OPr /

kzjyixr OP

/

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

POSITION, VELOCITY, ACCELERATION

i

j

k

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

kzjyixr OP

/

According to the parallelogram law, the relation between the position vectors is:

OPOOOP rrr ///

For point Q OQOOOQ rrr ///

rrr O/PO/Q

O/Or

r

OP/OQ/ rrrδor

In both equations 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 is,

This displacement is independent of the origin of the coordinate system since

- - OPOQOPOOOQOO rrrrrrr //////

For an instantaneous time difference the direction of will be tangent to the path.

r

The velocity of a body is defined as the time rate of change of its position.

t

rr

dt

rdv

tOP

OPP

0/

/ lim

r

Pv

Pv

r

Since the displacement is independent of the origin of the coordinate system, the velocity vector will also be independent of the origin

of the coordinate system.

Also the direction of will be the same as the direction of , that is tangent to the path of the body

The velocity vector in terms of its components,

zyx

zyxP

vdt

dzzv

dt

dyyv

dt

dxx

kzjyixkvjvivv

The acceleration of a body is defined as the time rate of change of its velocity,

pp a

dt

vd

The acceleration vector in terms of its components is given as

dt

dv

dt

zda

dt

dv

dt

yda

dt

dv

dt

xda

kzjyixa

kvjvivkajaiaa

z2

2

z

y

2

2

y

x2

2

x

P

zyxzyxP

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+t the particle has moved to P’ and its coordinate becomes s+s. The change in the position coordinate during the interval t is called the displacement s. The displacement would be negative if the particle moved in the negative s-direction.

svaasvvsxr PPOP , , or /

The average velocity of the particle during the interval t is the displacement divided by the time interval or

t

svav

As t becomes smaller and approaches zero in the limit, the average velocity approaches the instantaneous velocity of the particle

st

sv

t

dt

ds lim

0

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 t is the change in its velocity divided by the time interval or

t

vaav

As t becomes smaller and approaches zero in the limit, the average acceleration approaches the instantaneous acceleration of the particle

svdt

sd

dt

dv

t

va

t

2

2

0 lim

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.

sdsdssorvdvads

vds

dv

dt

dva

v

dsdt

,

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 2

3

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 .

020

20

20

2v

v

200

0

00

0

s

s

00

0

v

v

2 2

1

2

1

00

0

0

ssavvssavvdsavdv

adsvdv

attvssdtatvss

vdtdsvdt

ds

atvvatvv

dtadvadt

dv

s

s

t

t

t

)( )(ds

)( )(

)(dv )(

0

0

0

s

s

)(

0

0

0

0

0

v

v

0

0

tt

tgv

tt

t

dttgssdttg vdt

ds

dttfvvdttfvv

dttftfadt

dv

)(

0

v

vf(v)

dvt

dt f(v)

dvvfa

dt

dv

v

v

v

v

f(v)

vdvss

ds f(v)

vdv ds

f(v)

vdvdsvfadsvdv

0

00

)(

0

s

s

This result gives t as a function of v.Another approach is

)(

)(

)(2

)(2

)( )(

0

0

0

0

00

0

)(

s

s

20

2

s

s

20

2

s

s

s

s

ts

s

sgv

v

v

g(s)

dst

dtg(s)

dsdt

sg

dssg

dt

dsv

dssfvv

dssfvv

dssf vdv dssfadsvdv

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).

Similarly, the slope dv/dt of the v-t curve at any instant gives the acceleration at that instant (Figure c).

adt

dv

vdt

dx

vdt

ds

vdt

dx

12

2

1

2

1

ssvdtds

t

t

s

s

(area under v-t curve)

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

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

12

2

1

2

1

vvadtdv

t

t

v

v

(area under a-t curve)

21

222

12

1

2

1

vvadsvdv

s

s

v

v

(area under a-s curve)

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

As another example:

The slope of the v-t graphic gives the magnitude of the acceleration for that instant.

adt

dv

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.

vdt

dx

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.

piston

oil

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.

7. A particle moves along the y axis with an acceleration given by a(t)=5sint cm/s2 where =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.