POSITION, VELOCITY, ACCELERATION
A particle is a body whose physical dimensions are very small
compared with the radius of curvature of its path. The figure shows a
particle moving along some general curvilinear path in space and is
located at point P at a certain instant.
The position vector of particle at any time t can be described by
fixed rectangular coordinates measured from point the origin O to
point P as

k

rP / O

j

i

 

rP / O  xi  yj  zk
Let us construct a coordinate system x'y'z‘, parallel
to the xyz coordinate system and let the origin of
this system be point O'.
The position vector of point P with respect to O‘ is




rP / O  xi  y j  z k
According to the parallelogram law,
the relation between the position
vectors is:



rP / O  rO / O  rP / O
For point Q



rQ / O  rO / O  rQ / O

In both equations rO / O is a fixed vector. The difference
between the position vectors of a body at two different
instants is named “displacement” (yer değiştirme). At time
t if the body is at point P and
 at time t+dt it it is at point
Q, then the displacement r is,
 

 r  rQ / O  rP / O
or
 

δ r  rQ/O  rP/O
This displacement is independent of the origin of the coordinate system
since
 





r  rO / O  rQ / O  - rO / O  rP / O   rQ / O - rP / O
For an instantaneous time difference the direction of
tangent to the path.

r
will be
The velocity of a body is defined as the time rate of change of its position.


drP / O 
r

vP 
 rP / O  lim
t 0 t
dt

Since the displacement r is independent of the origin of the coordinate
system, the velocity vector v P will also be independent of the origin
of the coordinate system.

Also the direction of v P
will be the same as the direction of
tangent to the path of the body

r
, that is
The velocity vector in terms of its components,




 

v P  v x i  v y j  v z k  x i  y j  zk
dx
x 
 vx
dt
dy
y 
 vy
dt
dz
z 
 vz
dt
The acceleration of a body is defined as the time rate of
change of its velocity,

dv p
dt

 ap
The acceleration vector in terms of
its components is given as







a P  a x i  a y j  a z k  v x i  v y j  v z k
  

a P  x i  y j  zk
d 2 x dv x
ax  2 
dt
dt
2
d y dv y
ay  2 
dt
dt
d 2 z dv z
az  2 
dt
dt
If the position, velocity and acceleration of a body can be described
by only their x components, the y and z components are zero, such a
motion is named as “rectilinear motion” (doğrusal hareket) . In this
case, the x axis may be taken as the axis of the motion and the body
may move along a straight line with varying velocity and acceleration.
If only the z components of the position, velocity and acceleration
vectors are zero, x and y components are different from zero, such a
motion is named as “planar curvilinear motion” (düzlemde eğrisel
hareket) .
If all the components of the position, velocity and acceleration vectors
related to the motion of a body are different from zero, such a motion
is named as “general curvilinear motion or space curvilinear motion”
(genel eğrisel hareket veya uzayda eğrisel hareket).
Consider a particle P moving along a straight line. The position of P at any
instant of time is s measured from some convenient reference point O
fixed on the line. At time t+Dt the particle has moved to P’ and its
coordinate becomes s+Ds. The change in the position coordinate during
the interval Dt is called the displacement Ds. The displacement would be
negative if the particle moved in the negative s-direction.
rP / O  x or s , vP  v  s, aP  a  v  s
The average velocity of the particle during the interval Dt is the displacement
divided by the time interval or
vav
Ds

Dt
As Dt becomes smaller and approaches zero in the limit, the average velocity
approaches the instantaneous velocity of the particle
Ds ds
v  lim

 s
Dt 0 Dt
dt
Thus, the velocity is the time rate of change of the position coordinate s.
The velocity is positive or negative depending on whether the corresponding
displacement is positive or negative. v is (+) if the particle is moving right
and v is (-) if the particle is moving left.
The average acceleration of the particle during the interval Dt is
the change in its velocity divided by the time interval or
aav
Dv

