d1 1ex rectilinear
TRANSCRIPT
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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
+( )=