dynamics gla-1c.pdf

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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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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)


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.

dynamics gla-1c.pdf

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