Dt
As Dt becomes smaller and approaches zero in the limit, the average
acceleration approaches the instantaneous acceleration of the particle
Dv dv d 2 s
a  lim

 2  v  s
Dt 0 Dt
dt dt
The acceleration is positive or negative depending on
whether the velocity is increasing or decreasing.
If the signs of velocity and acceleration are the same
then velocity is increasing and it is said that the
particle is accelerating (pozitif ivmelenme). If the
signs of the velocity and acceleration are opposite,
then velocity is decreasing and it is said that the
particle is decelerating (negatif ivmelenme).
Note that the acceleration would be positive if the
particle had a negative velocity which becoming less
negative. If the particle is slowing down, the particle
is said to be decelerating.
By eliminating the time dt between velocity and
acceleration equations, we obtain a differential
equation
relating,
displacement,
velocity
and
acceleration.
ds
dt 
v
ads  vdv
dv dv
a

v
dt ds
,
or
sds  sds
Although position, velocity and acceleration are
vector expressions, for rectilinear motion, where
the direction of the motion is that of the given
straight-line path, they may be expressed by
their magnitudes indicating the sense of the
vectors along the path described by a plus (+) or
minus (-) sign.
The displacement of the particle may not always
be equal to the path taken. Displacement is the
vector difference between the initial and final
positions of the particle along the path. If the
particle completes its motion at exactly the same
point it starts its motion then the displacement is
zero. But the path taken will be the distance it has
traveled along the path and will not be equal to
zero.
1
3
2
If the position coordinate s is known for all values of
the time t, successive mathematical and graphical
differentiation with respect to t gives the velocity
and acceleration.
In many problems, however, the relationship between
position coordinate and time is unknown, and it must
be determined by successive integration from the
acceleration.
Depending on the nature of the forces, the
acceleration may be a function of time, velocity or
position coordinate, or a combined function of these
quantities.
The procedure for integrating the differential
equation in each case is indicated as follows.
For all cases let the initial conditions
be assumed as, t=0, s=s0 ,v=v0 .
dv
a
dt

v  v0  at 
t
v0
0
 dv  a  dt

v  v0  at
ds
v
dt

v

s
t
s0
0
 ds   vdt
t
s  s0 
 v0  at dt
 s  s 0  v0 t 
0

v
1 2
at
2
vdv  ads
s

vdv  a ds 

v0
s0


1 2
v  v02  a s  s0  
2
v 2  v02  2a s  s0 
dv
 a  f (t )
dt

 f (t )dt
t
v0
0
 dv   f (t )dt

t
v  v0 
v
t

v  v0 
 f (t )dt

0 

0
v  g (t )

ds
v
dt
s

t
 ds   g (t )dt
s0
0
t

s  s0 
 g (t )dt
0
dv
 a  f (v )
dt

v

t


dv
 dt
f(v)
dv
f(v)
v0
This result gives t as a function of v.
Another approach is
vdv  ads  f (v)ds
v
 s  s0 

v0
vdv
f(v)

vdv
 ds
f(v)
v


v0
vdv

f(v)
s
 ds
s0

vdv  ads  f ( s ) ds

v
s
v0
s0
 vdv   f (s)ds
s
 v 2  v02  2
 f (s)ds
s0
s
v 
2
v02
2
 f (s)ds
0
s


v g (s)

ds
v
 g (s)
dt
s

t

s0
ds
g(s)

ds
 dt
g (s)
s


s0
ds

g(s)
t
 dt
0
Figure a is a schematic plot of the variation of s
with t from time t1 to time t2 for some given
rectilinear motion.
By constructing the tangent to the curve at any
time t, we obtain the slope, which is the velocity
v=ds/dt (Figure b).
ds
v
dt
Similarly, the slope dv/dt of the v-t curve at any
instant gives the acceleration at that instant
(Figure c).
dv
a
dt
dx
v
dt
We see from Figure b that the area under the v-t
curve during time dt is vdt, which is the
displacement ds. The net displacement of the
particle during the interval from t1 to t2 is the
corresponding area under the curve, which is
s2
t2
 ds   vdt
s1

