Mechanics the ``Kinematics Equations''

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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 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 44 Chapter 2 Motion in VelocityOne Dimension vs. Time Graphs

Figure 2.15 Velocity versus time The area of the shaded rectangle for a particle moving along the is equal to the displacement in x axis. The total area under the vx the time interval ⌬t . curve is the total displacement of n the particle. vxn,avg

t t i t f

⌬t n

tf under the curve in the velocity–time∆x = lim graph.vn ∆Therefore,t = v dt in the limit n S `, or Dt S 0, ∆t→0 n n ti the displacement is X Z where ∆x represents the change in position (displacement) in the Dx 5 lim vxn,avg Dtn (2.18) time interval t to t . Dt S 0 i f n an

If we know the vx–t graph for motion along a straight line, we can obtain the displacement during any time interval by measuring the area under the curve corre- sponding to that time interval. The limit of the sum shown in Equation 2.18 is called a definite integral and is written

Definite integral lim vxn,avg Dtn 5 vx t dt (2.19) Dtn S 0 a 3 n ti vx 1 2 vx ϭ vxi ϭ constant where vx(t) denotes the velocity at any time t. If the explicit functional form of vx(t) is known and the limits are given, the integral can be evaluated. Sometimes the ⌬ t v –t graph for a moving particle has a shape much simpler than that shown in Fig- vxi x ure 2.15. For example, suppose an object is described with the particle under constant velocity model. In this case, the v –t graph is a horizontal line as in Figure 2.16 v x xi and the displacement of the particle during the time interval Dt is simply the area of the shaded rectangle: t ti tf Dx 5 vxi Dt (when vx 5 vxi 5 constant)

Figure 2.16 The velocity–time curve for a particle moving with Kinematic Equations constant velocity vxi. The displacement of the particle during the We now use the defining equations for acceleration and velocity to derive two of time interval tf 2 ti is equal to the our kinematic equations, Equations 2.13 and 2.16. area of the shaded rectangle. The defining equation for acceleration (Eq. 2.10), dv a 5 x x dt may be written as dvx 5 ax dt or, in terms of an integral (or antiderivative), as t 2 5 vxf vxi 3 ax dt 0

For the special case in which the acceleration is constant, ax can be removed from the integral to give t 2 5 5 2 5 vxf vxi ax 3 dt ax t 0 axt (2.20) 0 which is Equation 2.13 in the particle under constant1 2 acceleration model. Now let us consider the defining equation for velocity (Eq. 2.5): dx v 5 x dt Velocity vs. Time Graphs 44 Chapter 2 Motion in One Dimension

t t i t f

⌬t n Or we can write t 0 under the curve in the velocity–timex(t graph.) = vTherefore,dt in the limit n S `, or Dtn S 0, the displacement is Zti if the object starts at position x = 0 when t = ti . Dx 5 lim vxn,avg Dtn (2.18) Dt S 0 n an

What does the area under an acceleration-time graph represent? 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 accelerations 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 speed 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

This is a set of very useful equations for the case of constant acceleration.

Very often in real life accelerations are not constant, but one important case where acceleration is constant1 is for falling objects.

1At least nearly constant, neglecting air resistance and small variations in g near the Earth’s surface. The Kinematics Equations

For constant acceleration:

v = v0 + at 1 ∆x = v t + at2 0 2 1 ∆x = vt − at2 2 v + v ∆x = 0 t 2 2 2 v = v0 + 2 a ∆x

For zero acceleration:

∆x = vt However, we are looking only at 1-dimension for the moment. We know the displacement will be either in the positive or negative x direction (±i direction).

What we can do, is write this equation instead as a scalar equation by factoring out the unit vectors from each side:

∆x =

In that last expression, ∆x and v are the signed magnitudes of the ∆x and v vectors.

That is, ∆x and v can be positive or negative.

Vector Equations vs Scalar Equations I will write the kinematics equations in vector form, for example:

∆x = vt

since displacement and velocity are vector quantities. In that last expression, ∆x and v are the signed magnitudes of the ∆x and v vectors.

That is, ∆x and v can be positive or negative.

Vector Equations vs Scalar Equations I will write the kinematics equations in vector form, for example:

∆x = vt

since displacement and velocity are vector quantities.

However, we are looking only at 1-dimension for the moment. We know the displacement will be either in the positive or negative x direction (±i direction).

What we can do, is write this equation instead as a scalar equation by factoring out the unit vectors from each side:

∆xi = (vi)t In that last expression, ∆x and v are the signed magnitudes of the ∆x and v vectors.

That is, ∆x and v can be positive or negative.

Vector Equations vs Scalar Equations I will write the kinematics equations in vector form, for example:

∆x = vt

since displacement and velocity are vector quantities.

However, we are looking only at 1-dimension for the moment. We know the displacement will be either in the positive or negative x direction (±i direction).

