11 mech ijme kinematics of two bodies muneeb pakistan

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International Journal of Mechanical Engineering (IJME) ISSN(P): 2319-2240; ISSN(E): 2319-2259 Vol. 2, Issue 5, Nov 2013, 111-124 Š IASET

KINEMATICS OF TWO BODIES IN TERMS OF GEOMETRIC SERIES SH. M. MUNEEB ZAHID Mechanical Engineer, N.E.D University of Engineering and Technology, Karachi, Pakistan

ABSTRACT In this paper the motion of two bodies moving along straight lines with uniform velocities have been considered and studied with the help of Geometric series. When two bodies move in same straight line with different uniform velocities then the distance between them varies continuously and follows a geometric progression. Three different cases have been discussed in this paper. In first case both the bodies move along a straight line in same direction and in the second case bodies move along parallel lines and in third case they move along non-parallel lines.

KEYWORDS: Two Bodies, Straight, Geometric Series INTRODUCTION Geometric progression is the progression in which every term is a multiple of its previous term. The ratio of two consecutive terms in a constant and is known as common ratio. Following is an example of Geometric progression: đ?’‚, đ?’‚ đ?’“, đ?’‚đ?’“đ?&#x;? , đ?’‚đ?’“đ?&#x;‘ _ _ _ _ _ _ _ _ _ _ đ?’‚đ?’“đ?’?−đ?&#x;? where nth term is ar n−1 . ‘r’ is the common ratio of that geometric progression. A Geometric series is given by đ?’‚ + đ?’‚ đ?’“ + đ?’‚đ?’“đ?&#x;? _ _ _ _ _ _ _ _ _ _ đ?’‚đ?’“đ?’?−đ?&#x;? The sum of 1st n geometric series is đ?‘şđ?’? =

đ?’‚(đ?&#x;? − đ?’“đ?’? ) đ?&#x;?−đ?’“

If r >1 then it will be an increasing geometric series and if đ?&#x;Ž < đ?‘&#x; < 1 then it will be a decreasing geometric series the sum of infinite decreasing geometric series is đ?‘ş =

đ?’‚ đ?&#x;?−đ?’“

RESULTS AND DISCUSSIONS The Kinematics of Two Bodies in Terms of Geometric Series CASE 1 (When the Bodies are Moving in a Same Straight Line and Same Direction) Let body „Aâ€&#x; and „Bâ€&#x; are at rest and the distance between them is „d1 â€&#x;. Both bodies are placed in a straight line. Body „Aâ€&#x; is at the front and body „Bâ€&#x; is at the back. In the same time both the bodies start moving with uniform velocities. Body „Aâ€&#x; start moving with velocity „Uâ€&#x; and body „Bâ€&#x; start moving with velocity „Vâ€&#x;. Body „Bâ€&#x; takes time „t1 â€&#x; to travel distance „d1 â€&#x; and reaches to the 1st position of body „Aâ€&#x; and in time „t1 â€&#x; body „Aâ€&#x; travels distance „d2 â€&#x;. Then body „Bâ€&#x; takes time „t 2 â€&#x;to travel distance „d2 â€&#x; and reaches the 2nd position of body „Aâ€&#x;. In the same time body „Aâ€&#x; travels distance „d3 â€&#x; then body „Bâ€&#x; takes time „t 3 â€&#x;to travel distance „d3 â€&#x; and reaches the 3rd position of body „Aâ€&#x;. In the same time body „Aâ€&#x; travels distance „d4 â€&#x; and so on. Follow figure 1 and figure 2.

112

Sh. M. Muneeb Zahid đ???đ?&#x;?

đ??­đ?&#x;? =

đ??•

đ???đ?&#x;? = u đ??­ đ?&#x;? , where đ??­ đ?&#x;? =

đ???đ?&#x;? đ??•

đ??Ž

đ???đ?&#x;? = đ???đ?&#x;? đ??•

đ??­đ?&#x;?=

đ???đ?&#x;?

đ??Ž

, where đ???đ?&#x;? = đ???đ?&#x;?

đ??•

đ??­đ?&#x;? =

đ??•

đ???đ?&#x;? đ??Ž

.

đ??•

đ??•

đ???đ?&#x;‘ = u đ??­ đ?&#x;? , where đ??­ đ?&#x;? = đ??Ž đ?&#x;?

đ???đ?&#x;‘ = đ??­đ?&#x;‘ = đ??­đ?&#x;‘ =

đ??• đ???đ?&#x;‘ đ??• đ???đ?&#x;? đ??•

.

đ??•

đ??Ž đ??•

. đ???đ?&#x;?

, where đ???đ?&#x;‘ = .

đ???đ?&#x;?

đ??Ž đ?&#x;? đ??•

. đ???đ?&#x;?

đ??Ž đ?&#x;? đ??•

Figure 1: Initially the Bodies are at Rest and at their 1st Position

Figure 2: In Same Time Both the Bodies Start Moving with Uniform Velocities đ???đ?&#x;’ = u đ??­ đ?&#x;‘ , where đ??­ đ?&#x;‘ = đ??Ž đ?&#x;‘

đ???đ?&#x;’ = đ??­đ?&#x;? =

đ??• đ???đ?&#x;? đ??•

đ???đ?&#x;? đ??•

.

đ??Ž đ?&#x;? đ??•

. đ???đ?&#x;?

, đ??­đ?&#x;?=

đ???đ?&#x;? đ??Ž đ??• đ??•

, đ??­đ?&#x;‘=

đ???đ?&#x;? đ??•

.

đ??Ž đ?&#x;? đ??•

_ _ _ _ _ _ _ _ _ _ _ đ??­đ??§ =

đ???đ?&#x;? đ??•

.

