Kinematic Study in One Dimension (Rectilinear Motion)

This course will introduce “Freshman Sciences” students to the following concepts in classical mechanics: Physical quantities, standards and units, Vectors and scalars, Velocity and acceleration, Motion in one, two and three dimensions, Newton’s laws, falling bodies, uniform circular motion, Work and energy, power, Kinetic energy theorem, Conservation of total energy, Rectilinear sinusoidal motion, angular sinusoidal motion, Linear and angular momentum collisions, Gravitation

1. Chap 2
Motion Along a Straight Line
PHYS 100
Dr. Bassem HMOUDA

2. Introduction
•One-dimensional motion.
•Kinematics study : Position, velocity, acceleration, time equations and
trajectory.
•We are interested in motion without taking into consideration the causes
affecting this motion and the shape and the dimension of the moving body.

3. 1-Position, Displacement & Distance.
•The location or the positionof a particle is defined by the abscissa relative to a reference point (Origin) O.
Ex:�
0=0(at The origin)
�
�=30�(at 30 meters from the origin in the positive side)
�
�=−40�(at 40 meters from the origin in the negative side)
•The change from position 1 to position 2 in some time interval is
called a displacement∆�, where :
∆�=�
2−�
1
Ex:∆�
��=�
�−�
�=50−30=20�>0
(Poistivemotion or moving forward)
∆�
��=�
�−�
&#3627408437;=0−30=−30&#3627408474;<0
(Negative motion or moving backward)
∆&#3627408485;
&#3627408436;&#3627408437;
∆&#3627408485;
&#3627408437;&#3627408439;

4. 1-Position, Displacement & Distance.
•Thedistancetraveled(orsimplythedistance)isthelengthofthepathfollowedbythemovingpoint,itis
thenumberofmetersmovedindependentofdirection.
Ex:Between A and D :
The displacement is : ∆&#3627408485;
&#3627408436;&#3627408439;=&#3627408485;
&#3627408439;−&#3627408485;
&#3627408436;=0−30=−30&#3627408474;
The traveled distance is : &#3627408465;
&#3627408436;&#3627408439;=20+50=70&#3627408474;
&#3627408465;
&#3627408436;&#3627408437;
∆&#3627408485;
&#3627408436;&#3627408439;
&#3627408465;
&#3627408436;&#3627408437;= ∆&#3627408485;
&#3627408436;&#3627408437;+∆&#3627408485;
&#3627408437;&#3627408439;=&#3627408485;
&#3627408437;−&#3627408485;
&#3627408436;+&#3627408485;
&#3627408439;−&#3627408485;
&#3627408437;=50−30+0−50=20+50=70&#3627408474;

5. 2-Average Velocity & Speed
•Theaveragevelocityistheratioofthedisplacement∆&#3627408485;that
occursduringaparticulartimeinterval∆&#3627408481;tothatinterval:
•Onagraphof&#3627408485;versus&#3627408481;,??????
????????????istheslopeofthestraightlinethat
connectstwoparticularpointsonthex(t)curve(Secant).
•Thevelocityingeneralisexpressedinm/swhenthedisplacement
isexpressedinmeters(m)andthetimeinseconds(s).
•Likedisplacement,theaveragevelocityisavector(Ithasa
magnitudeandadirection.
•When the slope slants upward, it tells us that ??????
????????????&#3627408468;>0so the displacement ∆&#3627408485;>0so the motion is in the positive direction.
•When the slope slants downward, it tells us that ??????
????????????&#3627408468;<0so the displacement ∆&#3627408485;<0so the motion is in the negative direction.
??????
&#3627408436;&#3627408437;=
∆&#3627408485;
&#3627408436;&#3627408437;
∆&#3627408481;
=
50−30
10
=2&#3627408474;/&#3627408480;

6. 2-Average Velocity & Speed
•Theaveragespeedistheratioofthetraveleddistancethatoccurs
duringaparticulartimeintervaltothatinterval:
&#3627408454;
????????????&#3627408468;=
&#3627408455;&#3627408479;????????????&#3627408466;&#3627408473;&#3627408466;&#3627408465;&#3627408465;??????&#3627408480;&#3627408481;??????&#3627408475;&#3627408464;&#3627408466;
∆&#3627408481;
•Theaveragespeedrepresentsthenumberofmetersmoved
independentofdirection.
Ex:Between A and E:
∆&#3627408481;
∆&#3627408485;
•Theaveragespeedhasnodirectionassociatedwithit,soitisnota
vectorquantity.
The average speed : &#3627408454;
&#3627408436;&#3627408440;=
??????????????????
∆??????
=
20+90
40−0
=
110
40
=2.75&#3627408474;/&#3627408480;
&#3627408465;
&#3627408436;&#3627408440;=∆&#3627408485;
&#3627408436;&#3627408437;+∆&#3627408485;
&#3627408437;&#3627408440;=&#3627408485;
&#3627408437;−&#3627408485;
&#3627408436;+&#3627408485;
&#3627408440;−&#3627408485;
&#3627408437;=50−30+−40−50=20+90=110&#3627408474;
&#3627408465;
&#3627408436;&#3627408440;
The average velocity : ??????
&#3627408436;&#3627408440;=
∆??????????????????
∆??????
=
−40−30
40−0
=−1.75&#3627408474;/&#3627408480;