What we can do, is write this equation instead as a scalar equation by factoring out the unit vectors from each side:

∆xi£ = (vi£)t Vector Equations vs Scalar Equations I will write the kinematics equations in vector form, for example:

∆x = vt

since displacement and velocity are vector quantities.

However, we are looking only at 1-dimension for the moment. We know the displacement will be either in the positive or negative x direction (±i direction).

What we can do, is write this equation instead as a scalar equation by factoring out the unit vectors from each side:

∆x = vt

In that last expression, ∆x and v are the signed magnitudes of the ∆x and v vectors.

That is, ∆x and v can be positive or negative. v−v0 For constant acceleration aavg = a, so a = t

v(t) = v0 + at (1)

where v0 is the velocity at t = 0 and v(t) is the velocity at time t.

The Kinematics Equations: the “no-displacement” equation From the deﬁnition of average acceleration: ∆v a = avg ∆t

∆v = v − v0

and starting at time t = 0 means ∆t = t − 0 = t. The Kinematics Equations: the “no-displacement” equation From the deﬁnition of average acceleration: ∆v a = avg ∆t

∆v = v − v0

and starting at time t = 0 means ∆t = t − 0 = t.

v−v0 For constant acceleration aavg = a, so a = t

v(t) = v0 + at (1)

where v0 is the velocity at t = 0 and v(t) is the velocity at time 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 1 2 1 21 2 Average Velocity

IF the acceleration of an object is constant, then the velocity-timeEXERCISE 2–2 graph is 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 1 21 2 vavg = (v0 + v) v 2 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 vav = = ¢t tf - ti

vav Using the same identifications given previously for initial and final times, and letting xi = x0 and xf = x, we have v0 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 velocities. 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: the “no-acceleration” equation

From the deﬁnition of average velocity: ∆x v = avg t

v0+v and vavg = 2

Equating them, and multiplying by t:

v + v ∆x = 0 t (2) 2 The Kinematics Equations: the “no-ﬁnal-velocity” equation Using the equation v + v ∆x = 0 t 2 and the equation v = v0 + at replace v in the ﬁrst equation.

v + (v + at) ∆x = 0 0 t 2 1 = v t + at2 0 2 For constant acceleration:

1 x(t) = x + v t + at2 (3) 0 0 2 WALKMC02_0131536311.QXD 12/9/05 3:27 Page 33

2–5 MOTION WITH CONSTANT ACCELERATION 33

EXAMPLE 2–6 Put the Pedal to the Metal A drag racer starts from rest and accelerates at 7.40 m/s2. How far has it traveled in (a) 1.00 s, (b) 2.00 s, (c) 3.00 s? Picture the Problem We set up a coordinate system in which the drag racerSketch: starts at the origin and accelerates in the positive x direction. With 2 this choice, it follows that x0 = 0 and a =+7.40 m/s . Also, since the racer starts from rest, its initial velocity is zero, t = 0.00 t = 1.00 s t = 2.00 s t = 3.00 s v0 = 0. Incidentally, the positions of the racer in the sketch have been drawn to scale. Strategy Since this problem gives the acceleration, which is constant, O x and asks for a relationship between position and time, we use Equation 2–11. Solution Part (a) 2 1 2 1 2 1 2 1. Evaluate Equation 2–11 with a = 7.40 m/s and t = 1.00 s: x = x0 + v0t + 2 at = 0 + 0 + 2 at = 2 at 1 2 2 x = 2 7.40 m/s 1.00 s = 3.70 m Part (b) 1 21 2 1 2 2. From the calculation in part (a), Equation 2–11 reduces x = 2 at 1 2 1 2 to x = 2 at in this situation. Evaluate x = 2 at at t = 2.00 s: 1 2 2 = 2 7.40 m/s 2.00 s = 14.8 m = 4 3.70 m Part (c) 1 21 2 1 2 1 2 3. Repeat with t = 3.00 s: x = 2 at 1 2 2 = 2 7.40 m/s 3.00 s = 33.3 m = 9 3.70 m Insight 1 21 2 1 2 This Example illustrates one of the key features of accelerated motion—position does not change uniformly with time when an object accelerates. In this case, the distance traveled in the first two seconds is 4 times the distance traveled in the first second, and the distance traveled in the first three seconds is 9 times the distance traveled in the first second. This kind of behavior is a direct result of the fact that x depends on t2 when the acceleration is nonzero.

