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Dynamics MDB 2043
Rectilinear Kinematics: Erratic Motion
Guided Learning Activity
May 2016 Semester
Lesson Outcomes
At the end of this lecture you should be able to:
Determine position, velocity, and acceleration of a
particle using graphs.
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Example #1 (continued)
Similarly, the a-t graph can be constructed by finding the slope at various points
along the v-t graph.
when 0 < t < 5 s; a0-5 = dv/dt = d(6t)/dt = 6 m/s2
when 5 < t < 10 s; a5-10 = dv/dt = d(30)/dt = 0 m/s2
a-t graph
a(m/s2)
t(s)
6
5 10
Example #2
Find slopes of the v-t curve and draw the a-t graph.
Find the area under the curve. It is the distance traveled.
Finally, calculate average speed (using basic definitions!).
Given: The v-t graph shown.
Find: The a-t graph, average
speed, and distance
traveled for the 0 - 90 s
interval.
Plan:
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Example #2 (continued)
Solution:
Find the at graph:
For 0 t 30 a = dv/dt = 1.0 m/s
For 30 t 90 a = dv/dt = -0.5 m/s
a-t graph
-0.5
1
a(m/s)
30 90 t(s)
Example #2 (continued)
Now find the distance traveled:
Ds0-30 = v dt = (1/2) (30)2 = 450 m
Ds30-90 = v dt= (1/2) (-0.5)(90)2 + 45(90) (1/2) (-0.5)(30)2 45(30)
= 900 m
s0-90 = 450 + 900 = 1350 m
vavg(0-90) = total distance / time
= 1350 / 90
= 15 m/s
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Example #3
A motorcycle starts from rest and travels on a straight road with a constant
acceleration of 5 m/s2 for 8 sec, after which it maintains a constant speed for 2
sec. Finally it decelerates at 7 m/s2 until it stops. Plot a-t, v-t diagrams for theentire motion.Determine the total distance travelled.
Sketch a-t diagram from the known accelerations, thus
5
-7
)'10(7
)108(0
)80(5
tt
st
st
a
(segment I)
(segment II)
(segment III)
Since dv=adt, the v-t diagram is determined by integratingthe straight line segments of a-t diagram. Using the initial
condition t=0, v=0 for segment I, we have
st 80
tv
dtdv005 tv 5
When t =8 s, v =58= 40m/s. Using this as the initial condition
for segment II, thus
st 108
tv
dtdv8400 smv /40
Similarly, for segment III
'10 tt
tv
dtdv1040
)7( 1107tv
a(m/s2)
t (s)
8 10 t' (=15.71)
a-t Diagram
How can you determine t?
When v=0 (i.e. motorcycle stops)
110'70 t
)71.1510(1107
)108(40
)80(5
stt
st
stt
v
st 71.15'
Thus, the velocity as the function of time can
be expressed as
The total distance travelled (using the area under v-t diagram)
mssss 2.3544071.52
1
4024082
1321
v (m/s)
t (s)
40
8 10 15.71
v-t Diagram
s1 s2 s3
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Example #4
A test projectile is fired horizontally into a viscous liquid with a velocity
v0.The retarding force is proportional to the square of the velocity, so
that the acceleration becomes a=-kv2. Derive expressions for distanceD travelling in the liquid and the corresponding time t required to reduce
the velocity to v0/2.Neglect any vertical motion.
Note the acceleration a is non-constant.
Using dskvadsvdv 2
22
20
0
0
0
0
v
v
v
v
D
kv
dv
kv
vdvds
kkv
v
kk
vD
v
v
693.02ln2ln1ln
0
0
2
0
0
Using2
kv
dt
dva
tv
v
dtkv
dv
0
22
0
0
0
2 1110
0kvvk
t
v
v
Example #5
The acceleration of a particle which moves in thepositive s-direction varies with its position as
shown. If the velocity of the particle is 0.8 m/s
when s=0, determine the velocities v of the particle
when s=0.6 and 1.4 m.
ax (m/s2
)
s (m)
0.4
0.2
0.4 0.8 1.2
Using
22
2
0
22
0
0
0
vvvvdvads
v
v
v
v
s
smadsvv /17.102.04.04.0)4.02.0(2
1)4.04.0(28.02 2
4.1
0
2
0
For x=1.4m
Where v0=0.8 m/sArea under ax-x curve
(0x 1.4)
For x=0.6m
1.40.6
smadsvv /05.12.0)4.03.0(2
1)4.04.0(28.02 2
6.0
0
2
0
Area under ax-x curve(0x 0.6)
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Example #6
The v-s diagram for a testing vehicle travelling on a
straight road is shown. Determine the acceleration
of the vehicle at s=50 m and s=150 m. Draw thea-s diagram.
v (m/s)
s (m)
100 200
8
Since the equations for segments of v-s diagram are given,
we can use ads=vdv to determine a-s diagram.
ms 1000
ssds
ds
ds
dvva 0064.0)08.0()08.0(
ms 200100
28.10064.0)1608.0()1608.0( ssds
dsa
1608.0 sv
sv 08.0
When s=50 m, then (acceleration in segmentI)2
/32.0500064.0 sma
When s=150 m, then (deceleration in segmentII)2
/32.028.11500064.0 sma
a(m/s2)
100 200 s (m)
0.64
-0.64
1. The slope of a v-t graph at any instant represents instantaneous
A) velocity. B) acceleration.
C) position. D) jerk.
2. Displacement of a particle in a given time interval equals the
area under the ___ graph during that time.
A) a-t B) a-s
C) v-t D) s-t
Summary Questions
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3. If a particle starts from rest and
accelerates according to the graph
shown, the particles velocity at
t = 20 s is
A) 200 m/s B) 100 m/s
C) 0 D) 20 m/s
4. The particle in Problem 3 stops moving at t = _______.
A) 10 s B) 20 s
C) 30 s D) 40 s
Summary Questions (continued)
5. If a car has the velocity curve shown, determine the time t
necessary for the car to travel 100 meters.
A) 8 s B) 4 s
C) 10 s D) 6 s
t
v
6 s
75
t
v
6. Select the correct a-t graph for the velocity curve shown.
A) B)
C) D)
a
t
a
t
a
t
a
t
Summary Questions (continued)
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References:
R.C. Hibbeler, Engineering Mechanics: Dynamics,
SI 13th Edition, Prentice-Hall, 2012.