Derive an expression using the method of dimensions, the distance travelled by a body during an interval of time t, if its initial velocity is u and uniform acceleration is a.

Derive an expression using the method of dimensions, the distance trav

1.	Derive an expression using the method of dimensions, the distance travelled by a body during an interval of time t, if its initial velocity is u and uniform acceleration is a.
Answer» Solution :Let the distance travelled by the body S be GIVEN by `S=ku^(x)d^(y)t^(z)` Substituting the dimensions on both sides `L=[LT^(-1)]^(x)[LT^(-2)]^(y)[T]^(z)` `M^(2)T^(0)L=L^(x+y)T^(-x-2y+z)` Equating the dimensions of L and T on both sides `x+y=1`..........i `-x-2y+z=0` `x+2y=z`....ii We have three variables x,y and z but to solve we have only TWO equations, so we split the problem into two parts `S=S_(1)+S_(2)` Where `S_(1)` is the distance covered with a=0 and `S_(2)` is the distance with U=0 Then `S_(1)=k_(1)u^(a)t^(B)`.......iii Substituting the dimensions `L=(LT^(-1))^(a)(T)^(b)=L^(a)T^(-a)T^(b)=L^(a)T^(-a+b)` Equating the dimensions of L and T on both sides. `a=1, -a+b=0, b=a=1` Substituting in equation (iii) `S_(1)=k_(1)ut`............iv For the second part `S_(2)=k_(2)a^(c)t^(d)`.........v Substituting the dimensionis `L=[LT^(-2)]^(c)[T]^(d)=L^(c)T^(-2c+d)` Equating the dimensions of L and T on both sides `c=1, -2c+d=0, d=2c=2` Substituting in equation (v) `S_(2)=k_(2)at^(2)` `S=S_(1)+S_(2)=k_(1)ut+k_(2)at^(2)`................vi