Kinematics Kinematic Equations

Lana Sheridan

De Anza College

Jan 9, 2020 Last

• kinematic quantities

• graphs of kinematic quantities Overview

• relating graphs

• derivation of kinematics equations 2.4 33 down! It is very useful to equate the direction of the acceleration to the direction of a because it is easier from our everyday experience to think about what Pitfall Prevention 2.4 effect a force will have on an object than to think only in terms of the direction of Negative Acceleration Keep in the acceleration. mind that negative acceleration does not necessarily mean that an object is slowing down. If the acceleration is Q uick Quiz 2.4 If a car is traveling eastward and slowing down, what is the direc- negative and the is nega- tion of the force on the car that causes it to slow down? (a) eastward (b) west- tive, the object is speeding up! ward (c) neither eastward nor westward Pitfall Prevention 2.5 Deceleration The word deceleration has the common popular connota- From now on, we shall use the term acceleration to mean instantaneous accelera- tion of slowing down. We will not tion. When we mean average acceleration, we shall always use the adjective average. use this word in this book because Because v 5 dx/dt, the acceleration can also be written as it confuses the definition we have x given for negative acceleration. dv d dx d 2x a 5 x 5 5 (2.12) x dt dt dt dt 2 That is, in one-dimensional , the accelerationa b equals the second derivative of x with respect to time.

Conceptual Example 2.5 Graphical Relationships Between x, vx, and ax The of an object moving along the x axis varies with time as in Figure 2.8a. Graph the velocity versus time and the acceleration versus time for the object. Acceleration vs. Time Graphs SOLUTION x The velocity at any instant is the slope of the to the x–t graph at that instant. Between t 5 0 and t 5 tᎭ, the slope of the x–t graph increases uniformly, so the velocity increases linearly as shown in Figure 2.8b.

Between tᎭ and tᎮ, the slope of the x–t graph is con- stant, so the velocity remains constant. Between tᎮ and t tᎭ tᎮ t Ꭿ t൳ t൴ t൵ t൳, the slope of the x–t graph decreases, so the value of a the velocity in the v –t graph decreases. At t , the slope x ൳ vx of the x–t graph is zero, so the velocity is zero at that instant. Between t൳ and t൴, the slope of the x–t graph and therefore the velocity are negative and decrease uni- t tᎭ tᎮ tᎯ t൳ t൴ t൵ formly in this interval. In the interval t൴ to t൵, the slope of the x–t graph is still negative, and at t൵ it goes to zero. b Finally, after t൵, the slope of the x–t graph is zero, mean- ax ing that the object is at rest for t . t൵. The acceleration at any instant is the slope of the tan- gent to the v –t graph at that instant. The graph of accel- t x t t t eration versus time for this object is shown in Figure 2.8c. tᎭ Ꭾ ൴ ൵ The acceleration is constant and positive between 0 and c tᎭ, where the slope of the vx–t graph is positive. It is zero Figure 2.8 (Conceptual Example 2.5) (a) Position–time graph between tᎭ and tᎮ and for t . t൵ because the slope of the v –t graph is zero at these . It is negative between for an object moving along the x axis. (b) The velocity–time graph x for the object is obtained by measuring the slope of the position– tᎮ and t൴ because the slope of the vx–t graph is negative time graph at each instant. (c) The acceleration–time graph for during this interval. Between t൴ and t൵, the acceleration the object is obtained by measuring the slope of the velocity–time is positive like it is between 0 and tᎭ, but higher in value graph at each instant. because the slope of the vx–t graph is steeper. Notice that the sudden changes in acceleration shown in Figure 2.8c are unphysical. Such instantaneous changes cannot occur in reality.