Practice Problem In one second the racer travels 3.70 m. How long does it take for the racer to travel 2 3.70 m = 7.40 m? [Answer: t = 2 s = 1.41 s] 1 2 Some related homework2 problems: Problem 44, Problem 59

Figure 2–15 shows a graph of x versus t for Example 2–6. Notice the parabolic 1 2 40

shape of the x-versus-t curve, which is due to the 2 at term, and is characteristic of (m) x constant acceleration. In particular, if acceleration is positive a 7 0 , then a plot 30 of x-versus-t curves upward; if acceleration is negative a 6 0 , a plot of x-versus- 20 t curves downward. The greater the magnitude of a, the greater1 the 2curvature. In

Position, 10 1 2 2 contrast, if a particle moves with constant velocity a = 0 the t dependence van- ishes, and the x-versus-t plot is a straight line. O 0.5 11.5 2 32.5 3.5 Our ﬁnal equation of motion with constant acceleration1 2 relates velocity to Time, t (s) position. We start by solving for the time, t, in Equation 2–7: ▲ FIGURE 2–15 Position versus time for Example 2–6 v - v0 v = v + at or t = 0 a The upward-curving, parabolic shape of this x-versus-t plot indicates a positive, Next, we substitute this result into Equation 2–10, thus eliminating t: constant acceleration. The dots on the curve show the position of the drag racer 1 1 v - v0 in Example 2–6 at the times 1.00 s, 2.00 s, x = x0 + v0 + v t = x0 + v0 + v 2 2 a and 3.00 s.

1 2 2 12 2a 2 b 2 Noting that v0 + v v - v0 = v0v - v0 + v - vv0 = v - v0 , we have 2 2 1 21 2 v - v0 x = x0 + 2a

Example 2-6, page 34

A drag racer starts from rest and accelerates at 7.40 m/s2. How far has it traveled in (a) 1.00 s, (b) 2.00 s, (c) 3.00 s?

1Walker “Physics”, pg 33. WALKMC02_0131536311.QXD 12/9/05 3:27 Page 33

Example 2-6, page 34

2–5 MOTION WITH CONSTANT ACCELERATION 33

EXAMPLE 2–6 Put the Pedal to the Metal 2 A2 drag racer starts from rest and accelerates at 7.40 m/s . How A drag racer starts from rest and accelerates at 7.40 m/sfar. hasHow itfar traveled has it traveled in (a) in (a) 1.001.00 s, s, (b) 2.00 2.00 s, (c) s,3.00 (c) s? 3.00 s? Picture the Problem We set up a coordinate system in which the drag racerSketch: starts at the origin and accelerates in the positive x direction. With 2 this choice, it follows that x0 = 0 and a =+7.40 m/s . Also, since the racer starts from rest, its initial velocity is zero, t = 0.00 t = 1.00 s t = 2.00 s t = 3.00 s v0 = 0. Incidentally, the positions of the racer in the sketch have been drawn to scale. Strategy Since this problem gives the acceleration, which is constant, O x and asks for a relationship between position and time, we use Equation 2–11. Solution Part (a) 2 1 2 1 2 1 2 1. Evaluate Equation 2–11 with a = 7.40 m/s and t = 1.00 s: x = x0 + v0t + 2 at = 0 + 0 + 2 at = 2 at 1 2 2 x = 2 7.40 m/s 1.00 s = 3.70 m 1Walker “Physics”, pg 33. Part (b) 1 21 2 1 2 2. From the calculation in part (a), Equation 2–11 reduces x = 2 at 1 2 1 2 to x = 2 at in this situation. Evaluate x = 2 at at t = 2.00 s: 1 2 2 = 2 7.40 m/s 2.00 s = 14.8 m = 4 3.70 m Part (c) 1 21 2 1 2 1 2 3. Repeat with t = 3.00 s: x = 2 at 1 2 2 = 2 7.40 m/s 3.00 s = 33.3 m = 9 3.70 m Insight 1 21 2 1 2 This Example illustrates one of the key features of accelerated motion—position does not change uniformly with time when an object accelerates. In this case, the distance traveled in the first two seconds is 4 times the distance traveled in the first second, and the distance traveled in the first three seconds is 9 times the distance traveled in the first second. This kind of behavior is a direct result of the fact that x depends on t2 when the acceleration is nonzero.

Figure 2–15 shows a graph of x versus t for Example 2–6. Notice the parabolic 1 2 40 shape of the x-versus-t curve, which is due to the 2 at term, and is characteristic of (m) x constant acceleration. In particular, if acceleration is positive a 7 0 , then a plot 30 of x-versus-t curves upward; if acceleration is negative a 6 0 , a plot of x-versus- 20 t curves downward. The greater the magnitude of a, the greater1 the 2curvature. In

1 2 2 12 2a 2 b 2 Noting that v0 + v v - v0 = v0v - v0 + v - vv0 = v - v0 , we have 2 2 1 21 2 v - v0 x = x0 + 2a Using the Kinematics Equations

Process: 1 Identify which quantity we need to ﬁnd and which ones we are given. 2 Is there a quantity that we are not given and are not asked for? 1 If so, use the equation that does not include that quantity. 2 If there is not, more that one kinematics equation may be required or there may be several equivalent approaches. 3 Input known quantities and solve. Strategy: Use equation 1 ∆x = x(t) − x = v t + at2 0 0 2 (a) Letting the x-direction in my sketch be positive:

0 1 ∆x = v t + at2 0 2 1 = (7.40 m/s2)(1.00 s)2 2 = 3.70 m

Example 2-6, page 34 A drag racer starts from rest and accelerates at 7.40 m/s2. How far has it traveled in (a) 1.00 s, (b) 2.00 s, (c) 3.00 s?

