Physics

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

##### SOLUTION

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, $a=L$

$[bt]=[L]$

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

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

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

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

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Subjective Medium Published on 18th 08, 2020
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