đ??Ž đ??§âˆ’đ?&#x;? đ??•

It means that all the times from ‘đ??­ đ?&#x;? ’to „đ??­ đ??§ ’are in geometric progression so we can apply all the formulas of geometric series in it. Also, đ???đ?&#x;? , đ???đ?&#x;? , đ???đ?&#x;‘ , _ _ _ _ _ _ _ đ???đ??§ are in geometric progression so we can apply all the formulas of geometric series in it. If velocity of body „Bâ€&#x; is greater than the velocity of body „Aâ€&#x;(V > U) then in time ‘T’ body „Bâ€&#x; will collide with body „Aâ€&#x;. Then we know that T =

đ???đ?&#x;? đ??•âˆ’đ??”

where (V- U) is the relative velocity of body „Bâ€&#x; with respect to body „Aâ€&#x;.

Kinematics of two bodies in terms of geometric series also explain it We know that if V > U then

Kinematics of Two Bodies in Terms of Geometric Series

113

đ???đ?&#x;? , đ???đ?&#x;? ,_ _ _ _ _ _ đ???đ??§ are in decreasing geometric progression. Also đ??­ đ?&#x;? , đ??­ đ?&#x;? , đ??­ đ?&#x;‘ _ _ _ _ _ _ _ _ đ??­ đ??§ are in decreasing geometric progression. Also the time in which „Bâ€&#x; will collide with „Aâ€&#x; is the sum of all the times from t1 to infinity T = đ??­ đ?&#x;? +đ??­ đ?&#x;? +đ??­ đ?&#x;‘ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Also we know that the sum of geometric series a + ar +đ??šđ??Ť đ?&#x;? _ _ _ _ _ _ _ _ _ _to infinity is S=

đ??š đ?&#x;?−đ??Ť

Where S = T, a = đ??­ đ?&#x;? =

đ???đ?&#x;? đ??•

,r=

đ??­đ?&#x;? đ??­đ?&#x;?

=

đ??Ž đ??•

Substitute these values in above equation đ???đ?&#x;? đ??• đ??” đ?&#x;?− đ??•

T=

T=

đ???đ?&#x;? đ??•âˆ’đ??”

Proved.

(CASE 2) When the Bodies are Moving in Parallel Lines Supposed body „Aâ€&#x; is placed in line ‘đ??‹đ?&#x;? ’and Body „Bâ€&#x; is placed in line ‘đ??‹đ?&#x;? ’ as shown.‘đ??‹đ?&#x;? ’ and ‘đ??‹đ?&#x;? ’are parallel to each other and the smallest and perpendicular distance between ‘đ??‹đ?&#x;? ’ and ‘đ??‹đ?&#x;? ’is dy . Distance between „Aâ€&#x; and „Bâ€&#x; is đ???đ?&#x;? =

đ???đ?&#x;?đ??ą đ?&#x;? + đ???đ??˛ đ?&#x;? as shown in figure 3. In the same time both the bodies start moving with uniform velocities.

Body „Aâ€&#x; start moving with velocity „Uâ€&#x; and „Bâ€&#x; start moving with velocity „Vâ€&#x;. In time ‘đ??­ đ?&#x;? ’ body „Bâ€&#x; travels distance „d1x â€&#x; and body „Aâ€&#x; travels distance „d2x â€&#x;. Now „Aâ€&#x; and „Bâ€&#x; are in their 2nd position and distance between them is „d2 â€&#x;. Then in time ‘đ??­ đ?&#x;? ’body „Bâ€&#x; travels distance „d2x â€&#x; and „Aâ€&#x; travels distance „d3x â€&#x;. Now „Aâ€&#x; and „Bâ€&#x; are in their 3rd position and the distance between them is „d3 â€&#x; and so on follow figure „3â€&#x; and figure „4â€&#x;. đ???đ?&#x;?đ??ą đ?&#x;? + đ???đ??˛ đ?&#x;?

đ???đ?&#x;? = đ??­đ?&#x;? =

đ???đ?&#x;?đ??ą đ??•

đ???đ?&#x;?đ??ą = u đ??­ đ?&#x;? đ???đ?&#x;?đ??ą

Where đ??­ đ?&#x;? = đ???đ?&#x;?đ??ą = đ???đ?&#x;?đ??ą .

đ??” đ??•

đ???đ?&#x;?đ??ą đ?&#x;? + đ???đ??˛ đ?&#x;?

đ???đ?&#x;? =

đ???đ?&#x;? = đ??­đ?&#x;? =

đ??•

đ???đ?&#x;?đ??ą . đ???đ?&#x;?đ??ą đ??•

đ??” đ?&#x;? đ??•

+ đ???đ??˛đ?&#x;?

114

Sh. M. Muneeb Zahid

Figure 3: Initially the Bodies are at Rest and at their 1st Position

Figure 4: In Same Time Both the Bodies Start Moving with Uniform Velocities Where 𝐝𝟐𝐱 =𝐝𝟏𝐱 . 𝐭𝟐 =

𝐝𝟏𝐱 𝐕

.

𝐔 𝐕

𝐔 𝐕

𝐝𝟑𝐱 = U 𝐭 𝟐 𝐝𝟏𝐱

𝐝𝟑𝐱 = U

𝐕

𝐝𝟑𝐱 = 𝐝𝟏𝐱 .

𝐝𝟑 =

𝐔𝟐 𝐕𝟐

𝐝𝟏𝐱 . 𝐝𝟑𝐱 𝐕

𝐝𝟏𝐱

𝐭𝟑 =

𝐕

𝐝𝟑𝐱 𝟐 + 𝐝𝐲 𝟐

𝐝𝟑 =

𝐭𝟑 =

𝐔

.

