Question

The velocity V of a particle depends on time t as V=At²+Bt.Find the dimension units of A and B​

Answer 1

Given :

• The velocity V of a particle depends on time t as V=At²+Bt.

To Find :

• Find the dimension units of A and B

Solution :

Concept

• A dimension is a measure of a physical variable (without numerical values), while a unit is a way to assign a number or measurement to that dimension.

For example :

• length is a dimension, but it is measured in units of feet (ft) or meters (m). ... the British Gravitational System of Units (BG).

____________________

According to the Question :

Vectors of only same dimensions can be added or subtracted

Dimension of v = dimension of At²

[A] = [L T-¹]/[T²]

dimension of v = dimension of Bt

[B] = [L T-¹]/[T]

• we can find the dimensions in physic by converting any physical quantities into the simpler basic units after that as we know that the basic dimensions are of three units length, mass, and time(L,M,T).

Example : dimension of velocity

• V = m/s
• V = (L/T)
• V = LT-1

A = [ L¹ T-³ ]and B = [L¹ T-²]

• Hence The dimensions of A are [ L¹ T-³ ] and the dimensions of B are [ L ¹T-²].

Answer 2

Answer:-

Dimension unit of A is $\sf{[L^1T^{-3}]}$

Dimension unit of B is $\sf{[L^1T^{-2}]}$

Explanation:-

Given,

V = At² + Bt

where V is the velocity of the particle, which depends on time t

We need to find,

• Dimensions of units A and B

Solution,

Since, the unit of V is m/s therefore,

Dimensionally

$\sf{V = [L^1T^{-1}]}$ and $\sf{t = [T^1]}$

So,

Using the principle of homogeneity for dimensional equations

$\implies\sf{V = A t^2 = [L^1T^{-1}]}$

$\implies\sf{ A t^2 = [L^1T^{-1}]}$

$\implies\sf{ A\;.\; [T^2] = [L^1T^{-1}]}$

$\implies\sf{A=\dfrac{[L^1T^{-1}]}{[T^2]}}$

$\implies\sf{A = [L^1T^{-3}]}$

also,

$\implies\sf{V = Bt = [L^1T^{-1}]}$

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

$\implies\sf{ B\;.\;[T] = [L^1T^{-1}]}$

$\implies\sf{B = \dfrac{[L^1T^{-1}]}{[T]}}$

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

Therefore,

Dimension unit of A: $\sf{[L^1T^{-3}]}$

Dimension unit of B: $\sf{[L^1T^{-2}]}$