Question

check the dimenional correctness of equation v=u+at​

Answer 1

Answer:

Given equation:

V = u + at

Dimensions on the Left hand side:

[v] = [M^0 L^1 T^-1] (1)

Dimensions on the RHS:

According to principle of homogeneity, the quantities on the LHS and RHS that are being added or subtracted must have the same dimensional formula.

Therefore,

[u] = [at] = [v] = [M^0 L^1 T^-1] (from 1)

Now,

[u] = M^0 L^1 T^-1

Hence, [u] = [v]

Now,

[at] = [(M^0 L^1 T^-2)/(T^1)] = [M^0 L^1 T^-1]

Since,

dimensions of all the terms are equal, hence the given eq is dimensionally correct.

Answer 2

Given equation :-

v = u + at

We know,

• dimensional formula of v = (M⁰LT⁻¹)

• dimensional formula of u = (M⁰LT⁻¹)

• dimensional formula of a = (M⁰LT⁻²)

• dimensional formula of t = (M⁰L⁰T)

→ LHS  is the dimensional formula of v = (M⁰LT⁻¹)

→ RHS = dimensional formula of u + dimensional formula of a × dimensional formula of t

= (M⁰LT⁻¹) + (M⁰LT⁻²) × (M⁰L⁰T)

= (M⁰LT⁻¹) + (M⁰LT⁻¹)

= 2(M⁰LT⁻¹)

2 is a dimensionless constant, so LHS = RHS

Hence,the given relationship(equation) is dimensionally correct.