𝐕

𝐔𝟐

2

𝐕𝟐

+ 𝐝𝐲 𝟐

where 𝐝𝟑𝐱 = 𝐝𝟏𝐱 .

𝐔𝟐 𝐕𝟐

𝐔𝟐

.

𝐕𝟐

𝐝𝟒𝐱 = U 𝐭 𝟑 𝐝𝟒𝐱 = U.

𝐝𝟏𝐱 𝐕

𝐝𝟒𝐱 = 𝐝𝟏𝐱 . 𝐝𝟒 =

𝐝𝟒 =

𝐔𝟐

.

𝐕𝟐

𝐔𝟑 𝐕𝟑

𝐝𝟒𝐱 𝟐 + 𝐝𝐲 𝟐

𝐝𝟏𝐱 .

𝐔𝟑 𝐕𝟑

2

+ 𝐝𝐲 𝟐

𝐭𝟏, 𝐭𝟐, 𝐭𝟑_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ , 𝐭𝐧

115

Kinematics of Two Bodies in Terms of Geometric Series

đ???đ?&#x;?đ??ą đ??• đ??­đ?&#x;? đ??­đ?&#x;?

,

=

đ???đ?&#x;?đ??ą đ??”

. ,

đ??• đ??­đ?&#x;‘ đ??­đ?&#x;?

đ???đ?&#x;?đ??ą đ??” đ?&#x;?

đ??•

=

.

đ??•

đ??­ đ??§+đ?&#x;?

,__________

đ???đ?&#x;?đ??ą đ??•

đ??” đ??§âˆ’đ?&#x;? đ??•

đ??Ž

=

đ??­đ??§

đ??•đ?&#x;?

đ??•

It means that t1 , t 2 , _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ t n are in geometric progression. Therefore, đ??­ đ??§ =

đ???đ?&#x;?đ??ą

.

đ??•

đ??” đ??§âˆ’đ?&#x;? đ??•

____________ 1st Result

đ???đ?&#x;?đ??ą , đ???đ?&#x;?đ??ą , đ???đ?&#x;‘đ??ą , đ???đ?&#x;’đ??ą , _ _ _ _ _ _ _ _ _ _ _ đ???đ??§đ??ą đ??Ž

đ??”đ?&#x;?

đ??•

đ??•đ?&#x;?

đ???đ?&#x;?đ??ą , đ???đ?&#x;?đ??ą . , đ???đ?&#x;?đ??ą .

, đ???đ?&#x;?đ??ą .

đ??”đ?&#x;‘ đ??•đ?&#x;‘

, _ _ _ _ _ , đ???đ?&#x;?đ??ą .

đ??” đ??§âˆ’đ?&#x;? đ??•

đ???đ?&#x;?đ??ą , đ???đ?&#x;?đ??ą _ _ _ _ _ _ _ _ _ _ _ đ???đ??§đ??ą are also in geometric progression. Therefore, đ???đ??§đ??ą = đ???đ?&#x;?đ??ą .

đ??” đ??§âˆ’đ?&#x;? đ??•

____________ 2nd Result

đ???đ??§ = đ???đ??§đ??ą đ?&#x;? + đ???đ??˛ đ?&#x;?

đ???đ??§ = đ??­ đ??§+đ?&#x;? đ??­đ??§

đ???đ?&#x;?đ??ą . đ??? đ??š+đ?&#x;? đ??ą đ???đ??šđ??ą

=

đ?&#x;? đ??” đ??§âˆ’đ?&#x;? đ??•

+ đ???đ??˛ đ?&#x;? ____________ 3rd Result

đ??Ž

= ____________ 4th Result đ??•

In time „t n â€&#x; body „Aâ€&#x; travels distance „d(a+1)x â€&#x;and reaches to its (n +1)th position from its nth position.____________ 5th Result In time „t n â€&#x; body „Bâ€&#x; travels distance „dnx â€&#x; and reaches to its (n + 1)th position from its nth position.____________ 6thResult When the body „Aâ€&#x; and „Bâ€&#x; are at their nth position the distance between them is đ???đ??§ đ???đ??§đ??ą đ?&#x;? + đ???đ??˛ đ?&#x;? ____________ 7thResult

đ???đ??§ = đ??­đ?&#x;? = đ??­đ??§ =

đ???đ?&#x;?đ??ą đ??• đ???đ??§đ??ą đ??•

, đ??­đ?&#x;? =

đ???đ?&#x;?đ??ą đ??•

, đ??­đ?&#x;‘ =

đ???đ?&#x;‘đ??ą đ??•

____________ 8thResult

đ???đ?&#x;?đ??ą = U đ??­ đ?&#x;? , đ???đ?&#x;‘đ??ą = U đ??­ đ?&#x;? , đ???đ?&#x;’đ??ą = U đ??­ đ?&#x;‘ đ??? đ??§+đ?&#x;? đ??ą = U đ??­ đ??§ ____________ 9thResult The total time required by the body „Aâ€&#x; and body „Bâ€&#x; to reach at their (n +1) th position from their 1st position is denoted by „tâ€&#x;. „tâ€&#x; is the sum of all the times from t1 to t n t = đ??­đ?&#x;? + đ??­đ?&#x;? _ _ _ _ _ _ _ _ _ _ _ _ + đ??­đ??§ We know that,

116

Sh. M. Muneeb Zahid đ??š

đ??’đ??§ =

(1 - đ??Ť đ??§ ) ___________ (A)

đ?&#x;?−đ??Ť

đ??’đ??§ = ‘t’ đ???đ?&#x;?đ??ą

a = đ??­đ?&#x;? = t=

t=

đ??•

đ???đ?&#x;?đ??ą đ??• đ??” đ?&#x;?− đ??•

đ???đ?&#x;?đ??ą đ??•âˆ’đ??”

