91Ó°ÊÓ

Given data:

The equation of speed,$v=A{t}^3-Bt.$

Here, t is the time.

You have already studied thatthe total dimension and SI unit of quantities on the left-hand side and the right-hand side must be equal in an equation.

In other words, you can add or subtract the quantities having the same dimension (units), and the result will also have the same dimension of these quantities.

You know that speed is the distance divided by time, and the unit of speed is $\mathrm{m}·{\mathrm{s}}^{-1}$. The dimension of speed will be the ratio of the dimension of distance and time. So,

$\begin{array}{c}\left[v\right]=\frac{\left[d\right]}{\left[t\right]}\\ =\frac{\left[\mathrm{L}\right]}{\left[\mathrm{T}\right]}\\ =\left[{\mathrm{M}}^0 {\mathrm{L}}^1 {\mathrm{T}}^{-1}\right]\end{array}$

Thus, the dimension and SI unit of the values $A{t}^3$and Btmust be $\left[{\mathrm{M}}^0 {\mathrm{L}}^1 {\mathrm{T}}^{-1}\right]$and $\mathrm{m}·{\mathrm{s}}^{-1}$, respectively.

You know that the unit and dimension of time are s and $\left[\mathrm{T}\right]$, respectively.

Thus, the unit of A will be

$\frac{\mathrm{m}·{\mathrm{s}}^{-1}}{{\mathrm{s}}^3}=\mathrm{m}·{\mathrm{s}}^{-4}$.

Similarly, the unit of B will be

$\frac{\mathrm{m}·{\mathrm{s}}^{-1}}{\mathrm{s}}=\mathrm{m}·{\mathrm{s}}^{-2}$.

The SI unit of A is $\mathrm{m}·{\mathrm{s}}^{-4}$, and that of B is $\mathrm{m}·{\mathrm{s}}^{-2}$.

The dimensional formula for A will be

$\frac{\left[{\mathrm{M}}^0 {\mathrm{L}}^1 {\mathrm{T}}^{-1}\right]}{\left[{\mathrm{M}}^0 {\mathrm{L}}^0{\mathrm{T}}^3\right]}=\left[{\mathrm{M}}^0 {\mathrm{L}}^1 {\mathrm{T}}^{-4}\right]$.

Similarly, the dimensional formula for A will be

$\frac{\left[{\mathrm{M}}^0 {\mathrm{L}}^1 {\mathrm{T}}^{-1}\right]}{\left[{\mathrm{M}}^0 {\mathrm{L}}^0{\mathrm{T}}^1\right]}=\left[{\mathrm{M}}^0 {\mathrm{L}}^1 {\mathrm{T}}^{-2}\right]$.

The dimension formula of A is $\left[{\mathrm{M}}^0 {\mathrm{L}}^1 {\mathrm{T}}^{-4}\right],$and that of Bis $\left[{\mathrm{M}}^0 {\mathrm{L}}^1 {\mathrm{T}}^{-2}\right]$.