Dynamics
Thermodynamics–I
Dr. Ihab Alsurakji
Chapter 12
Kinematics of a Particle
Objectives
o To introduce the concepts of position, displacement, velocity, and acceleration.
o To study particle motion along a straight line.
Areas of Mechanics
Areas of Mechanics
Introduction
• Scalars (e.g.)
• Vectors (e.g.)
– distance
– speed
– mass
– temperature
– pure numbers
– time
– pressure
– area, volume
– charge
– energy
– displacement
– velocity
– acceleration
– force
– weight (force)
– momentum
Vectors
• Can be represented by an arrow (called the “vector”)
• Length of the vector represents the magnitude of the quantity
• Symbols for vectors:
- (e.g. force) 𝐹Ԧ ,or F (bold type)
Distance and Displacement
Distance is a scalar quantity that refers to "how
much ground an object has covered" during its
motion.
Displacement is a vector quantity that refers to
"how far out of place an object is"; it is the
object's overall change in position.
7 m = Distance
5 m = Displacement
Distance and Displacement
Distance
Displacement
➢ Total length of path travelled ➢ Direct distance
➢ Must be greater than (or ➢ Shortest distance between
equal to) magnitude of
two points
displacement
➢ Distance between start and
➢ Only equal if path is straight
finish points
➢ Symbol “d “
➢ Can describe with only one
direction
➢ Symbol “s”
𝑠 = 𝑥𝑓𝑖𝑛𝑎𝑙 − 𝑥𝑖𝑛𝑖𝑡𝑖𝑎𝑙
7 m = Distance
5 m = Displacement
Average Speed and Average Velocity
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑑
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =
=
𝑡𝑖𝑚𝑒
𝑡
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑠
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
=
𝑡𝑖𝑚𝑒
𝑡
𝑠 ∆𝑥 𝑥 − 𝑥𝑜
𝑣= =
=
𝑡
𝑡
𝑡
Section 12.2 : Rectilinear Kinematics
❖Rectilinear : Straight line motion
❖Kinematics : Study the geometry of the motion dealing with s, v, a.
❖Rectilinear Kinematics : To identify at any given instant, the particle’s
position, velocity, and acceleration.
All objects such as rockets, projectiles, or vehicles will be considered as
particles “has negligible size and shape”
Particles : has mass but negligible size and shape
Velocity & Speed
✓ Speed : Distance per unit time
✓ Average speed :
𝑆𝑇 𝑑 (Always positive scalar )
𝑣𝑠𝑝 𝑎𝑣𝑔 =
=
∆𝑡 𝑡
✓ Velocity: displacement per unit time
✓ Average velocity :
∆𝑠
𝑣𝑎𝑣𝑔 =
∆𝑡
𝑑𝑠
✓ Instantaneous velocity :
𝑣=
𝑑𝑡
Acceleration
✓ Acceleration : The rate of change in velocity (𝑚Τ𝑠 2 )
∆𝑽 = 𝑉 ′ − 𝑉
✓ Average Acceleration:
∆𝑽
𝒂𝑎𝑣𝑔 =
∆𝑡
✓ Instantaneous acceleration :
∆𝑣 𝑑𝑣 𝑑 2 𝑠
𝒂 = lim
=
= 2
∆𝑡→0 ∆𝑡
𝑑𝑡 𝑑𝑡
• If 𝑉 ′ > 𝑉 “Acceleration”
• If 𝑉 ′ < 𝑉 “Deceleration”
Relation involving s, v, and a
✓ Position “s”
𝑑𝑠
𝒗=
𝑑𝑡
𝑑𝑠
→ 𝑑𝑡 = 𝒗
𝑑𝑠 𝑑𝑣
=
𝒗
𝒂
✓ Velocity “v”
✓ Acceleration “a”
𝑑𝑣
𝒂=
𝑑𝑡
𝑑𝑣
→ 𝑑𝑡 =
𝒂
𝒂𝑑𝑠 = 𝒗𝑑𝑣
For Constant Acceleration
𝒎
𝒇𝒕
𝒂 = 𝒂𝒄 = −𝟗. 𝟖𝟏 𝟐 = 𝟑𝟐. 𝟐 𝟐
𝒔
𝒔
Constant Acceleration
v
t
 dv =  a dt
c
vo
o
s
t
 ds =  v dt
so
yields
v = vo + act
yields
s = s o + v ot + (1/2)a ct 2
o
v
s
vo
so
 v dv =  a c ds yields
v2 = (vo )2 + 2ac(s - so)
Velocity as a Function of Time
𝑑𝑣
𝒂𝑪 =
𝑑𝑡
𝑑𝑣 = 𝒂𝑪 𝑑𝑡
𝑣
𝑣 = 𝑣𝑜 + 𝒂𝑪 𝑡
𝑡
න 𝑑𝑣 = න 𝒂𝑪 𝑑𝑡
𝑣𝑂
0
so and vo are the initial position and velocity at t = 0.