đ??”

, r = substitute these values in equation (A) đ??•

. đ?&#x;?−

đ?&#x;?−

đ??” đ??§ đ??•

đ??” đ??§ đ??•

____________ 10th Result

đ???đ??€ = U t đ???đ??€ =

đ???đ?&#x;?đ??ą .đ??” đ??•âˆ’đ??”

đ?&#x;?−

đ??” đ??§ đ??•

____________ 11th Result

đ???đ?? = V t dB =

đ???đ?&#x;?đ??ą .đ??• đ??•âˆ’đ??”

đ?&#x;?−

đ??” đ??§ đ??•

____________ 12th Result

Where „dA â€&#x; and „dB â€&#x; are the total distances travelled by the body „Aâ€&#x; and body „Bâ€&#x; in time „tâ€&#x;. 3rd CASE (When the Bodies are Moving in Non Parallel Lines) 

Kinematics of Two Bodies in Terms of Geometric Series Along X-Axis Suppose body „Aâ€&#x; and „Bâ€&#x; are at rest and at their 1st position the distance of body „Bâ€&#x; to body „Aâ€&#x;

is d1 = d1 x i + d1 y j as shown in figure „5â€&#x;. In the same time both the bodies start moving with uniform velocities. Body „Aâ€&#x; start moving with velocity U = Ux i + Uy j in line „L1 â€&#x; and body „Bâ€&#x; start moving with velocity V = Vx i + Vy j in line „L2 â€&#x;. Where d1x ,d1y ,Ux , Uy , Vx and Vy are positive. Line „L1 â€&#x; and „L2 â€&#x; are non-parallel. Body „Bâ€&#x; takes time „t1 â€&#x; to reach at the x-co-ordinate of body „Aâ€&#x;. In time „t1 â€&#x; body „Aâ€&#x; will travel distance d1A and „Bâ€&#x; will travel the distance d1B . Now „Aâ€&#x; and „Bâ€&#x; are in their 2nd position and distance of „Bâ€&#x; to „Aâ€&#x; is d2 as shown in figure „6â€&#x;.Then body „Bâ€&#x; takes time „t 2 â€&#x; to reach at the 2nd x co-ordinate of body „Aâ€&#x;. In time „t 2 â€&#x; body „Aâ€&#x; will travel distance „d2A â€&#x; and body „Bâ€&#x; will travel distance „d2B â€&#x; now both the bodies are at their 3rd position and the distance of body „Bâ€&#x; to „Aâ€&#x; is „d3 â€&#x; and so on. Follow figure „5â€&#x; and „6â€&#x;. đ??­đ?&#x;? =

đ???đ?&#x;?đ??ą đ??•đ??ą

đ???đ?&#x;?đ??€ = đ??” đ??­ đ?&#x;? Where, đ??” = đ??”đ??ą đ??˘ + đ??”đ??˛ đ??Ł , đ??­ đ?&#x;? = đ???đ?&#x;?đ??€ = đ??”đ??ą .

đ???đ?&#x;?đ??ą đ??•đ??ą

đ??˘ + đ??”đ??˛ .

đ???đ?&#x;?đ??ą đ??•đ??ą

đ???đ?&#x;?đ?? = đ??• đ??­ đ?&#x;? đ???đ?&#x;?đ?? = đ??•đ??ą đ??˘ + đ??•đ??˛ đ??Ł . đ???đ?&#x;?đ?? = đ???đ?&#x;?đ??ą đ??˘ + đ??•đ??˛ .

đ???đ?&#x;?đ??ą đ??•đ??ą đ???đ?&#x;?đ??ą đ??•đ??ą

đ??Ł

đ??Ł

đ???đ?&#x;?đ??ą đ??•đ??ą

Kinematics of Two Bodies in Terms of Geometric Series

Figure 5: Initially Both the Bodies are at Rest and at Their 1st Position

Figure 6: In Same Time Both the Bodies Start Moving with Uniform Velocities and in Non Parallel Lines đ???đ?&#x;? = đ???đ?&#x;?đ??˛ − đ??•đ??˛ .

đ???đ?&#x;?đ??ą đ??•đ??ą

đ??Ł + đ???đ?&#x;?đ??€

117

118

Sh. M. Muneeb Zahid 𝐝𝟏𝐱

𝐝𝟐 = 𝐝𝟏𝐲 − 𝐕𝐲 . 𝐝𝟐 = 𝐔 𝐱 . 𝐝𝟐𝐱 = 𝐔𝐱 .

𝐝𝟏𝐱

𝐢 + 𝐝𝟏𝐲 − 𝐕𝐲 .

𝐕𝐱

𝐢 + 𝐔𝐲 .

𝐕𝐱 𝐝𝟏𝐱 𝐕𝐱

+ 𝐔𝐲 .

𝐝𝟏𝐱

𝐝𝟏𝐱 𝐕𝐱

𝐕𝐱

𝐣

𝐣

𝐕𝐱 𝐝𝟏𝐱

+ 𝐔𝐲 .

𝐕𝐱

𝐝𝟏𝐱 𝐕𝐱

𝐝𝟐𝐱 𝐕𝐱

Where 𝐝𝟐𝐱 = 𝐔𝐱 . 𝐭𝟐 =

𝐝𝟏𝐱

𝐝𝟏𝐱

𝐝𝟐𝐲 = 𝐝𝟏𝐲 − 𝐕𝐲 . 𝐭𝟐 =

𝐣 + 𝐔𝐱 .

