Question

Check the correctness of the given formula:

1). v = u + at.
2). s = ut + 1/2 at².
3). v² = u² + 2as.

Answer 1

Correct Question:

Check the correctness of the given formula dimensionally:

1). v = u + at.

2). s = ut + 1/2 at².

3). v² = u² + 2as.

ANSWERS:

1). v = u + at

Firstly we write the dimensions of all the variables,

$\sf{\implies v = [M^{0}L^{1}T^{-1}]}$

$\sf{\implies u = [M^{0}L^{1}T^{-1}]}$

$\sf{\implies a = [M^{0}L^{1}T^{-2}]}$

$\sf{\implies t = [M^{0}L^{0}T^{1}]}$

Put the following dimensions in given formula,

$\sf{\implies[M^{0}L^{1}T^{-1}]=[M^{0}L^{1}T^{-1}] + [M^{0}L^{1}T^{-2}] \times [M^{0}L^{0}T^{1}]}$

$\sf{\implies[M^{0}L^{1}T^{-1}]=[M^{0}L^{1}T^{-1}] + [M^{0}L^{1}T^{-1}]}$

$\sf{\implies[M^{0}L^{1}T^{-1}]=2[M^{0}L^{1}T^{-1}]}$

By rule 2 has removed.

$\sf{\implies[M^{0}L^{1}T^{-1}]=[M^{0}L^{1}T^{-1}]}$

LHS = RHS

So, this formula is correct.

2). s = ut + 1/2 at²

Firstly we write the dimensions of all the variables,

$\sf{\implies s = [M^{0}L^{1}T^{0}]}$

$\sf{\implies u = [M^{0}L^{1}T^{-1}]}$

$\sf{\implies a = [M^{0}L^{1}T^{-2}]}$

$\sf{\implies t = [M^{0}L^{0}T^{1}]}$

Put the following dimensions in given formula,

$\sf{\implies [M^{0}L^{1}T^{0}]=[M^{0}L^{1}T^{-1}] \times [M^{0}L^{0}T^{1}] + \dfrac{1}{2} [M^{0}L^{1}T^{-2}] \times [M^{0}L^{0}T^{1}]^{2}}$

By Rule 1/2 is removed.

$\sf{\implies [M^{0}L^{1}T^{0}]=[M^{0}L^{1}T^{0}]+[M^{0}L^{1}T^{-2}]\times [M^{0}L^{0}T^{2}]}$

$\sf{\implies [M^{0}L^{1}T^{0}]=[M^{0}L^{1}T^{0}]+[M^{0}L^{1}T^{0}]}$

$\sf{\implies [M^{0}L^{1}T^{0}]=2[M^{0}L^{1}T^{0}]}$

By rule 2 is removed.

$\sf{\implies [M^{0}L^{1}T^{0}]=[M^{0}L^{1}T^{0}]}$

LHS = RHS

So, this formula is correct.

3). v² = u² + 2as

Firstly we write the dimensions of all the variables,

$\sf{\implies v = [M^{0}L^{1}T^{-1}]}$

$\sf{\implies u = [M^{0}L^{1}T^{-1}]}$

$\sf{\implies a = [M^{0}L^{1}T^{-2}]}$

$\sf{\implies s = [M^{0}L^{1}T^{0}]}$

Put the following dimensions in given formula,

$\sf{\implies [M^{0}L^{1}T^{-1}]^{2}= [M^{0}L^{1}T^{-1}]^{2}+2 [M^{0}L^{1}T^{-2}] \times  [M^{0}L^{1}T^{0}]}$

$\sf{\implies [M^{0}L^{2}T^{-2}]= [M^{0}L^{2}T^{-2}]+2 [M^{0}L^{1}T^{-2}] \times  [M^{0}L^{1}T^{0}]}$

By rule 2 is removed.

$\sf{\implies [M^{0}L^{2}T^{-2}]= [M^{0}L^{2}T^{-2}]+[M^{0}L^{2}T^{-2}]}$

$\sf{\implies [M^{0}L^{2}T^{-2}]= 2[M^{0}L^{2}T^{-2}]}$

By rule 2 is removed.

$\sf{\implies [M^{0}L^{2}T^{-2}]= [M^{0}L^{2}T^{-2}]}$

LHS = RHS

So, this formula is correct.

Extra Info:

Rules for correctness of formula:

1). All the numerical values are dimensionless.

2). All the constant, trigonometric functions, exponential functions and log functions are also dimensionless.