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x n = u n + 1 2 a n 2 x n - 1 = u ( n - 1 ) + 1 2 a ( n - 1 ) 2

The displacement in the n th second, therefore, is :

s n = x 2 - x 1 = u n + 1 2 a n 2 - u n + u - 1 2 a n 2 - 1 2 a + a n s n = x 2 - x 1 = u - 1 2 a + a n

s n = x 2 - x 1 = u + a 2 ( 2 n - 1 )

Problem : The equation of motion for displacement in n th second is given by :

x n = u + a 2 ( 2 n - 1 )

This equation is dimensionally incompatible, yet correct. Explain.

Solution : Since “n” is a number, the dimension of the terms of the equation is indicated as :

[ x n ] = [ Displacment ] = [ L ] [ u ] = [ velocity ] = [ L T - 1 ] [ a 2 ( 2 n - 1 ) ] = [ acceleration ] = [ L T - 2 ]

Clearly, dimensions of the terms are not same and hence equation is apparently incompatible in terms of dimensions.

In order to resolve this apparent incompatibility, we need to show that each term of right hand side of the equation has dimension that of displacement i.e. length. Now, we know that the relation is derived for a displacement for time equal to “1” second. As multiplication of “1” or " 1 2 " with any term is not visible, the apparent discrepancy has appeared in otherwise correct equation. We can, therefore, re-write the equation with explicit time as :

x n = u x 1 + a 2 ( 2 n - 1 ) x 1 2 x n = u x t + a 2 ( 2 n - 1 ) x t 2

Now, the dimensions of the first and second terms on the right side are :

[ u x t ] = [ L T - 1 T ] = [ L ] [ a 2 ( 2 n - 1 ) x t 2 ] = [ L T - 2 T 2 ] = [ L ]

Thus, we see that the equation is implicitly correct in terms of dimensions.

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If a particle, moving in straight line, covers distances “x”, “y” and “z” in p th , q th and r th seconds respectively, then prove that :

( p - q ) z + ( r - p ) y + ( q - r ) x = 0

This is a motion with constant acceleration in one dimension. Let the particle moves with a constant acceleration, “a”. Now, the distances in particular seconds as given in the question are :

x = u + a 2 ( 2 p - 1 ) y = u + a 2 ( 2 q - 1 ) z = u + a 2 ( 2 r - 1 )

Subtracting, second from first equation,

x - y = a 2 ( 2 p - 1 - 2 q + 1 ) = a ( p - q ) ( p - q ) = ( x - y ) a ( p - q ) z = ( x - y ) z a

Similarly,

( r - p ) y = ( z - x ) y a ( q - r ) x = ( y - z ) x a

Adding three equations,

( p - q ) z + ( r - p ) y + ( q - r ) x = 1 a ( x y - y z + y z - x y + x y - x z ) ( p - q ) z + ( r - p ) y + ( q - r ) x = 0

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Average acceleration

Average acceleration is ratio of change in velocity and time. This is an useful concept where acceleration is not constant throughout the motion. There may be motion, which has constant but different values of acceleration in different segments of motion. Our job is to find an equivalent constant acceleration, which may be used to determine attributes for the whole of motion. Clearly, the single value equivalent acceleration should be such that it yields same value of displacement and velocity for the entire motion. This is the underlying principle for determining equivalent or average acceleration for the motion.

Problem : A particle starting with velocity “u” covers two equal distances in a straight line with accelerations a 1 and a 2 . What is the equivalent acceleration for the complete motion?

Solution : The equivalent acceleration for the complete motion should yield same value for the attributes of the motion at the end of the journey. For example, the final velocity at the end of the journey with the equivalent acceleration should be same as the one calculated with individual accelerations.

Here initial velocity is "u". The velocity at the end of half of the distance (say “x”) is :

v 1 2 = u 2 + 2 a 1 x

Clearly, v 1 is the initial velocity for the second leg of motion,

v 2 2 = v 1 2 + 2 a 2 x

Adding above two equations, we have :

v 2 2 = u 2 + 2 ( a 1 + a 2 ) x

This is the square of velocity at the end of journey. Now, let “a” be the equivalent acceleration, then applying equation of motion for the whole distance (2x),

v 2 2 = u 2 + 2 a ( 2 x )

Comparing equations,

a = a 1 + a 2 2

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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