Question

What does the area under an acceleration-time graph represent? 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 Constant Accelerationx Graphs 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 The if we graph know is thea straight object’s line, initial the velocityslope of vwhichxi and is its 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 a at 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. x metic mean of the initial velocity v xi and1 v xfthe final velocity v : alonggraph, the and x axis:(c) the (a) acceleration– the position– Because velocity at constant accelerationxi varies linearlyxf in time according to vx,avg 5 for constant ax (2.14) timeFigure graph,graph. 2.11 (b) A the particle velocity–time under Equation 2.13, we can express the average2 velocity in any time interval as the arith- vxi 1 vxf graph,constant and acceleration (c) the acceleration– ax moving metic mean of the initialvx ,avgvelocity5 v and thefor 1final constant velocity ax v 2: (2.14) timealong graph. the x axis: (a) the position– 2xi xf time graph, (b) the velocity–time v 1 v 1 2 graph, and (c) the acceleration– xi xf vx,avg 5 for constant ax (2.14) time graph. 2 1 2 38 Chapter 2 Motion in One Dimension

Q uick Quiz 2.6 In Figure 2.12, match each vx–t graph on the top with the ax–t graph on the bottom that best describes the motion. Matching Velocity to Acceleration Graphs

Figure 2.12 (Quick Quiz 2.6) vx vx vx Parts (a), (b), and (c) are vx–t graphs of objects in one-dimensional motion. The possible of each object as a function of time are shown in scrambled order in (d), t t t (e), and (f). a b c

ax ax ax

t t t d e f

Analysis Model Particle Under Constant Acceleration Imagine a moving object that can be modeled as a particle. If it Examples begins from position x and initial velocity v and moves in a straight i xi r BDBSBDDFMFSBUJOHBUBDPOTUBOUSBUF line with a constant acceleration a , its subsequent position and x along a straight freeway velocity are described by the following kinematic equations: r BESPQQFEPCKFDUJOUIFBCTFODFPGBJS

vxf 5 vxi 1 axt (2.13) resistance (Section 2.7) r BOPCKFDUPOXIJDIBDPOTUBOUOFUGPSDF v 1 v xi xf acts (Chapter 5) vx,avg 5 (2.14) 2 r BDIBSHFEQBSUJDMFJOBVOJGPSNFMFDUSJD 1 xf 5 xi 1 2 vxi 1 vxf t (2.15) field (Chapter 23)

1 1 2 2 xf 5 xi 1 vxit 1 2axt (2.16)

2 2 vxf 5 vxi 1 2ax(xf 2 xi) (2.17) v a

Example 2.7 Carrier Landing AM

A jet lands on an aircraft carrier at a of 140 mi/h (< 63 m/s). (A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the jet and brings it to a stop?

SOLUTION You might have seen movies or television shows in which a jet lands on an aircraft carrier and is brought to rest sur- prisingly fast by an arresting cable. A careful reading of the problem reveals that in addition to being given the initial speed of 63 m/s, we also know that the final speed is zero. Because the acceleration of the jet is assumed constant, we model it as a particle under constant acceleration. We define our x axis as the direction of motion of the jet. Notice that we have no information about the change in position of the jet while it is slowing down. The Kinematics Equations

The kinematic equations describe motion in terms of position, velocity, acceleration, and time. There are 5 that apply specifically to the case of constant acceleration.

To solve kinematics problems: identify what equations apply, then do some algebra to find the quantity you need. The Kinematics Equations For zero acceleration: # » #» ∆r = v t

Always: # » #» ∆r = v avg t

For constant acceleration: #» #» #» vf = vi + a t # » #» 1 #» ∆r = v t + a t2 i 2 # » #» 1 #» ∆r = v t − a t2 f 2 #» #» # » v + v ∆r = i f t 2 2 2 vf ,x = vi,x + 2 ax ∆x In general, when velocity is not constant, this is always true: # » #» ∆r = v avgt

The Kinematics Equations

#» For constant velocity ( a = 0):

# » #» ∆r = v t (1) The Kinematics Equations

#» For constant velocity ( a = 0):

# » #» ∆r = v t (1)

In general, when velocity is not constant, this is always true: # » #» ∆r = v avgt The Kinematics Equations

Always true:

