Question

The position of a particle moving along x-axis depends on time according to the equation x=at2+bt3, where x is in metre and t is in sec. What are the units and dimensions of a and b : what do they represent?

Answer 1

❈ Given ❈

◘ x = at² + bt³

◘ x is in metre and t is in second.

❈ To Find ❈

◘ The units and dimensions of a and b.

◘ What a and b represent.

❈ Solution ❈

We know,

x = [L¹]

[t] = {T¹]

Now,

[x] = [at² + bt³] = [at²] = [bt³]

→ [a] = [x] / [t²]

→ [a] = [L¹] / [T²]

→ [a] = [M⁰ L¹] / T²

→ [a] = [M⁰ L¹ T⁻²]

Power of L is 1, so it will be meter. Power of T is -2, so it will be second^-2.

[a] = [M⁰ L¹ T⁻²] ⇒ ms⁻²

As ms⁻² or (m/s²) is the unit of acceleration, so a represents acceleration.

_________________

→ [b] = [x] / [t³]

→ [b] = [L¹] / [T³]

→ [b] = [M⁰ L¹] / T³

→ [b] = [M⁰ L¹ T⁻³]

Power of L is 1, so it will be meter. Power of T is -3, so it will be second^-3.

[b] = [M⁰ L¹ T⁻³] ⇒ ms⁻³

As ms⁻³ or m/s³ is the unit of acceleration / time (m/s² ÷ s), so b represents rate of change of acceleration.

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given that position of a particle moving along x-axis
• x = at² + bt³
• Where x is given in metre and time is given in second

To Find:

• We have to find the unit and dimension of a and b

Concept Used:

$\underline{\large\sf{\orange{Principle \: of \: Homogeneity}}}$

The Principle states that dimensions of each of the terms of a dimensional equation on both sides should be the same.

Solution:

Given that position of a particle moving along x-axis in terms of time is

$\large\boxed{\text{x = at² + bt³}}$

_______________________________

Using Principle of Homogeneity

$\implies \sf{at^2 = x }$

$\implies \boxed{\sf{a = \dfrac{x}{t^2}}}$

• Using Dimensional Analysis

• $\sf{a = \dfrac{[ \: L \: ]}{[ \: T \: ]^2}}$

• $\sf{a = \dfrac{[ \: L \: ]}{[ \: T^2 \: ]}}$

• $\sf{a = [ \: L \: T^{-2}\: ]}$

Dimension of a is [ L T¯² ]

• Since x is in metre and time in sec

• $\sf{a = \dfrac{x}{t^2}}$

• $\sf{a = \dfrac{m}{(s)^2}}$

• $\sf{a = m \: s^{-2}}$

Unit of a is ms¯²

______________________________

Using Principle of homogeneity

$\implies \sf{bt^3 = x }$

$\implies \boxed{\sf{b = \dfrac{x}{t^3}}}$

• Using Dimensional Analysis

• $\sf{b = \dfrac{[ \: L \: ]}{[ \: T \: ]^3}}$

• $\sf{b = \dfrac{[ \: L \: ]}{[ \: T^3 \: ]}}$

• $\sf{b = [ \: L \: T^{-3}\: ]}$

Dimension of a is [ L T¯³ ]

• Since x is in metre and time in sec

• $\sf{b = \dfrac{x}{t^3}}$

• $\sf{b = \dfrac{m}{(s)^3}}$

• $\sf{b = m \: s^{-3}}$

Unit of b is ms¯³

_______________________________

$\huge\underline{\sf{\red{A}\orange{n}\green{s}\pink{w}\blue{e}\purple{r}}}$

$\odot \:$ For Quantity a :

• Dimension = [ L T¯² ]
• Unit = ms¯²
• Represents acceleration

$\odot \:$ For Quantity b :

• Dimension = [ L T¯³ ]
• Unit = ms¯³
• Represents rate of change of acceleration

_______________________________