Question

How to check the correctness of the equation v^2-u^2=2as dimensionally

Answer 1

Answer:

v² – u² = 2as

v = final velocity

(Unit = m/s)

Dimension = [L / T]

u = initial velocity

(Unit = m/s)

Dimension = [L / T]

a = acceleration

(Unit = m/s²)

Dimension = [L / T²]

s = displacement

(Unit = m)

Dimension = [L]

The equation —

v² – u² = 2as

Dimension-wise —

[L / T]² – [L / T]² = 2 [L / T²] [L]

When considering dimensions,

》Subtraction of similar dimensions results in the same dimension

》Multiplication with a constant doesn't change the dimension

So,

[L / T]² – [L / T]² = 2 [L / T²] [L]

=> [L² / T²] – [L² / T²] = 2 [L / T²] [L]

=> [L² / T²] = [L² / T²]

=> LHS = RHS

Hence, the equation is correct.

Hope this helps you :)

Answer 2

Answer:

The dimensional correctness of the equation $v^{2} -u^{2}$ = $2aS$ can be checked by checking the dimensions of each term.

Dimension of $v^{2}$ = $M^{0} L^{2} T^{-2}$

Dimension of $u^{2}$ = $M^{0} L^{2} T^{-2}$

Dimension of 2aS = $M^{0} L^{2} T^{-2}$

Explanation:

We know velocity = distance ÷ time

Dimension of distance = $M^{0} L^{1} T^{0}$

Dimension of time = $M^{0} L^{0} T^{1}$

∴ Dimension of velocity = $M^{0} L^{1} T^{0}$ ÷ $M^{0} L^{0} T^{1}$

                                       = $M^{0} L^{1} T^{-1}$

∴Dimension of $v^{2} = u^{2}$ = $(M^{0} L^{1} T^{-1})^{2}$

                                     = $M^{0} L^{2} T^{-2}$

We know acceleration a = velocity ÷ time

∴ Dimension of a = $M^{0} L^{1} T^{-1}$ ÷ $M^{0} L^{0} T^{1}$

                             = $M^{0} L^{1} T^{-2}$

Dimension of displacement S = $M^{0} L^{1} T^{0}$

∴Dimension of 2aS = $M^{0} L^{1} T^{-2}$ × $M^{0} L^{1} T^{0}$

                                = $M^{0} L^{2} T^{-2}$

Since dimension of $v^{2} = u^{2}$ = 2aS, the equation is dimensionally correct.