7. 3-Instantaneous Velocity & Speed
The instantaneous velocity(or simply the velocity) at a given instant t is the average velocity between two
instants t1 and t2 around t and very close to each other:
Derivative of xw.r.t
time
x x x

8. 3-Instantaneous Velocity & Speed
The instantaneous velocity(or simply the velocity) at a given instant t is the average velocity between two
instants t1 and t2 around t and very close to each other:
Derivative of xw.r.t
time
Ex:
The positions are given by : x= 5t
2
+ 3t+1
The instantaneous velocity is :
??????=
????????????
????????????
=10&#3627408481;+3(varies with time)
At t= 0 s →??????
1=100+3=3&#3627408474;/&#3627408480;
At t= 1 s →??????
1=101+3=13&#3627408474;/&#3627408480;
At t= 2 s →??????
1=102+3=23&#3627408474;/&#3627408480;

9. 3-Instantaneous Velocity & Speed
•The Instantaneous speed (Or simply the speed) is equal to the magnitude of the instantaneous velocity.
•With average speed, The Instantaneous speed has no direction associated with it, so it is not a vector
quantity.
Ex:
If a particle A has an instantaneous velocity of 25 m/s along a line
And a particle B has an instantaneous velocity of -25m/s along the same line
Both of them has the same speed 25 m/s
Ԧ??????
&#3627408436;
Ԧ??????
&#3627408436;

10. 4-Acceleration
•When the velocity of a particle changes, it said to undergo acceleration.
•The time rate change of the velocity is the average acceleration:
•The instantaneous acceleration (Or simply the acceleration) at a given instant t is the average acceleration
between to close instants around t :
??????=&#3627408525;??????&#3627408526;
∆&#3627408533;→??????
∆??????
∆&#3627408533;
=
&#3627408517;??????
&#3627408517;&#3627408533;
•The acceleration is expressed in &#3627408474;/&#3627408480;
2
when the velocity is expressed in m/sand the time in second (s)
•Like velocity, the acceleration is vector quantity.
Ex:
If a= 10 &#3627408474;/&#3627408480;
2
= constant, it means that the velocity varies 10 m/s every 1 sec.
Derivative of vw.r.t
time

11. 4-Acceleration
??????=&#3627408525;??????&#3627408526;
∆&#3627408533;→??????
∆??????
∆&#3627408533;
=
&#3627408517;??????
&#3627408517;&#3627408533;
v v v
Derivative of vw.r.t
time

12. 4-Acceleration
Position x
Velocity??????=
&#3627408465;&#3627408485;
&#3627408465;&#3627408481;
Acceleration??????=
&#3627408465;??????
&#3627408465;&#3627408481;
Derivative
w.r.t time
Derivative
w.r.t time
Ex:
Position:
x= 5t
2
+ 3t+1
Derivative
w.r.t time
Velocity:
v= 10t+ 3
Derivative
w.r.t time
Acceleration:
a= 10 m/s
2
= constant

13. No
Motion
No
Motion
URM
UARM
UDRM
URM

14. 4-Acceleration
V
t
V0
V>0
V<0
t1
t2
t3
0 < t < t1
V > 0 so the motion is in the positive direction
a = constant < 0 so the velocity varies uniformly
UDRM
t1 < t < t2
V < 0 so the motion is in the negative direction
a = constant < 0 so the velocity varies uniformly
UARM
t2 < t < t3
V < 0 so the motion is in the negative direction
a’ = constant > 0 so the velocity varies uniformly
UDRM
t > t3
V > 0 so the motion is in the positive direction
a’ = constant > 0 so the velocity varies uniformly
UARM

