Question

Suppose you are given the following equation, where xf and xi represent positions at two instants of time, vxi is a velocity, ax is an acceleration, t is an instant of time, and a, b, and c are integers. xf = xita + vxitb + ½axtc.

Required:
For what values of a, b, and c is this equation dimensionally correct?

Answer 1

For the equation to be dimensionally consistent, the values of a, b, and c are a = any integer, b = 1, and c = 0.

Step-by-step explanation:

To determine the values of a, b, and c for which the equation xf = xita + vxitb + ½axtc is dimensionally correct, we need to consider the dimensions of each term in the equation.

Breaking down each term, we have:

xita has dimensions of length L.

vxitb has dimensions of length L * time T raised to the power b.

½axtc has dimensions of length L * time T raised to the power 1 + c.

For the equation to be dimensionally correct, all three terms must have the same dimension. This gives us the following equation:

L = L * Tᵇ = L * T⁽¹⁺c⁾

For the equation to be dimensionally consistent, we must have b = 1 and c = 0. Therefore, the values of a, b, and c that make the equation dimensionally correct are a = any integer, b = 1, and c = 0.

Answer 2

Step-by-step explanation:

xf = xita + vxitb + ½axtc.

xf is displacement , dimensional formula L .

Xi initial displacement , dimensional formula L

t is time , dimensional formula T ,

vxi is velocity , dimensional formula LT⁻¹

ax is acceleration , dimensional formula = LT⁻²

xf = xi t a + vxi t b + ½ ax t c.

L = aLT + b LT⁻¹ T + c LT⁻² T

From the law of uniformity , dimensional formula of each term of RHS must be equal to term on LHS

aLT = L

a = T⁻¹

b LT⁻¹ T = L

b = 1 ( constant )

c LT⁻² T = L

c = T

so a = T⁻¹ , b = constant and c = T .