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    10/27/08 1:28 PMChapter D2 Problem 3 Solution

    MECH 2110 - Statics & Dynamics

    Chapter D2 Problem 3 Solution

    Page 27, Engineering Mechanics - Dynamics, 4th Edition, Meriam and Kraige

    Given: Particle moving along a straight line (s-axis) with velocity, v, given in terms of time, t, by:

    v = A - B t + C t3/2 A = 2 m/s, B = 4 m/s2 C = 5 m/s5/2The position of the particle at time t = 0 is given by s0 equal to 3 m.

    Find: The position, s, velocity, v, and acceleration, a, when the time, t, is equal to 3 s.

    0. Observations:1. Interested in motion only without regard to the forces causing the motion, free body diagram is not ofinterest.

    2. The motion is along a single straight line. The motion diagram is simple enough that it can be omitted.

    1. Mechanical System - Particle during the interval from t = 0 to t = 3 s.

    3. Equations

    v = A - B t + C t3/2Relationship between velocity, acceleration and time:

    a = dv/dt = -B + 3/2 C t1/2Relationship between velocity, position, and time:

    ds/dt = v = A - B t + C t3/2

    Separating variables and integrating:()s0

    s dx = ()0t v dt

    s|s0s = ()0

    t { A - B t + C t3/2 } dt

    s - s0 = { A t - B t2/2 + 2/5 C t5/2 }|0

    t

    s = s0 + A t - B t2/2 + 2/5 C t5/2

    4. SolveEvaluating each of the three expressions at t equal to 3 s:

    v = A - B t + C t3/2

    v(t=3s) = 2 m/s - 4 m/s2 3 s + 5 m/s5/2 (3 s)3/2= 15.98 m/s

    a = -B + 3/2 C t1/2

    a(t=3s) = -4 m/s2 + 3/2 5 m/s5/2 (3 s)1/2

    = 8.99 m/s2

    s = s0 + A t - B t2/2 + 2/5 C t5/2

    ( 3 ) 3 2 / 3 4 / 2 1/2 (3 )2 2/5 5 / 5/2 (3 )5/2

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    (t 3 ) 3 2 / 3 4 / 2 1/2 (3 )2 2/5 5 / 5/2 (3 )5/2

    10/27/08 1:28 PMChapter D2 Problem 3 Solution

    = 22.2 m

    ResultsPosition = s(t=3s) = 22.2 m

    Velocity = v(t=3s) = 15.98 m/sAcceleration = a(t=3s) = 8.99 m/s2

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    Sample Problem 2/2

    A particle moves along the x-axis with an initial velocity vx 50 ft/sec at theorigin when t 0. For the first 4 seconds it has no acceleration, and thereafter it

    is acted on by a retarding force which gives it a constant acceleration ax 10

    ft/sec2. Calculate the velocity and the x-coordinate of the particle for the condi-

    tions of t 8 sec and t 12 sec and find the maximum positive x-coordinate

    reached by the particle.

    Solution. The velocity of the particle after t 4 sec is computed from

    and is plotted as shown. At the specified times, the velocities are

    Ans.

    The x-coordinate of the particle at any time greater than 4 seconds is the dis-

    tance traveled during the first 4 seconds plus the distance traveled after the dis-

    continuity in acceleration occurred. Thus,

    For the two specified times,

    Ans.

    The x-coordinate for t 12 sec is less than that for t 8 sec since the motion is

    in the negative x-direction after t 9 sec. The maximum positive x-coordinate is,

    then, the value ofx for t 9 sec which is

    Ans.

    These displacements are seen to be the net positive areas under the v-t graph upto the values oft in question.

    xmax5(92) 90(9) 80 325 ft

    t 12 sec, x5(122) 90(12) 80 280 ftt 8 sec, x5(8

    2

    ) 90(8) 80 320 ft

    ds vdt x 50(4) t

    4

    (90 10t) dt5t2 90t 80 ft

    t 12 sec, vx 90 10(12) 30 ft/sec

    t 8 sec, vx 90 10(8) 10 ft/sec

    dv adt vx

    50

    dvx10

    t

    4

    dt vx 90 10t ft/sec

    28 C ha pt er 2 K in em at ic s o f P a rt ic le s

    Helpful Hints

    Learn to be flexible with symbols.The position coordinate x is just as

    valid as s.

    Note that we integrate to a generaltime t and then substitute specific

    values.

    Show that the total distance traveledby the particle in the 12 sec is 370 ft.