15. Constant Acceleration
??????
????????????&#3627408468;=
??????
??????+??????
&#3627408467;
2
??????=
∆??????
∆&#3627408481;
=
??????
&#3627408467;−??????
??????
&#3627408481;−0
⟹??????
&#3627408467;=??????
??????+??????&#3627408481;
??????
????????????&#3627408468;=
∆&#3627408485;
∆&#3627408481;
=
&#3627408485;
&#3627408467;−&#3627408485;
??????
&#3627408481;−0
⟹&#3627408485;
&#3627408467;=&#3627408485;
??????+??????
????????????&#3627408468;&#3627408481;=&#3627408485;
??????+
??????
??????+??????
&#3627408467;
2
&#3627408481;=&#3627408485;
??????+
1
2
??????
??????+??????
??????+??????&#3627408481;&#3627408481;=&#3627408485;
??????+??????
??????&#3627408481;+
1
2
??????&#3627408481;
2
⟹&#3627408485;
&#3627408467;=&#3627408485;
??????+??????
??????&#3627408481;+
1
2
??????&#3627408481;
2
??????
&#3627408467;=??????
??????+??????&#3627408481;⇒&#3627408481;=
??????
&#3627408467;−??????
??????
??????
⇒&#3627408485;
&#3627408467;=&#3627408485;
??????+??????
??????
??????
&#3627408467;−??????
??????
??????
+
1
2
??????
??????
&#3627408467;−??????
??????
??????
2
⟹??????
&#3627408467;
2
−??????
??????
2
=2??????(&#3627408485;
&#3627408467;−&#3627408485;
??????)

16. Constant Acceleration
UVRM(??????=&#3627408516;&#3627408528;&#3627408527;&#3627408532;&#3627408533;≠??????) URM(??????=??????)
&#3627408485;
&#3627408467;=&#3627408485;
??????+??????
??????&#3627408481;+
1
2
??????&#3627408481;
2 &#3627408485;
&#3627408467;=&#3627408485;
??????+??????
??????&#3627408481;
??????
&#3627408467;=??????
??????+??????&#3627408481; ??????
&#3627408467;=??????
??????
??????
&#3627408467;
2
−??????
??????
2
=2??????(&#3627408485;
&#3627408467;−&#3627408485;
??????)
??????
&#3627408467;
2
−??????
??????
2
=0⇒??????
&#3627408467;=??????
??????

17. •Ifwedropanobjectinvacuum,the
objectacceleratesdownwardina
constantratecalled“Free-Fall
acceleration”.
•The magnitude of that acceleration is
represented by g and it is , at sea level in
Earth’s midlatitudes, g = 9.8 m/s
2
.
•Theaccelerationisindependentofthe
object’scharacteristics,suchasmass,
density,orshape;itisthesameforall
objects.
Free-Fall Acceleration

18. •Inafree-fallthedirectionofmotionis
nowalongtheverticaly-axisinsteadofx-
axis.
•Ifthepositivedirectionofthey-axisis
upwardsothefree-fallaccelerationis:a
=-g=-9.8m/s
2
,otherwisea=g=9.8
m/s
2
•Theequationsofmotionisthiscaseare
asfollows:
&#3627408486;
&#3627408467;=&#3627408486;
??????+??????
??????&#3627408481;±
1
2
??????&#3627408481;
2
??????
&#3627408467;=??????
??????±??????&#3627408481;
??????
&#3627408467;
2
−??????
??????
2
=±2??????(&#3627408486;
&#3627408467;−&#3627408486;
??????)
Free-Fall Acceleration

19. Free-Fall Acceleration

20. Because acceleration a is defined in terms of velocity as :
??????=
&#3627408465;??????
&#3627408465;&#3627408481;
So the Fundamental Theorem of Calculus tells us that:
??????
1−??????
0=න
??????0
??????1
??????&#3627408465;&#3627408481;
Which means graphically:
න
??????0
??????1
??????&#3627408465;&#3627408481;=areabetweenaccelerationcurveandtimeaxis,fromt
0tot
1
Then the corresponding unit of area on the graph is: (1m/s
2
)(1s) = 1m/s, which is (properly) a unit of velocity.
Graphical integration in motion analysis

21. Because velocity v is defined in terms of the position as :
??????=
&#3627408465;&#3627408485;
&#3627408465;&#3627408481;
So the Fundamental Theorem of Calculus tells us that:
&#3627408485;
1−&#3627408485;
0=න
??????0
??????1
??????&#3627408465;&#3627408481;
Which means graphically:
න
??????0
??????1
??????&#3627408465;&#3627408481;=areabetweenvelocitycurveandtimeaxis,fromt
0tot
1
Then the corresponding unit of area on the graph is: (1m/s)(1s) = 1m, which is (properly) a unit of distance.
Graphical integration in motion analysis

22. Position x Velocity??????=
&#3627408465;&#3627408485;
&#3627408465;&#3627408481;
Acceleration??????=
&#3627408465;??????
&#3627408465;&#3627408481;
Derivative
w.r.t time
Derivative
w.r.t time
Graphical integration in motion analysis
Integral
w.r.t time
Integral
w.r.t time
Graphically
Position x Velocity??????=
&#3627408465;&#3627408485;
&#3627408465;&#3627408481;
Acceleration??????=
&#3627408465;??????
&#3627408465;&#3627408481;
Slope
of xversus t
Slope
of vversus t
Area
Between aand t
Area
Between vand t
By calculation