𝐕𝐱

𝐝𝟏𝐱 𝐕𝐱

.

𝐝𝟏𝐱 𝐕𝐱

𝐔𝐱 𝐕𝐱

𝐝𝟐𝑨 = 𝐔𝐭 𝟐 𝐝𝟏𝐱

𝐝𝟐𝑨 = 𝐔𝐱 𝐢 + 𝐔𝐲 𝐣 𝐝𝟐𝐀 = 𝐝𝟏𝐱 .

𝐔𝐱𝟐

𝐔𝐱 𝐕𝐱

𝐝𝟏𝐱 .𝐔 𝐱 .𝐔 𝐲

𝐢+

𝐕𝐱 𝟐

.

𝐕𝐱

𝐣

𝐕𝐱 𝟐

𝐝𝟐𝐁 = 𝐕𝐭 𝟐 𝐝𝟏𝐱

𝐝𝟐𝐁 = 𝐕𝐱 𝐢 + 𝐕𝐲 𝐣 𝐝𝟐𝐁 = 𝐝𝟏𝐱 . 𝐝𝟑 = 𝐝𝟐𝐲 −

𝐔𝐱

𝐕𝐱

𝐔𝐱 𝐕𝐱

𝐝𝟏𝐱 .𝐔 𝐱 .𝐕𝐲

𝐢+

𝐕𝐱

.

𝐕𝐱 𝟐

𝐝𝟏𝐱 .𝐔 𝐱 .𝐕𝐲 𝐕𝐱 𝟐

𝐣

𝐣 + 𝐝𝟐𝐀

Substitute the values of d2y and d2A in above equation. 𝐝𝟑 = 𝐝𝟏𝐲 − 𝐕𝒚 . 𝐔𝐱 𝟐

𝐝𝟑 = 𝐝𝟏𝐱 . 𝐝𝟑𝐱 = 𝐝𝟏𝐱 .

𝐝𝟏𝐱 𝐕𝐱

+

𝐕𝐱

𝐕𝐱

𝐝𝟑𝐀 = 𝐔 𝐭 𝟑

𝐕𝐱

−

𝐝𝟏𝐱 .𝐔 𝐱 .𝐕𝐲

𝐝𝟏𝐱 𝐕𝐱

𝐕𝐱

−

𝟐

𝐣 + 𝐝𝟏𝐱 .

𝐝𝟏𝐱 .𝐔 𝐱 .𝐕𝐲 𝐕𝐱 𝟐

+ 𝐔𝐲 .

𝐔𝐱𝟐 𝐕𝐱 𝟐

𝐝𝟏𝐱 𝐕𝐱

𝐕𝐱 𝟐

𝐝𝟑𝐱

𝐝𝟏𝐱

𝐝𝟏𝐱

𝐔𝐱 𝟐

Where 𝐝𝟑𝐱 = 𝐝𝟏𝐱 . 𝐭𝟑 =

𝐕𝐱

+ 𝐔𝒚 .

𝐢 + 𝐝𝟏𝐲 − 𝐕𝒚 .

𝐕𝐱 𝟐

𝐝𝟑𝐲 =𝐝𝟏𝐲 - 𝐕𝒚 . 𝐭𝟑 =

𝐝𝟏𝐱

.

𝐔𝐱𝟐 𝐕𝐱 𝟐

𝐔𝐱 𝟐 𝐕𝐱 𝟐

𝐝𝟏𝐱 .𝐔 𝐱 .𝐕𝐲 𝐕𝐱 𝟐

+ 𝐔𝒚 .

𝐝𝟏𝐱 𝐕𝐱

+

𝐝𝟏𝐱 .𝐔 𝐱 .𝐔 𝐲 𝐕𝐱 𝟐

+

𝐢+

𝐝𝟏𝐱 .𝐔 𝐱 .𝐔 𝐲 𝐕𝐱 𝟐

𝐝𝟏𝐱 .𝐔 𝐱 .𝐔 𝐲 𝐕𝐱 𝟐

𝐣

𝐣

119

Kinematics of Two Bodies in Terms of Geometric Series

𝐝𝟏𝐱

𝐝𝟑𝐀 = 𝐔𝐱 𝐢 + 𝐔𝐲 𝐣 𝐝𝟑𝐀 = 𝐝𝟏𝐱 .

𝐔𝐱𝟑

𝐕𝐱

𝐔𝐱 𝟐 𝐕𝐱 𝟐

𝐝𝟏𝐱 .𝐔 𝐲 .𝐔 𝐱 𝟐

𝐢+

𝐕𝐱 𝟑

.

𝐕𝐱 𝟑

𝐣

𝐝𝟑𝐁 = 𝐕 𝐭 𝟑 𝐝𝟏𝐱

𝐝𝟑𝐁 = 𝐕𝐱 𝐢 + 𝐕𝐲 𝐣 𝐝𝟑𝐁 = 𝐝𝟏𝐱 .

𝐔𝐱𝟐 𝐕𝐱

𝐕𝐱

𝐕𝐱 𝟐

𝐕𝐱 𝟑

𝐝𝟏𝐱 .𝐔 𝐱 𝟐 .𝐕𝐲

𝐝𝟒 = 𝐝𝟑𝐲 −

𝐔𝐱𝟐

𝐝𝟏𝐱 .𝐔 𝐱 𝟐 .𝐕𝐲

𝐢+

𝟐

.

𝐕𝐱 𝟑

𝐣

𝐣 + 𝐝𝟑𝐀

Substitute the values of d3y and d3A in above equation we get 𝐝𝟒 = 𝐝𝟏𝐱 .

𝐔𝐱 𝟑 𝐕𝐱

𝐢+ 𝐝𝟏𝐲 − 𝐕𝐲 .