    1

    vx, ft/sec

    t, sec

    10

    50

    00 4 8 12

    30

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    10/27/08 1:29 PMChapter D2 Problem 29 Solution

    MECH 2110 - Statics & Dynamics

    Chapter D2 Problem 29 Solution

    Page 31, Engineering Mechanics - Dynamics, 4th Edition, Meriam and Kraige

    Given: The acceleration of an arrow decreases linearly with distance, s, from a maximum of a 0 equal to

    16,000 ft/s2 upon release of the arrow to zero after a distance of travel L equal to 2 ft.

    Find: The maximum velocity of the arrow.

    0. Observations:1. Interested exclusively in the motion of the arrow independent of the forces producing that motion, thus nofree body diagram is of interest.

    2. The motion is along a single straight line. The motion diagram is simple enough that it can be omitted.

    3. The arrow will travel nearly in a straight line during that brief interval between release of the aroow andthe launch point.

    4. As the arrow continues accelerating until it reaches the distance L, the maximum velocity will occur atthat point.

    1. Mechanical System - Arrow from release until it has traveled a distance L.

    3. EquationsAcceleration, a, is linear with distance, s:

    a = m s + bThe acceleration is known at two points:a(s=0) = -a0/L

    a(s=L) = 0The "intercept", b, is the value of the acceleration at s = 0, that is a 0. The "slope", m, is the change in

    acceleration, a, divided by the change in distance, s, between two points where both of those quantities areknown:

    m = (0 - a0) / ( L - 0 ) = -a0/LThe dependence of the acceleration on position can be expressed as:a = -a0/L s + a0 = a0 { 1 - s/L }

    The relationship between acceleration, velocity, and position is:a = v dv/dsv dv/ds = a0 { 1 - s/L }

    Separating variables and integrating:

    ()0vmax v dv = ()0L a0 { 1 - s/L }ds

    1/2 v2|0vmax = a0 { s - 1/2 s

    2/L }|0L

    1/2 vmax2 = a0 { L - 1/2 L

    2/L }

    vmax2 = a0 L

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    10/27/08 1:29 PMChapter D2 Problem 29 Solution

    4. Solve

    vmax2 = a0 L

    vmax = (a0 L)1/2

    = (16,000 ft/s2 2 ft)1/2= 178.9 ft/s

    ResultsMaximum velocity = vmax = 178.9 ft/s

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    Sample Problem 2/3

    The spring-mounted slider moves in the horizontal guide with negligiblefriction and has a velocity v0 in the s-direction as it crosses the mid-position

    wheres 0 and t 0. The two springs together exert a retarding force to the

    motion of the slider, which gives it an acceleration proportional to the displace-

    ment but oppositely directed and equal to a k2s, where k is constant. (The

    constant is arbitrarily squared for later convenience in the form of the expres-

    sions.) Determine the expressions for the displacement s and velocity v as func-

    tions of the time t.

    Solution I. Since the acceleration is specified in terms of the displacement, the

    differential relation v dv a ds may be integrated. Thus,

    Whens 0, v v0, so that C1 and the velocity becomes

    The plus sign of the radical is taken when v is positive (in the pluss-direction).

    This last expression may be integrated by substitutingv ds/dt. Thus,

    With the requirement oft 0 whens 0, the constant of integration becomes

    C2 0, and we may solve the equation fors so that

    Ans.

    The velocity is v which gives

    Ans.

    Solution II. Since a the given relation may be written at once as

    This is an ordinary linear differential equation of second order for which the so-

    lution is well known and is

    whereA,B, andKare constants. Substitution of this expression into the differ-

    ential equation shows that it satisfies the equation, provided thatKk. The ve-

    locity is v which becomes

    The initial condition v v0 when t 0 requires thatA v0/k, and the condition

    s 0 when t 0 givesB 0. Thus, the solution is

    Ans.sv0

    ksinkt and vv0 coskt

    vAk cosktBk sinkt

    s ,

    sA sinKtB cosKt

    s k2s 0

    s ,

    vv0 coskt

    s ,

    sv0

    k

    sinkt

    dsv0

    2k2s2 dtC2 a constant, or 1k sin1ksv0tC2

    vv02k2s2

    v02/2,

    vdv k2sdsC1 a constant, or v22 k2s2

    2C1

    A rt ic le 2/ 2 R e ct il in ea r M ot io n 29

    s

    This motion is called simple har-monic motion and is characteristic of

    all oscillations where the restoring

    force, and hence the acceleration, is

    proportional to the displacement but

    opposite in sign.

    Again try the definite integral hereas above.

    Helpful Hints

    We have used an indefinite integralhere and evaluated the constant of

    integration. For practice, obtain the

    same results by using the definite

    integral with the appropriate limits.