2 Given: a = 7.40 m/s , v0 = 0 m/s, t. Asked for: ∆x

1Walker “Physics”, pg 33. Example 2-6, page 34 A drag racer starts from rest and accelerates at 7.40 m/s2. How far has it traveled in (a) 1.00 s, (b) 2.00 s, (c) 3.00 s?

2 Given: a = 7.40 m/s , v0 = 0 m/s, t. Asked for: ∆x

Strategy: Use equation 1 ∆x = x(t) − x = v t + at2 0 0 2 (a) Letting the x-direction in my sketch be positive:

0 1 ∆x = v t + at2 0 2 1 = (7.40 m/s2)(1.00 s)2 2 = 3.70 m

1Walker “Physics”, pg 33. Example 2-6, page 34 A drag racer starts from rest and accelerates at 7.40 m/s2. How far has it traveled in (a) 1.00 s, (b) 2.00 s, (c) 3.00 s?

Use the same equation for (b), (c) 1 ∆x = x(t) − x = v t + at2 0 0 2 1 (b) ∆x = at2 2 1 = (7.40 m/s2)(2.00 s)2 2 = 14.8 m

1 (c) ∆x = at2 2 1 = (7.40 m/s2)(3.00 s)2 2 = 33.3 m Example 2-6, page 34

(a) 3.70 m, (b) 14.8 m, (c) 33.3 m

Analysis: It makes sense that the distances covered by the car increases with time, and it makes sense that the distance covered in each one second interval is greater than the distance covered in the previous interval since the car is still accelerating.

The distance covered over 3 seconds is 9 times the distance covered in 1 second.

The car covers ∼ 30 m in 3 s, giving an average speed of ∼ 10 m/s. We know cars can go much faster than this, so the answer is not unreasonable.

1Walker “Physics”, pg 33. The Kinematics Equations: the “no-initial-velocity” equation

Exercise for you: try to prove this equation.

For constant acceleration:

1 x(t) = x + vt − at2 (4) 0 2 v + v ∆x = 0 t 2 We could also write this as: v + v (∆x) = 0 t 2

where ∆x, vi , and vf could each be positive or negative. We do the same for equation (1):

v = (v0 + at)

Rearranging for t: v − v t = 0 a

The Kinematics Equations: the “no-time” equation

The last equation we will derive is a scalar equation. We do the same for equation (1):

v = (v0 + at)

Rearranging for t: v − v t = 0 a

The Kinematics Equations: the “no-time” equation

The last equation we will derive is a scalar equation.

v + v ∆x = 0 t 2 We could also write this as: v + v (∆x) i = 0 t i 2

where ∆x, vi , and vf could each be positive or negative. Rearranging for t: v − v t = 0 a

The Kinematics Equations: the “no-time” equation

The last equation we will derive is a scalar equation.

v + v ∆x = 0 t 2 We could also write this as: v0 + v (∆x) i£ = t i£ 2

where ∆x, vi , and vf could each be positive or negative. We do the same for equation (1):

v i£ = (v0 + at) i£ The Kinematics Equations: the “no-time” equation

The last equation we will derive is a scalar equation.

v + v ∆x = 0 t 2 We could also write this as: v + v (∆x) = 0 t 2

where ∆x, vi , and vf could each be positive or negative. We do the same for equation (1):

v = (v0 + at)

Rearranging for t: v − v t = 0 a The Kinematics Equations: the “no-time” equation

v − v v + v t = 0 ; ∆x = 0 t a 2

Substituting for t in our ∆x equation:

v + v v − v ∆x = 0 0 2 a

2a∆x = (v0 + v)(v − v0)

so,

2 2 v = v0 + 2 a ∆x (5) The Kinematics Equations Summary

For constant acceleration:

v = v0 + at 1 ∆x = v t + at2 0 2 1 ∆x = vt − at2 2 v + v ∆x = 0 t 2 2 2 v = v0 + 2 a ∆x

For zero acceleration: x = vt Summary

• acceleration • the “kinematics equations” • applying the kinematics equations

Homework • previous: Ch 2, Questions: 1, 2, 4, 5; Problems: 19, 21, 90 • new: Ch 2, Problems: 23, 25, 31, 35, 41, 69, (73 - can wait to do)