First Class-Second Course
Engineering Mechanics
Dynamic
2018-2019
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Examples
Example(1):The car shown in figure moves in a straight line such that for a short time its velocity is
defined by v=(3t2+2t) m/s, where t is in seconds. Determine its position and acceleration
when t=3s. When t=0, s=0.
Solution:𝐏𝐨𝐬𝐢𝐭𝐢𝐨𝐧 →+ 𝑣 =
𝑠
𝑑𝑠
= (3𝑡 2 + 2𝑡)
𝑑𝑡
𝑡
∫ 𝑑𝑆 = ∫ (3𝑡 2 + 2𝑡)𝑑𝑡
0
0
|𝑆|0𝑆 = |𝑡 3 + 𝑡 2 |𝑡0
𝑆 = 𝑡3 + 𝑡2
When t=3 s,
𝑺 = 𝟑𝟑 + 𝟑𝟐 = 𝟑𝟔 𝐦
𝐀𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧 →+ 𝑎 =
𝑑𝑣
𝑑
= (3𝑡 2 + 2𝑡) = 6𝑡 + 2
𝑑𝑡 𝑑𝑡
When t=3s,
𝐚 = 𝟔(𝟑) + 𝟐 = 𝟐𝟎
𝐦
→
𝐬𝟐
171
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
Dynamic
2018-2019
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(2):A small projectile is fired vertically down ward into a fluid medium with an initial velocity of 60m/s. due to
the drag resistance of the fluid the projectile experiences a deceleration of a=(- 0.4v3)m/s2, where v is in m/s.
Determine the projectile velocity and position after 4s it is fired.
Solution:Since the motion is downward, the position coordinate is positive downward with origin located at O.
Velocity +↓
a=
𝑑𝑣
𝑑𝑡
= −0.4𝑣 3
𝑣
𝑡
𝑑𝑣
∫
= ∫ 𝑑𝑡
3
60 −0.4𝑣
0
1
1 1 𝑣
|
( ) | =𝑡−0
−0.4 −2 𝑣 2 60
1 1
1
[ 2−
]=𝑡
0.8 𝑣
(60)2
𝑣 = {[
−1/2
1
+
0.8𝑡]
} 𝑚/𝑠
(60)2
When t=4s, v=0.559m/s↓
Position +↓
v=
dS
dt
1
= [(60)2 + 0.8t]
−1/2
−1/2
1
∫ 𝑑𝑆 = ∫ [
+ 0.8𝑡]
𝑑𝑡
2
0
0 (60)
𝑆
𝑡
1 𝑡
1
2
2
2
1
1
1
1
𝑆=
|[
+ 0.8𝑡] | =
{[
+ 0.8𝑡] − } 𝑚 𝑤ℎ𝑒𝑛 𝑡 = 4𝑠 𝑆 = 4.43𝑚
2
2
0.8 (60)
0.4 (60)
60
0
172
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
Dynamic
2018-2019
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(3):A particle travels along a straight line with a velocity v=(12-3t2)m/s, where t is in
seconds.When t=1s, the particle is located 10m to the left of the origin. Determine the
acceleration when t=4s, the displacement from t=0 to t=10s, and the distance the particle
travels during this time period.
Solution:v = 12 − 3t 2
a=
dv
= |−6t|t=4 = −24m/s 2
dt
S
t
t
∫ dS = ∫ vdt = ∫ (12 − 3t 2 )dt
−10
1
1
S + 10 = 12t − t 3 − 11
S = 12t − t 3 − 21
When t=0
S=-21m
When t=10 S=-901m
∆S = −901 − (−21) = −880m
From velocity equation
When t=2s
v=0
When t=2s
S=12(2)-23-21=-5
ST = (21 − 5) + (901 − 5) = 912m
173
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
Dynamic
2018-2019
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(4):A particle moves along a horizontal path with a velocity of v=(3t 2-6t)m/s, where t is the time
in seconds. If it is initially located at the origin O, determine the distance traveled in 3.5s,
and the particle average velocity and average speed during the time interval.
Solution:𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞 +→ dS = vdt = (3t 2 − 6t)dt
S
t
∫ dS = ∫ (3t 2 − 6t)dt
0
0
S = (t 3 − 3t 2 )m
When t=0s
S=0
When t=2s
S=-4m
When t=3.5s S=6.125m
The distance traveled in 3.5s is
ST=4+4+6.125=14.125m
Velocity
The displacement from t=0s to t=3.5s is
∆S = 6.125 − 0 = 6.125m
∆𝑆
6.125
=
= 1.75𝑚/𝑠
∆𝑡 3.5 − 0
ST 14.125
(vsp )avg =
=
= 4.04m/s
∆t 3.5 − 0
𝑣𝑎𝑣𝑔 =
174
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
Dynamic
2018-2019
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(5):The position of a particle along a straight line path is defined by S=(t 3-6t2-15t+7)m, where t
is in seconds. Determine the total distance traveled when t=10s. What are the particle
average velocity, average speed, and the instantaneous velocity and acceleration at this time.
Solution:𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲
S = t 3 − 6t 2 − 15t + 7
𝑑𝑆
= 3𝑡 2 − 12𝑡 − 15
𝑑𝑡
When t=10s v=165m/s
𝑣=
𝐀𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧
a=
dv
= 6t − 12
dt
When t=10s a=48m/s2
When v=0
0 = 3𝑡 2 − 12𝑡 − 15
When t=0
→ 𝑡 = 5𝑠
S=7m
When t=5s S=-93m
When t=10s S=257m
Total distance traveled
ST=7+93+93+257
∆𝑆 257 − 7
=
= 20𝑚/𝑠
∆𝑡
10 − 0
𝑆𝑇 450
(𝑣𝑠𝑝 )𝑎𝑣𝑔 =
=
= 45𝑚/𝑠
∆𝑡
10
𝑣𝑎𝑣𝑔 =
175
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬
First Class-Second Course
Engineering Mechanics
Dynamic
2018-2019
Highway and Transport. Engineering Department
College of Engineering
Mustansiriyah University
Lec.Rana Hashim
Example(6):The rocket travels in a straight line with acceleration motion such that a=21-12S2.The rocket
starts with no initial velocity at the position S=0.Determine the velocity when S=1.5m, the
position where the velocity is again zero and the position where the velocity is maximum.
‫عندما يصل الصاروخ اقصى ارتفاع له فالسرعة =صفر‬:‫مالحظة‬
Solution:𝑎 = 21 − 12𝑆 2 = 𝑣
v
𝑑𝑣
𝑑𝑠
S
∫ vdv = ∫ (21 − 12S 2 )dS
0
0
v
S
v2
S3
| | = |21S − 12 |
2 0
3 0
v2
= 21S − 4S 3
2
v = √42S − 8S 3
velocity when S = 1.5m
= √42(1.5) − 8(1.53 ) = ∓6𝑚/𝑠
The position where v=0
√42S − 8S 3 = 0
42S − 8S 3 = 0
S(42 − 8S 2 ) = 0 → S = 0, S = ∓2.29m
The position where the v is max
𝑎𝑡 𝑣𝑚𝑎𝑥 → 𝑎 = 0
21 − 12𝑆 2 = 0
𝑆=√
21
= ∓1.32𝑚
12
176
‫رنا هاشم‬.‫م‬
‫قسم الطرق والنقل‬
‫الجامعة المستنصرية‬. ‫كلية الهندسة‬