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                                            Lecture 2

                         Fundamental concepts of motion
Physics is classified either by subject: Mechanics, Thermodynamics, Optics, and
Electrodynamics and so on; or by methods; as experimental and theoretical Physics.
There are different theories to describe phenomena, like Classical (Newtonian) Physics,
Quantum Theory, Statistical Physics, Theory of Special and General Relativity.

Classical Mechanics:
       It investigates the motion of macroscopic bodies or system of bodies traveling at
       speeds much lower than that of light. The motion of a body is specified if we know the
       position of the body and their parts in space and time.
       The position in space is given with respect to a “frame of reference” : that means one
       fixed point and three fixed directions.
       Time: Classical Mechanics assumes a universal, unique time to which we are
       accustomed in everyday life.
       There is causality in the events; in principle, the position of a body can be determined
       at every instant of time if all the forces acting on the body now and in the past are
       given, and the state of the body is known at a certain instant of time.

Bodies can change their position, orientation and shape or volume. We can talk about rigid
bodies, elastic bodies, and fluids. The motion of rigid bodies is the combination of translation
and rotation. If the size of the body is negligible with respect to the extension of its orbit, the
body can be modeled with a material point, a particle with zero extension. A material point
performs only translational motion.


                            Kinematics of the material point




The position of a particle P in the
space is specified by its position vector
r, pointing from the origin O of the
frame of the reference to the particle
and with magnitude equal to the
distance of the particle from the origin.
The position vector is determined by
three coordinates. In the Cartesian
system of coordinates shown in the
picture these coordinates are the
projection lengths of r (x,y,z) to the
mutually perpendicular coordinate
axes.
                                                                                                  2

There can be other coordinate systems. The position of an object on the Earth, for example,
can be given by longitude, latitude and height above sea level.
Degree of freedom means the number of necessary data to specify the position of the body.
Sometime the body moves under constraints – on a surface (2 degrees of freedom) or on a line
(one degree of freedom).

       One-dimensional motion: continuous change of position along a specified line

A car moves along a road. Its position can be given with the milestones (kilometer-stones)
along the road.

                                               The function of motion x(t ) is the distance of
                                               the body from the reference point “0”along the
                                               curve in terms of time “t”.
                                               The displacement during the time interval t
                                               between t1 and t2 is
                                                x  x(t 2 )  x(t1 ) .
                                               x can be both negative and positive.

                                                                         x
The average velocity is defined as displacement per unit time:   vav       .
                                                                         t
The instantaneous velocity is defined as the time derivative of the position

                  x(t  t )  x(t ) dx(t )
v(t )  lim                                .
          t 0          t           dt
The speed is the magnitude (absolute value) of the velocity.

The path length or distance traveled (s) during a time interval t is the sum (integral) of the
absolute value of the elementary displacements. It can be calculated as the average speed
multiplied by the length of the time interval . It is always a positive quantity.

                                            [lenght ]
Unit of speed and velocity:   [ speed ]               m/ s
                                             [time]
1 km/h = (1000 m)/ (3600 s) = 1/3.6 m/s;
1 m/s = (0.001 km)/(1/3600 h)= 0.0013600 km/h = 3.6 km/h.


Example

The figure shows the time dependence of position x(t) of a car along a road. Determine the
displacement and the distance traveled and both the velocity and the speed of the car for each
linear segment of the plot. What is the displacement of the car and how far it has reached
during the whole time interval between 9-14 h, and how many km-s has it traveled (what is
the distance traveled)? What is its average speed during the route?
                                                                                              3

X(t) plot of a car




Time interval (h) Displacement          Distance traveled   Velocity (km/h)    Speed (km/h)
                  (km)                  (km)
9-10              100                   100                 100                100
10-10.5           0                     0                   0                  0
10.5-11           100                   100                 200                200
11-12             0                     0                   0                  0
12-14             -300                  300                 -150               150

The total displacement of the car is -100 km, it is at a 100 km distance from its original
position. It traveled 450 km and the average speed is 90 km/h.