Position as a Function of Time
𝑑𝑠
𝑣=
= 𝑣𝑜 + 𝒂𝑪 𝑡
𝑑𝑡
𝑠
𝑡
න 𝑑𝑠 = න 𝑣𝑜 + 𝒂𝑪 𝑡 𝑑𝑡
𝑠𝑂
0
𝟏
𝟐
𝑠 = 𝑠𝑜 + 𝑣𝑜 𝑡 + 𝒂𝑪 𝒕
𝟐
so and vo are the initial position and velocity at t = 0.
Velocity as a Function of Position
𝒗𝑑𝑣 = 𝒂𝒄 𝑑𝑠
𝑣
𝑠
න 𝒗𝑑𝑣 = න 𝒂𝒄 𝑑𝑠
𝑣𝑂
𝑠𝑂
1 2 1 2
𝑣 − 𝑣𝑜 = 𝑎𝑐 𝑠 − 𝑠𝑜
2
2
2
2
𝑣 = 𝑣𝑜 + 2𝑎𝑐 𝑠 − 𝑠𝑜
so and vo are the initial position and velocity at t = 0.
Example # 1
A particle moves along a straight line such that its position is defined by
𝑠 = 𝑡 3 − 3𝑡 2 + 2 m
Determine the velocity of the particle when t = 4 s.
Ans. v = 24 m/s.
Example # 2
A motorcyclist travels along a straight road at a speed of 27 m/s. When the
brakes are applied, the motorcycle decelerates at a rate of -6t m/s2. The distance
the motorcycle travels before it stops.
Ans. s = 54 m
Example # 3
A car moves in a straight line such that for a short time its velocity is defined by
v = 3𝑡 2 + 2𝑡
ft/s, where t is in seconds. Determine its position and
acceleration when t = 3 s. When t = 0, s = o.
Ans.
s = 36 m
a= 20 m/s2
Example # 4
A car starts from rest and moves along a straight line with an acceleration of a =
(3𝑠 −1/3 ) m/s2. where s is in meters. Determine the car’s acceleration when t = 4
sec.
Ans.
a= 1.06 m/s2
Example # 5
Ball A is released from rest at a height of 12m at the same
time that ball B is thrown upward, 1.5m from the ground.
The balls pass one another at a height of 6m. Find the speed
at which ball B was thrown upward.
Ans.
t= 1.106 sec
v= 9.5 m/s
Example # 6
Ali and Omar are standing at the top of a cliff of height “H”. Both throw a ball
with initial speed 𝑣𝑜 , Ali straight down and Omar straight up. The speed of the
balls when they hit the ground are 𝑣𝐴𝐿 and 𝑣𝑂𝑀 respectively. Which of the
following is true:
a) 𝑣𝐴𝐿 < 𝑣𝑂𝑀
b) 𝑣𝐴𝐿 = 𝑣𝑂𝑀
c) 𝑣𝐴𝐿 > 𝑣𝑂𝑀
Summary
Time dependent Acceleration
Constant Acceleration
𝑠(𝑡)
𝑑𝑠
𝑣=
𝑑𝑡
𝑣 = 𝑣𝑜 + 𝑎𝐶 𝑡
1
𝑠 = 𝑠𝑜 + 𝑣𝑜 𝑡 + 𝑎𝐶 𝑡 2
2
𝑑𝑣 𝑑 2 𝑠
𝑎=
= 2
𝑑𝑡 𝑑𝑡
𝑣 2 = 𝑣𝑜2 + 2𝑎𝑐 𝑠 − 𝑠𝑜
𝑎𝑑𝑠 = 𝑣𝑑𝑣
This applies to a freely falling object:
a = g = 9.81 m/sec2 = 32.2 ft/sec2