𝟑

𝐝𝟏𝐱 𝐕𝐱

−

𝐝𝟏𝐱 .𝐔 𝐱 .𝐕𝐲 𝐕𝐱

𝟐

−

𝐝𝟏𝐱 .𝐔 𝐱 𝟐 .𝐕𝐲

𝐝𝟏𝐀 = 𝐔 𝐭 𝟏 𝐝𝟐𝐀 = 𝐔 𝐭 𝟐 𝐝𝟑𝐀 = 𝐔 𝐭 𝟑 Therefore, 𝐝𝐧𝐀 = 𝐔 𝐭 𝐧 _________ 1st Result. Also, 𝐝𝟏𝐁 = 𝐕 𝐭 𝟏 𝐝𝟐𝐁 = 𝐕 𝐭 𝟐 𝐝𝟑𝐁 = 𝐕 𝐭 𝟑 Therefore, 𝐝𝐧𝐁 = 𝐕 𝐭 𝐧 _________ 2ndResult. 𝐭𝟏 = 𝐭𝟐 = 𝐭𝟑 = 𝐭𝟐 𝐭𝟏

=

𝐝𝟏𝐱 𝐕𝐱 𝐔𝐱 𝐕𝐱

.

𝐔𝐱 𝟐 𝐕𝐱 𝟐 𝐭𝟑 𝐭𝟐

=

𝐝𝟏𝐱 𝐕𝐱

.

𝐝𝟏𝐱 𝐕𝐱

𝐭 𝐧+𝟏 𝐭𝐧

=

Therefore, 𝐭 𝐧 = Put 𝐭 𝐧 =

𝐝𝟏𝐱 𝐕𝐱

𝐝𝐧𝐀 = 𝐔 𝐭 𝐧

.

𝐔𝐱 𝐕𝐱 𝐝𝟏𝐱 𝐕𝐱

.

𝐔 𝐱 𝐧−𝟏 𝐕𝐱

𝐔 𝐱 𝐧−𝟏 𝐕𝐱

_________ 3rd Result.

in 1st result.

𝐕𝐱

𝟑

+ 𝐔𝐲 .

𝐝𝟏𝐱 𝐕𝐱

+

𝐝𝟏𝐱 .𝐔 𝐱 .𝐔 𝐲 𝐕𝐱

𝟐

+

𝐝𝟏𝐱 .𝐔 𝐱 𝟐 . 𝐔 𝐲 𝐕𝐱 𝟑

𝐣

120

Sh. M. Muneeb Zahid

Where 𝐔 = 𝐔𝐱 𝐢 + 𝐔𝐲 𝐣 𝐝𝟏𝐱

𝐝𝐧𝐀 = 𝐔𝒙 .

𝐕𝐱

𝐕𝐱

𝐝𝟏𝐱 𝐕𝐱

𝐝𝟏𝐱

𝐣

𝐕𝐱

𝐔 𝐱 𝐧−𝟏

.

𝐕𝐱

𝐔 𝐱 𝐧−𝟏

𝐣 _________ 4th Result.

𝐕𝐱

𝐔 𝐱 𝐧−𝟏

.

𝐕𝐱

𝐔 𝐱 𝐧−𝟏

𝐝𝐧𝐁 = 𝐝𝟏𝐱 .

𝐕𝐱

.

in 2nd Result

𝐕𝐱

𝐝𝐧𝐁 = 𝐕𝐱 𝐢 + 𝐕𝐲 𝐣

𝐝𝟏𝐱

𝐢 + 𝐔𝒚 .

𝐔 𝐱 𝐧−𝟏

.

𝐝𝟏𝐱

𝐢 + 𝐔𝒚 .

𝐕𝐱

𝐔𝐱 𝐧

𝐝𝐧𝐀 = 𝐝𝟏𝐱 . Put 𝐭 𝐧 =

𝐔 𝐱 𝐧−𝟏

.

𝐕𝐱

𝐢+

𝐕𝐱

𝐝𝟏𝐱 .𝐕𝐲 𝐔 𝐱 𝐧−𝟏 𝐕𝐱

𝐣_________ 5th Result.

𝐕𝐱

We know that 𝐝𝟏 = 𝐝𝟏𝐱 𝐢 + 𝐝𝟏𝐲 𝐣 𝐝𝟐 = 𝐝𝟏𝐱 .

𝐔𝐱

𝐔𝐱 𝟐

𝐝𝟑 = 𝐝𝟏𝐱 . 𝐝𝟒 = 𝐝𝟏𝐱 .

𝐢 + 𝐝𝟏𝐲 − 𝐕𝐲 .

𝐕𝐱

𝐕𝐱

𝐢 + 𝐝𝟏𝐲 − 𝐕𝐲 .

𝐕𝐱 𝟐

𝐔𝐱 𝟑 𝐕𝐱

𝐝𝟏𝐱

𝐢+ 𝐝𝟏𝐲 – 𝐕𝐲 .

𝟑

+ 𝐔𝐲 .

𝐝𝟏𝐱 𝐕𝐱

𝐝𝟏𝐱 𝐕𝐱

+

+

𝐝𝟏𝐱 𝐕𝐱

𝐕𝐲 .𝐔 𝐱 .𝐝𝟏𝐱 𝐕𝐱 𝟐

𝐕𝐲 .𝐔 𝐱 .𝐝𝟏𝐱 𝐕𝐱

𝟐

+

𝐣 + 𝐔𝐲 .

𝐝𝟏𝐱 𝐕𝐱

𝐕𝐲 . 𝐔 𝐱 𝟐 .𝐝𝟏𝐱 𝐕𝐱

𝟑

+

𝐔 𝐲 .𝐔 𝐱 .𝐝𝟏𝐱 𝐕𝐱 𝟐

+ 𝐔𝐲 .