# » #» ∆r = v avgt (2)

This is just a rearrangement of the definition of average velocity: #» average velocity, v avg # » #» ∆r v = avg t # » where ∆r is the that occurs in a time interval t. WALKMC02_0131536311.QXD 12/9/05 3:26 Page 30

30 CHAPTER 2 ONE-DIMENSIONAL KINEMATICS

average acceleration, Equation 2–5, we have

vf - vi aav = = a tf - ti

where the initial and final times may be chosen arbitrarily. For example, let ti = 0 for the initial time, and let vi = v0 denote the velocity at time zero. For the final time and velocity we drop the subscripts to simplify notation; thus we let tf = t and vf = v. With these identifications we have

v - v0 aav = = a t - 0 Therefore,

v - v0 = a t - 0 = at or 1 2 Constant-Acceleration Equation of Motion: Velocity as a Function of Time

v = v0 + at 2–7 Note that Equation 2–7 describes a straight line on a v-versus-t plot. The line crosses the velocity axis at the value v0 and has a slope a, in agreement with the graphical interpretations discussed in the previous section. For example, in curve 2 I of Figure 2–9, the equation of motion is v = v0 + at = 1 m/s + -0.5 m/s t. Also, note that -0.5 m/s2 t has the units m/s2 s = m/s; thus each term in Equation 2–7 has the same dimensions (as it must to be a1 valid physical2 1 equation).2 The Kinematics Equations 1 2 1 21 2 #» If acceleration is constant ( a = const), the velocity-time graph is EXERCISE 2–2 a straight line A ball is thrown straight upward with an initial velocity of +8.2 m/s. If the acceleration of the ball is -9.81 m/s2, what is its velocity after v v a. 0.50 s, and b. 1.0 s? Solution 1 vav= 2 (v0 + v) a. Substituting t = 0.50 s in Equation 2–7 yields

v = 8.2 m/s + -9.81 m/s2 0.50 s = 3.3 m/s v0 b. Similarly, using t = 1.0 s in Equation1 2–7 gives21 2 t O t v = 8.2 m/s + -9.81 m/s2 1.0 s =-1.6 m/s (a) # » 1 21 2 The area under the graph is equal to ∆r, and using the area of a trapezoid: v Next, how far does a particle move in a given time if its acceleration is con- v stant? To answer this question, recall the definition of average velocity: # »  #» #»  ¢x xf - xi vi + vf vav = = ∆r = t (3) ¢t t - t 2 f i vav Using the same identifications given previously for initial and final times, and let- 1 ting xi = x0 and xf = x, we have Figure from Jamesv0 S. Walker “”. t x - x0 O t vav = t - 0 (b) Thus, ▲ FIGURE 2–13 The average velocity x x v t 0 v t (a) When acceleration is constant, the - 0 = av - = av velocity varies linearly with time. As a or 1 2 result, the average velocity, vav, is simply x x v t 2–8 the average of the initial velocity, v0, and = 0 + av the final velocity, v. (b) The velocity Now, Equation 2–8 is fine as it is. In fact, it applies whether the acceleration is curve for nonconstant acceleration is nonlinear. In this case, the average constant or not. A more useful expression, for the case of constant acceleration, is velocity is no longer midway between obtained by writing vav in terms of the initial and final . This can be done the initial and final velocities. by referring to Figure 2–13 (a). Here the velocity changes linearly (since a is The Kinematics Equations

For constant acceleration: #» #» # »  v + v  ∆r = i f t 2

# » #» ∆r From v avg = t , we also have:

#» #» #» v + v v = i f avg 2 (constant acceleration) For constant acceleration:

#» #» #» v (t) = vi + a t (4) #» #» where vi is the velocity at t = 0 and v (t) is the velocity at time t.

The Kinematics Equations

From the definition of acceleration: #» #» d v a = dt We can integrate this expression.

# » t #» ∆v = a dt0 Z0 The Kinematics Equations

From the definition of acceleration: #» #» d v a = dt We can integrate this expression.

