Question

Question ❓✔️
• The correctness of physical equation s = ut + ½ at2. In the equation, s is the displacement, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time in which change occurs.​

Answer 1

Answer:

Given equation :

$\implies s = ut + \dfrac{1}{2} {at}^{2}$

Solution :

LHS:

• s = Displacement

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

RHS :

• ut = initial velocity × time

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

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

And,

• at² = acceleration × (time)²
• ½ = Dimensionless quantity

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

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

LHS = RHS

Therefore, given formula is dimensionally correct.

Answer 2

ANSWER

We know that

L.H.S = s

and

R.H.S = ut + 1/2at2

The dimensional formula for the L.H.S can be written as s = [L¹M⁰T⁰]. (1)

We know that

R.H.S is ut + ½ at2 , simplifying we can write R.H.S as [u][t] + [a] [t]2

${[L¹M⁰T-¹] \: [L⁰M⁰T-¹] +[L¹M⁰T-²] \: [L⁰M⁰T⁰]}$

$=[L¹M⁰T⁰].(2)$

From (1) and (2), we have [L.H.S] = [R.H.S]

•Therefore ,by the principle of homogeneity, the given equation is dimensionally correct.

Applications of Dimensional Analysis

• Dimensional analysis means fundamental aspect of measurement which is applied in real-life physics.

•We make use of dimensional analysis for three prominent reasons:

• To check the consistency of a dimensional equation
• To derive the relation between physical quantities in physical phenomena
• To change units from one system to another

Demirts of Dimensional Analysis

• It doesn’t provide information about the dimensional constant.

• The formula containing trigonometric function, exponential functions, logarithmic function, etc. cannot be derived.

• It gives no information about whether a physical quantity is a scalar or vector.

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