𝐝𝟏 𝐱 𝐕𝐱

+

𝐣 𝐔 𝐲 .𝐔 𝐱 .𝐝𝟏𝐱 𝐕𝐱

𝟐

+

𝐔 𝐲 . 𝐔 𝐱 𝟐 . 𝐝𝟏𝐱 𝐕𝐱 𝟑

𝐣

Now I am finding dn 𝐕𝐲 . 𝐭𝟐 𝐭𝟏

=

𝐝𝟏𝐱 𝐕𝐱 𝐭𝟑 𝐭𝟐

,

=

𝐕𝐲 .𝐝𝟏𝐱 .𝐔 𝐱 𝐕𝐱 𝐔𝐱 𝐕𝐱

𝟐

,

𝐚 𝟏−𝐫

Put a = 𝐕𝐲 .

𝐒𝐧−𝟏 =

𝐒𝐧−𝟏 =

𝐚 𝟏−𝐫

------------------------- are in geometric progression in which

𝐝𝟏𝐱 𝐕𝐱

𝟏 − 𝐫𝐧

𝟏 − 𝐫 𝐧−𝟏 𝐝𝟏𝐱 𝐕𝐱

𝐝 𝐕𝒚 . 𝟏𝐱

𝟏−

𝐕𝐱 𝟑

= r, a = 𝐕𝐲 .

We know that 𝐒𝐧 = 𝐒𝐧−𝟏 =

𝐕𝐲 .𝐝𝟏𝐱 .𝐔 𝐱 𝟐

𝐕𝐱 𝐔𝐱 𝐕𝐱

𝐕𝐲 . 𝐝𝟏𝐱 𝐕𝐱 − 𝐔 𝐱

and r =

𝐔𝐱 𝐕𝐱

𝐔 𝐱 𝐧−𝟏

𝟏 −

𝐕𝐱 𝐔 𝐱 𝐧−𝟏

𝟏 −

𝐕𝐱

Also, 𝐔𝐲 .

𝐝𝟏𝐱 𝐕𝐱

,

𝐔 𝐲 . 𝐝𝟏𝐱 . 𝐔 𝐱 𝐕𝐱

𝟐

We know that 𝐒𝐧 = 𝑺𝒏−𝟏 =

𝐚 𝟏−𝐫

,

𝐔 𝐲 . 𝐝𝟏𝐱 . 𝐔 𝐱 𝟐 𝐕𝐱 𝟑 𝐚

𝟏−𝐫

𝟏 − 𝐫 𝐧−𝟏

, --------------------------- are in geometric progression in which a = 𝐔𝐲 .

𝟏 − 𝐫𝐧

𝐝𝟏𝐱 𝐕𝐱

,r=

𝐔𝐱 𝐕𝐱

121

Kinematics of Two Bodies in Terms of Geometric Series

Put a = đ??”đ??˛ .

đ?‘şđ?’?−đ?&#x;? =

đ?‘şđ?’?−đ?&#x;? =

đ???đ?&#x;?đ??ą

,r=

đ??•đ??ą

đ??? đ??” đ??˛ đ?&#x;?đ??ą

đ??•đ??ą đ??” đ?&#x;?− đ??ą đ??•đ??ą

đ??” đ??˛ . đ???đ?&#x;?đ??ą đ??•đ??ą − đ??” đ??ą

đ??”đ??ą đ??•đ??ą

in above equation

đ??” đ??ą đ??§âˆ’đ?&#x;?

đ?&#x;? −

đ??•đ??ą đ??” đ??ą đ??§âˆ’đ?&#x;?

đ?&#x;? −

đ??•đ??ą

đ???đ?&#x;? = đ???đ?&#x;?đ??ą đ??˘ + đ???đ?&#x;?đ??˛ đ??Ł đ???đ?&#x;? = đ???đ?&#x;?đ??ą . đ???đ??§ = đ???đ?&#x;?đ??ą .

đ??”đ??ą

đ??˘ + đ???đ?&#x;?đ??˛ − đ??•đ??˛ .

đ??•đ??ą

đ??” đ??ą đ??§âˆ’đ?&#x;?

đ??˘+ đ???đ?&#x;?đ??˛ −

đ??•đ??ą

đ???đ?&#x;?đ??ą đ??•đ??ą

+ đ??”đ??˛ .

đ??•đ??˛ . đ???đ?&#x;?đ??ą đ??•đ??ą − đ??” đ??ą

đ?&#x;?–

đ???đ?&#x;?đ??ą đ??•đ??ą

đ??Ł

đ??” đ??ą đ??§âˆ’đ?&#x;?

+

đ??•đ??ą

đ??” đ??˛ . đ???đ?&#x;?đ??ą đ??•đ??ą − đ??” đ??ą

đ?&#x;? −

đ??” đ??ą đ??§âˆ’đ?&#x;? đ??•đ??ą

đ??Ł--- 6th Result

Body „Aâ€&#x; takes time ‘tn â€&#x; and travel distance dnA to reach at its (n+1)th position from its nth position. ___________ 7th Result. Body „Bâ€&#x; takes time „tn â€&#x; and travel distance „dnB â€&#x; to reach at its (n+1)th position from its nth position. ___________ 8th Result. When the body „Aâ€&#x; and body „Bâ€&#x; are at their nth position the distance of body „Bâ€&#x; to body „Aâ€&#x; is dn . ___________ 9th Result. The total time required by the body „Aâ€&#x; and body „Bâ€&#x; to reach at their (n+1)th position from their 1st position is ‘t’. Also t is the sum of all the times from „t1 â€&#x; to „tn â€&#x;. t = đ??­ đ?&#x;? +đ??­ đ?&#x;? +đ??­ đ?&#x;‘ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ + đ??­ đ??§ We know that đ??’đ??§ = đ??’đ??§ = t, a = đ??­ đ?&#x;? =

