing bea constant. negative. Notice When that the theacceleration slope is posi is con- - a t velocity–time graph for this constant-acceleration motion is shown in Figure 2.11b. Slope ϭ 0 t x tive,stant, which the indicatesgraph of aacceleration positive acceleration. versus time If (Fig.the acceleration 2.11c) is a straightwere negative, line having the a The graph is a straight line, the slope of which is the acceleration ax; the (constant) c t slopeslope of of the zero. line in Figure 52.11b would be negative. When the acceleration is con- t slope is consistent with ax dvx/dt being a constant. Notice that the slope is posi- ax stant,tive, Because which the graph indicates velocity of acceleration aat positive constant acceleration.versus acceleration time (Fig. If variesthe 2.11c) acceleration linearly is a straight in were time line negative, according having the a to Figurec 2.11 A particle under slopeEquation of zero. 2.13, we can express the average velocity in any time interval as the arith- t slope of the line in Figure 2.11b would be negative. When the acceleration is con- constant acceleration ax moving Because velocity at constant acceleration varies linearly in time according to t stant,metic themean graph of the of initialacceleration velocity versus vxi and time the (Fig. final 2.11c) velocity is av xfstraight: line having a Figurealong the 2.11 x axis: A particle (a) the position–under Equation 2.13, we can express the average velocity in any time interval as the arith- constanttimec graph, acceleration (b) the velocity–time a moving slope of zero.