Question

The distance covered by a particle in time t is given by $x=a+bt+c{t}^2+d{t}^3$. Find the dimensions of a,b,c and d.

Answer 1

Given that,

$x=a+bt+c{t}^2+d{t}^3$

$x$ is the displacement.

Now, by principle of homogeneity

$a=bt=c{t}^2=d{t}^3=x$

Now, $a=L$

$\left[bt\right]=\left[L\right]$

$b=\left[L{T}^{-1}\right]$

$c=\left[L{T}^{-2}\right]$

$d=\left[L{T}^{-3}\right]$

Hence, the dimension of a, b, c and d is $\left[L\right],\left[L{T}^{-1}\right],\left[L{T}^{-2}\right]and\left[L{T}^{-3}\right]$