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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