# » t #» ∆v = a dt0 Z0 For constant acceleration:

#» #» #» v (t) = vi + a t (4) #» #» where vi is the velocity at t = 0 and v (t) is the velocity at time t. The Kinematics Equations

For constant acceleration: #» #» #» 1 #» r (t) = r + v t + a t2 i i 2

Equivalently,

# » #» 1 #» ∆r = v t + a t2 (5) i 2

How do we know this? The Kinematics Equations

For constant acceleration: #» #» #» 1 #» r (t) = r + v t + a t2 i i 2

Equivalently,

# » #» 1 #» ∆r = v t + a t2 (5) i 2

How do we know this? Integrate!

# » t #» ∆r = v dt0 Z0 The Kinematics Equations

Similarly: #» #» #» 1 #» r (t) = r + v t − a t2 i f 2

Equivalently,

# » #» 1 #» ∆r = v t − a t2 (6) f 2

#» #» #» Show by substituting vi = vf − a t into equation (5). #» #» # »  v + v  ∆r = i f t 2 We could also write this as: v + v  (∆x) = i,x f ,x t 2

where ∆x, vi,x , and vf ,x could each be positive or negative. We do the same for equation (4):

vf ,x = (vi,x + ax t)

Rearranging for t: v − v t = f ,x i,x ax

The Kinematics Equations

The last equation we will derive is a scaler equation. We do the same for equation (4):

vf ,x = (vi,x + ax t)

Rearranging for t: v − v t = f ,x i,x ax

The Kinematics Equations

The last equation we will derive is a scaler equation. #» #» # »  v + v  ∆r = i f t 2 We could also write this as: v + v  (∆x)ˆi = i,x f ,x t ˆi 2

where ∆x, vi,x , and vf ,x could each be positive or negative. Rearranging for t: v − v t = f ,x i,x ax

The Kinematics Equations

The last equation we will derive is a scaler equation. #» #» # »  v + v  ∆r = i f t 2 We could also write this as:   vi,x + vf ,x (∆x)ˆ£i£ = t ˆ£i£ 2

where ∆x, vi,x , and vf ,x could each be positive or negative. We do the same for equation (4):

vf ,x ˆ£i£ = (vi,x + ax t)ˆ£i£ The Kinematics Equations

The last equation we will derive is a scaler equation. #» #» # »  v + v  ∆r = i f t 2 We could also write this as: v + v  (∆x) = i,x f ,x t 2

where ∆x, vi,x , and vf ,x could each be positive or negative. We do the same for equation (4):

vf ,x = (vi,x + ax t)

Rearranging for t: v − v t = f ,x i,x ax The Kinematics Equations

v − v v + v  t = f ,x i,x ; ∆x = i,x f ,x t a 2 Substituting for t in our ∆x equation:

v + v  v − v  ∆x = i,x f ,x f ,x i,x 2 ax 2ax ∆x = (vi,x + vf ,x )(vf ,x − vi,x )

so,

2 2 vf ,x = vi,x + 2 ax ∆x (7) The Kinematics Equations For zero acceleration: # » #» ∆r = v t

Always: # » #» ∆r = v avg t

For constant acceleration: #» #» #» vf = vi + a t # » #» 1 #» ∆r = v t + a t2 i 2 # » #» 1 #» ∆r = v t − a t2 f 2 #» #» # » v + v ∆r = i f t 2 2 2 vf ,x = vi,x + 2 ax ∆x The Kinematics Equations Summary For zero acceleration: # » #» ∆r = v t

Always: # » #» ∆r = v avg t

For constant acceleration: #» #» vf = vi + a t # » #» 1 #» ∆r = v t + a t2 i 2 # » #» 1 #» ∆r = v t − a t2 f 2 #» #» # » v + v ∆r = i f t 2 2 2 vf ,x = vi,x + 2 ax ∆x Summary

• graphs • the kinematic equations

Assignment (now posted on website) Due Jan 16.

Quiz Tomorrow at the start of class.

(Uncollected) Homework Serway & Jewett, • Ch 2, onward from page 49. Conceptual Q: 7; Problems: 25, 29, 33, 39, 83