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    Sample Problem 2/4

    A freighter is moving at a speed of 8 knots when its engines are suddenlystopped. If it takes 10 minutes for the freighter to reduce its speed to 4 knots, de-

    termine and plot the distance s in nautical miles moved by the ship and its speed

    v in knots as functions of the time t during this interval. The deceleration of the

    ship is proportional to the square of its speed, so that a kv2.

    Solution. The speeds and the time are given, so we may substitute the expres-

    sion for acceleration directly into the basic definition a dv/dt and integrate.

    Thus,

    Now we substitute the end limits ofv 4 knots and t hour and get

    Ans.

    The speed is plotted against the time as shown.

    The distance is obtained by substituting the expression for v into the defi-

    nition v ds/dt and integrating. Thus,

    Ans.

    The distance s is also plotted against the time as shown, and we see that the ship

    has moved through a distance s ln ln 2 0.924 mi (nautical) dur-

    ing the 10 minutes.

    43

    (1 66)

    43

    8

    1 6t

    ds

    dt t

    0

    8 dt

    1 6t

    s

    0

    ds s43

    ln (1 6t)

    4 8

    1 8k(1/6) k3

    4mi1 v 8

    1 6t

    16

    1060

    1

    v

    1

    8kt v 8

    1 8kt

    kv2dv

    dt dv

    v2kdt v

    8

    dv

    v2k

    t

    0

    dt

    30 C ha pt er 2 K in em at ic s o f P a rt ic le s

    Helpful Hints

    Recall that one knot is the speed ofone nautical mile (6076 ft) per hour.

    Work directly in the units of nauti-

    cal miles and hours.

    We choose to integrate to a generalvalue ofv and its corresponding time

    t so that we may obtain the variation

    ofv with t.

    00

    2

    4

    6

    8

    2t, min

    v,

    knots

    4 6 8 10

    0

    1.0

    0.8

    0.6

    0.4

    0.2

    02

    t, min

    s,mi

    (nautical)

    4 6 8 10

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    Sample Problem 2/5

    The curvilinear motion of a particle is defined by vx 50 16t and y 100 4t2, where vx is in meters per second, y is in meters, and t is in seconds.

    It is also known that x 0 when t 0. Plot the path of the particle and deter-

    mine its velocity and acceleration when the position y 0 is reached.

    Solution. The x-coordinate is obtained by integrating the expression for vx,

    and the x-component of the acceleration is obtained by differentiatingvx. Thus,

    The y-components of velocity and acceleration are

    We now calculate corresponding values ofx and y for various values oft and

    plot x againsty to obtain the path as shown.

    When y 0, 0 100 4t2, so t 5 s. For this value of the time, we have

    The velocity and acceleration components and their resultants are shown on the

    separate diagrams for point A, where y 0. Thus, for this condition we may

    write

    Ans.

    Ans.a16i 8j m/s2

    v30i 40j m/s

    a(16)2 (8)2 17.89 m/s2

    v(30)2 (40)2 50 m/s

    vy8(5) 40 m/s

    vx 50 16(5)30 m/s

    ayd

    dt(8t) ay8 m/s

    2[ayvy]

    vy

    d

    dt (100 4t2

    )

    vy8t m/s[vyy]

    axd

    dt(50 16t) ax16 m/s

    2[axvx]

    x

    0

    dxt

    0

    (50 16t) dt x 50t 8t2 m dx vxdt

    46 C ha pt er 2 K in em at ic s o f P a rt ic le s

    Helpful Hint

    We observe that the velocity vector lies

    along the tangent to the path as it

    should, but that the acceleration vector

    is not tangent to the path. Note espe-cially that the acceleration vector has a

    component that points toward the in-

    side of the curved path. We concluded

    from our diagram in Fig. 2/5 that it is

    impossible for the acceleration to have a

    component that points toward the out-

    side of the curve.

    100

    80

    60

    40

    20

    00 20 40

    t = 5 s

    1 2

    3

    4

    A

    t = 0

    y,m

    x,m60 80

    = 53.1

    vx= 30 m/s

    vy= 40 m/s v= 50 m/s

    a = 17.89 m/s2 ay = 8 m/s

    2

    ax = 16 m/s2

    Path Path

    A A

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    Problem 12-11

    The acceleration of a particle as it moves along a straight line is given by a = b t + c. If s = s0 and v

    = v0

    when t= 0, determine the particle's velocity and position when t= t1. Also, determine the total

    distance the particle travels during this time period.