đ???đ?&#x;?đ??ą

đ??š

,r=

đ??•đ??ą

đ?&#x;? − đ??Ťđ??§

đ?&#x;?−đ??Ť

đ??”đ??ą đ??•đ??ą

Substitute all these values in above equation t=

t=

đ???đ?&#x;?đ??ą đ??•đ??ą đ??” đ?&#x;?− đ??ą đ??•đ??ą

đ???đ?&#x;?đ??ą đ??•đ??ą − đ??” đ??ą

đ??”đ??ą đ??§

đ?&#x;? −

đ?&#x;? −

đ??•đ??ą đ??”đ??ą đ??§ đ??•đ??ą

___________ 10th Result.

The total distance travelled by the body „Aâ€&#x; in time „tâ€&#x; is đ??ƒđ??€ . đ??ƒđ??€ = đ??” t đ??ƒđ??€ = đ??” đ??ą đ??˘ + đ??” đ??˛ đ??Ł đ??ƒđ??€ =

đ???đ?&#x;?đ??ą . đ??” đ??ą đ??•đ??ą − đ??” đ??ą

đ?&#x;?−

đ???đ?&#x;?đ??ą đ??•đ??ą − đ??” đ??ą đ??”đ??ą đ??§ đ??•đ??ą

đ?&#x;?− đ??˘+

đ??”đ??ą đ??§ đ??•đ??ą

đ???đ?&#x;?đ??ą . đ??” đ??˛ đ??•đ??ą − đ??” đ??ą

đ?&#x;?–

đ??”đ??ą đ??§ đ??•đ??ą

đ??Ł________ 11th Result.

The total distance travelled by the body „Bâ€&#x; in time „tâ€&#x; is đ??ƒđ?? = đ??• t

đ??ƒđ?? .

122

Sh. M. Muneeb Zahid

đ??ƒđ?? = đ??•đ??ą đ??˘ + đ??•đ??˛ đ??Ł đ??ƒđ?? =

đ???đ?&#x;?đ??ą .đ??•đ??ą đ??•đ??ą − đ??” đ??ą

đ?&#x;?–

đ???đ?&#x;?đ??ą đ??•đ??ą − đ??” đ??ą đ??”đ??ą đ??§ đ??•đ??ą

đ?&#x;?−

đ??˘+

đ??”đ??ą đ??§ đ??•đ??ą

đ???đ?&#x;?đ??ą đ??•đ??˛ đ??•đ??ą − đ??” đ??ą

đ?&#x;?–

đ??”đ??ą đ??§ đ??•đ??ą

đ??Ł_________ 12th Result.

CONCLUSIONS It is obvious that motion of two bodies with uniform velocity can be described in terms of Geometric Series. A tool is proved “Geometric Progression� for the easiness of system of two bodies moving in straight line with uniform velocities.

ADVANTAGES OF USING THIS RESEARCH 

This research is very important in point to point analysis for system of bodies moving with uniform velocities.



By using this research we can find the exact position of one body with respect to another body at any instant of time.



We know that when two bodies are moving with uniform velocities then their time and distance varies according to geometric progression hence we can easily apply all the formulas of geometric progression in it.



This research will give 100% accurate results.



This research can help scientists for the point to point analysis of the bodies which are moving in any curve path for example circular path, elliptical path etc.



This research can help scientists to find the position of other moving objects in space with respect to earth at any instant of time.



By using this research we can easily perform complex and lengthy calculations by putting the values in equations for example. 1) If we have to find the distance between two bodies when they are at their 1000000000 th position then we can easily calculate by simply putting n= 109. 2) If we have to find the time required by both the bodies to reach at their 10000th position from their 9999th position then we can easily calculate by putting n= 9999 in the equation of time as derived in results and discussions. 3) Many other parameters can be easily calculated by using this research.



By using this research we can easily find the direction of distance and velocities of objects moving with uniform velocities.

BACKGROUND Once I had watched a movie in which a group of boys was having a racing competition between them. When the race started a boy fell down due to which he was left behind. I got the idea that in order to win the race the velocity of the boy left behind should be greater than the velocity of boy at front. Then I assumed that the velocity of the boy at back is V and the velocity of boy at front is U and the distance between is d1. I used the approach that when the boy at back reaches to the position of boy at front then in the same time the boy at front travels the distance d2. When the boy at back reaches the 2nd position of the boy at front then in the same time the boy at front travels the distance d3. Then I found that d1, d2, d3, _______________, dn and t1, t2, t3, ________________ tn are in geometric progression. Then by using the formulas of geometric series I derived the equations.

Kinematics of Two Bodies in Terms of Geometric Series

123

ACKNOWLEDGEMENTS During this research I am very thankful to Dr. Muzaffar Mahmood (Pro Vice Chancellor NEDUET), Dr. Mahmood Khan Pathan, Dr. Saqib Anjum (Chairman Department of Physics NEDUET), Mr. Ghulam Mustafa (Lecturer Department of Physics NEDUET) and Dr Zaheer uddin (Assistant Professor University of Karachi (FSL), Pakistan)

REFERENCES 1.

Chapter 31 out of 37 from Discrete Mathematics for Neophytes: Number Theory, Probability, Algorithms, and Other Stuff by J. M. Cargal

2.

Chapter 16 and 20 from Engineering Mechanics Dynamics 10 th edition by R.C Hibbeler

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