Question

If the velocity of the particle is given by $v=\frac{a}{t}+b{t}^2$
The dimensions of a & b are given as ______

Answer 1

$v=\frac{a}{t}+{bt}^2$

$\left[v\right]=\left[\frac{a}{t}\right]=\left[{bt}^2\right]$ [from dimensional analysis]

$L{T}^{-1}=a{T}^{-1}=B{T}^2$

Comparing the sides we get,

$a=L^1$

$b=L{T}^{-3}$