Question

Using the dimensional analysis, check the correctness of the relation : v² = u²+2as

Answer 1

Answer:

Heya mate,

dimension of v^2= L^2/T2=L^2T^-2

dimension of u^2= L^2/T2=L^2T^-2

dimension of 2as= (L/T^2) (L)=L^2T^-2 -------(because a is acceleration and s is distance(L)) Also, we not considering 2 because it is dimensionless constant.

Now , by Principle of homogenity, it should be

dimension of v^2=dimension of u^2=dimension of 2as

therefore, putting the respective dimensions,

[L^2/T2=L^2T^-2]=[L^2/T2=L^2T^-2]=[L^2/T2=L^2T^-2]

Thus, it is a correct equation as it follows , the principle of homogenity i.e. LHS=RHS

or v^2=u^2+2as.

THANKYOU

Answer 2

Answer:

dimension of v^2= L^2/T2=L^2T^-2

dimension of u^2= L^2/T2=L^2T^-2

dimension of 2as= (L/T^2) (L)=L^2T^-2 -------(because a is acceleration and s is distance(L)) Also, we not considering 2 because it is dimensionless constant.

Now , by Principle of homogenity, it should be

dimension of v^2=dimension of u^2=dimension of 2as

therefore, putting the respective dimensions,

[L^2/T2=L^2T^-2]=[L^2/T2=L^2T^-2]=[L^2/T2=L^2T^-2]

Thus, it is a correct equation as it follows , the principle of homogenity i.e. LHS=RHS

or v^2=u^2