    Given: b 2m

    s3:= c 1

    m

    s2:= s0 1m:= v0 2

    m

    s:=t1 6s:=

    Solution:

    v0

    v

    v1

    d

    0

    t

    tb t c+( )

    d= v v0

    b t2

    2+ c t+=

    s0

    s

    s1

    d

    0

    t

    tv0

    b t2

    2+ c t+

    d= s s0

    v0

    t+b

    6t3

    +c

    2t2

    +=

    When t = t1 v1

    v0

    b t1

    2

    2+ c t1+:= v1 32.00

    m

    s=

    s1

    s0

    v0

    t1+

    b

    6t1

    3+

    c

    2t1

    2+:= s1 67.00 m=

    The total distance traveled depends on whether the particle turned around or not. To tell we will plot

    the velocity and see if it is zero at any point in the interval

    t 0 0.01t1, t1..:= v t( ) v0b t

    2

    2+c t+:= If v never goes to zero then

    0 2 4 60

    20

    40

    v t( )

    t

    d s1

    s0:= d 66.00 m=

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    Problem 12-15

    A particle travels to the right along a straight line with a velocity vp = a / (b + sp ). Determine its position

    when t = t1 ifsp = sp0when t = 0.

    Given: a 5m

    2

    s:= b 4 m:= sp0 5m:= t1 6s:=

    Solution:dsp

    dt

    a

    b sp+=

    sp0

    sp

    spb sp+( )

    d

    o

    t

    ta

    d=

    b spsp

    2

    2

    + b sp0sp0

    2

    2

    a t=

    Guess sp1 1m:=

    Given b sp1sp1

    2

    2+ b sp0

    sp02

    2 a t1= sp1 Find sp1( ):= sp1 7.87 m=

    Problem 12-39

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    Problem 12-39

    A freight train starts from rest and travels with a constant acceleration a. After a time t1 it maintains

    a constant speed so that when t = t2 it has traveled a distance d. Determine the time t1 and draw the

    v-tgraph for the motion.

    Given : a 0.5ft

    s2

    := t2

    160s:= d 2000ft:=

    Solution : Guesses t1 80s:= vmax 30

    ft

    s:=

    Given vmax a t1= d1

    2a t1

    2 vmax t2 t1( )+=

    vmax

    t1

    Find v

    maxt1

    ,

    ( ):= v

    max 13.67

    ft

    s= t

    1 27.34s=

    The equations of motion

    ta 0 0.01 t1, t1..:= tc t1 1.01 t1, t2..:=

    va ta( ) a ta sft

    :=vc tc( ) vmax

    s

    ft:=

    The plot

    0 20 40 60 80 100 120 140 1600

    10

    20

    Time in seconds

    Velocityinft/s

    va

    ta( )

    vc

    tc( )

    ta

    tc,

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    Problem 12-44

    A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v-t

    graph. Determine the motorcycle's acceleration and position when t= t4 and t = t5.

    s 1.00 s=

    Given:

    v0 5

    m

    s:=

    t1 4s:=

    t2 10s:=

    t3 15s:=

    t4 8s:=

    t5 12s:=

    Solution: At t t4

    := Because t1 t4< t2< then a4dv

    dt= 0=

    s4

    1

    2v0 t1 t4 t1( ) v0+:= s4 30.00 m=

    At t t5:= Because t2 t5< t3< then

    a5

    v0

    t3 t2:= a5 1.00

    m

    s2

    =

    s5

    1

    2t1 v0

    v0 t2 t1( )+

    1

    2v0 t3 t2

    ( )+1

    2

    t3

    t5

    t3 t2 v0

    t3 t5( ):=

    s5 48.00 m=

    P bl 12 48

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    Problem 12-48

    The velocity of a car is plotted as shown. Determine the total distance the car moves until it stops at

    time t = t2. Construct the a-t graph.

    Given :

    v 10m

    s

    :=

    t1 40s:=

    t2 80s:=

    Solution :

    d v t11

    2v t2 t1( )+:= d 600.00 m=

    The graph

    1 0 0.01 t1, t1..:= a1 1( ) 0s2

    m

    :=

    2

    t1 1.01 t1, t2..:= a2 2( )

    v

    t2 t1

    s2

    m:=

    0 10 20 30 40 50 60 70 800.4

    0.2

    0

    0.2

    Time in seconds

    Accelerationinm/s^2

    a1

    1( )

    a2

    2( )

    1

    2

    ,

    Problem 12-75

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    The path of a particle is defined byy2 = 4kx, and the component of velocity along they axis is

    vy = ct, where both kand c are constants. Determine thex andy components of acceleration.

    Solution :

    y2 4 k x=

    2 y vy 4 k vx=

    2 vy2

    2 y ay+ 4 k ax=

    vy c t=

    ay c=

    2 c t( )2

    2 y c+ 4 k ax=

    axc

    2 k

    y c t2

    +( )=