s2  s1  (area under v-t curve)
t1
Similarly, from Figure c we see that the area under
the a-t curve during time dt is adt, which is the
velocity dv. The net change in velocity between t1
and t2 is the corresponding area under the curve,
which is
v2
t2
 dv   adt
v1
t1

v2  v1  (area under a-t curve)
dx
v
dt
When the acceleration a is plotted as a
function of the position coordinate s, the
area under the curve during a displacement ds
is ads is
v2

v1
s2

vdv  ads
s1



1 2
v2  v12  (area under a-s
2
curve)
As another example:
The slope of the v-t graphic gives the
magnitude of the acceleration for that
instant.
dv
a
dt
If the magnitude of the acceleration is (+)
then either the velocity is increasing along
the +s direction or the velocity is decreasing
along –s direction. If acceleration is (–)
either the velocity is decreasing along the +s
direction or it is increasing along the –s
direction.
The slope of the x-t graphic gives the
magnitude of the velocity for that
instant.
dx
dt
v
If v is + the particle moves in +s direction, it
v is – the particle moves in –s direction.
1. A ball is thrown vertically upward
with an initial speed of 25 m/s from
the base A of a 15-m cliff.
Determine the distance h by which
the ball clears the top of the cliff
and the time t after release for
the ball to land at B. Also, calculate
the impact velocity vB. Neglect air
resistance and the small horizontal
motion of the ball.
2. A particle starts its motion along a
horizontal axis at t=0 with s=0 and v0=50
m/s. For the first 4 seconds its
acceleration is zero; afterwards it
comes into effect of a force which gives
the particle a deceleration of a= -10
m/s2. Determine the velocity and
position of the particle for t=8 s. and
t=12 s. What is the maximum position
reached by the particle?
3. The steel ball A of diameter D slides freely on
the horizontal rod which leads to the pole face
of the electromagnet. The force of attraction
obeys an inverse-square law, and the resulting
acceleration of the ball is a=K/(L – x)2, where K
is a measure of the strength of the magnetic
field. If the ball is released from rest at x=0,
determine the velocity v with which it strikes the
pole face.
4. The driver of a car, which is initially at rest at
the top A of the grade, releases the brakes and
coasts down the grade with an acceleration in
meter per second squared given by a=0.981 –
0.013v2, where v is the velocity in meters per
second. Determine the velocity vB at the bottom
B of the grade.
5. A bumper, consisting of a nest
of three springs, is used to
arrest the horizontal motion of a
large mass which is traveling at
40 m/s as it contacts the
bumper. The two outer springs
cause a deceleration proportional
to the spring deformation. The
center spring increases the
deceleration rate when the
compression exceeds 0.5 m as
shown on the graph. Determine
the maximum compression x of
the outer springs.
6. The brake mechanism shown in the figure is composed
of a piston moving in a fixed cylinder filled with oil. When
the brake pedal is pressed while the vehicle moves with a
speed v0, the piston moves, oil passes through the
channels inside the piston and the vehicle slows down in
proportion to its speed, a=-kv. Determine a) v in terms of
t, b) x in terms of t, c) v in terms of x. Also construct the
related graphics.
piston
oil
7. A particle moves along the y axis with an acceleration
given by a(t)=5sinwt cm/s2 where w=0.7 rad/s. Initially
when t=0, the particle is 2 cm above the origin and is
moving downward with a speed of 5 cm/s.
a) Determine the velocity and position of the particle as
functions of time.
b) Show the position, velocity and acceleration on a graph
for the interval of t=0 and t=4 s.
c) Determine the displacement  of the particle between
t=0 and t=4 s.
d) Determine the total distance s traveled by the
particle between t=0 and t=4 s.