MECHANICS
KINEMATICS
Kinematics is the study of motion, without the investigation of the cause for the motion
MOTION
•Motion of what?
Material point
(An object with an irrelevant dimension for the purposes of a particular problem)
•Development of models
Example: linear movement v. rotation
•Motion is relative to the object of reference
Examples: the motion of an airplane passenger relative to the air plane, or the
motion of an air plane passenger relative to the ground.
•System of reference:
The object of reference
z
System of coordinates (measures distance)
Watch (measures time)
   

r  r2  r1  r t 2   r t1 

r

r1

r2
y
x
ONE DYMENTIONAL (1D) UNIFORM MOTION
(Motion along a straight line with a constant velocity)
x  x2  x1
x
0
x1
x2
x  xt 
x1  xt1 
x2  xt2 
Uniform - motion with a constant ratio:
if t2  t1 then x2  x1
x1 x2

v
t1 t 2
For uniform motion, velocity (v) is the displacement of
an object over the time passed.
VELOCITY OF 1D UNIFORM MOTION
x  x0  vt  t 0 

x
x
v
t
x2
t2

x
 tan 
t
x1
t1
t3
t2
t1
t4
t
v
v
x
t1
t
x  v  t
t2
if t 0  o
x  x0  v  t
AVERAGE VELOCITY
(Total distance covered, divided by total time passed)
x
v
t
x
x
t
t
-speed is the distance covered over time passed
-velocity is the displacement of an object over time passed
-velocity has a direction, while speed does not
Example 1
A
x 2
x 1
Formula used:
Given: AB  x  100mi
B
x
x1  x 2 
 50mi
2
v1  40mi/h
v 2  60mi/h
Δx
v
Δt
v ?
Solution:
t1 
x1
v1
t2 
x2
v2
x1 x2 x  1 1  x  v1  v2 
   


t  t1  t2 


v1
v2
2  v1 v2  2  v1v2 
x
x
2v1v2
2  40  60
v



 48
t x  v1  v2  v1  v2
40  60


2  v1v2 
Analytical answer:
2v1v2
v
v1  v2
Numerical answer:
v  48mi/h
Example 1 (comments):
a)v1  v 2  v  v  v
v1  v 2
v1  v 2
2v1  v 2
b)v 


2
2
v1  v 2
v1  v 2 2  4v1  v 2
v1  v 2   0  v1  v 2
Example 2
Given:
Δt1  Δt 2  1h
v1  40km/h
Formula used:
x
v
t
v 2  60km/h
v ?
Solution:
x1  40km
x 2  60km
x  x1  x 2  40km  60km  100km
t  2h
x 100km
v

 50km/h
t
2h
Answer:
v  50km/h
INSTANTANIOUSE VELOCITY
(Velocity at a given point)
x dx
v  lim

 x t   x t 
t 0 t
dt
B
x(t)
x
A
t
t
NONUNIFORM MOTION
x  x0  x   x n   vt n t
v(t)
n
n
t
x  x0  lim
x1 x2 x3 x4
t 0
 vt t   vt dt
n
n
t0
t
t0 t
t
t
t
t
t t
t
x  x0   vt dt
t0
dx
vt  
dt
dx  vt dt
if vt  - const
x
t
x0
t0
 dx   vt dt
t
x  x0   vt dt
x  x0  v  dt  x0  vt  t 0 
t0
t
t0