One-dimensional motion with non-uniform velocity: acceleration and deceleration

The acceleration is defined as the time derivative of the velocity:

                  v(t  t )  v(t ) dv
a(t )  lim                          .
          t 0          t          dt
                                   [ speed ]
The unit of acceleration   [a]              =m/s2
                                    [time]

                                        v v2  v1
Uniformly accelerating motion:     a              const during the whole motion.
                                        t t2  t1
The velocity of an uniformly accelerating body is a linear function of time.

                                     v(t )  v0  a(t  t0 )

Deceleration means negative acceleration.
                                                                                                  4



The function of motion x(t) of an
uniformly accelerating body can be
derived from the area under the v(t)
plot.

x  vavt

                v(t )  v0
x(t )  x0                (t  t0 )
                     2


                                                       a
                            x(t )  x0  v0 (t  t0 )  (t  t0 ) 2 .
                                                       2

Examples:

a. A Formula-1 car reaches its final speed of 216 km/h in 5 seconds. What is its acceleration?
How far it has travelled from the starting point when it reached this speed?(Assume uniform
acceleration.)

216 km/h = 60 m/s; a = 60/5=12 m/s2; x = a/2  t2 = 652 = 150 m.

b. The acceleration of the falling bodies is g=9.81m/s2 here in Hungary. A boy throws a stone
upward with v0=3.27 m/s speed from the top of a 19.62 m high building. How long does the
stone fall to the ground? What is its velocity at that instant when it reaches the ground?

Let us denote the height from the ground by y:
y = y0 + v0t - g/2t2 0 = 19.62 + 3.27t + 9.81/2 t2 ; v = v0-gt=5-9.81t1.5 t2 - t - 6 = 0. t =
2.36 s. v=3.27-9.812.36 = - 19.9 m/s.

c. A burglar escapes from the police in a car, and drives with 144 km/h. It is 200 m far when
the policeman starts his car after him. The maximum speed of the police car is 180 km/h. He
must catch the burglar before he reaches the border at 2 km distance. How fast should the
policemen accelerate up his car so as to reach the burglar in time?

The burglar drives with the constant speed of 40 m/s. We count the time from that instant
when the police car starts. The police car accelerates for t1 seconds till it reaches its final 50
m/s speed: a=50/t1. The distance traveled by the burglar is 200 m shorter than the distance
traveled by the police. He would reach the border in t=1800/40=45 s.
The policeman accelerates his car for t1 seconds, after that he drives with 50 m/s.
2000 m = 25/t1 t12 + 50(45-t1)  t1 = 10 s. The policeman has to speed up his car within 10 s.
                                                                                                5

                               Motion in two or three dimensions


                                                       The motion of the body is described by
                                                       a time dependent vector r(t) which
                                                       means three scalar functions for three
                                                       coordinates, x(t), y(t), z(t). The
                                                       displacement in the time interval
                                                       t  t2  t1 is r  r(t2 )  r(t1 ) .
                                                       The velocity and acceleration vectors
                                                       are defined
                                                                      r (t  t )  r (t )
                                                        v(t )  lim                        ,
                                                                t 0         t
                                                                      v(t  t )  v(t )
                                                       a(t )  lim                          .
                                                                t 0         t

The speed is the magnitude of the velocity,   v  vx2  v y2  vz2

               .
Projectile motion.

The two-dimensional motion of the projectile can be resolved into a vertical motion with
uniform acceleration equal to g, and a horizontal motion with constant velocity.


vx  vx 0
v y  v y 0  gt
x(t )  x0  vx 0t
                         g 2
y (t )  y0  v y 0t      t
                         2
Example: A stone is thrown at 45o angle
with 7.07 m/s speed from the top of a 20
m high building. How far from the
building will the stone arrive to the
ground? (Use the approximate value g =
10m/s2)

  vx0 = vy0 = 5 m/s
0=20+5t-5t2t=2.56 s
x = 5t =12.8 m.

				
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