Question

For an equation X
$x = a + bt + ct {}^{2}$
what will be the dimension of b and c wher X is in meter and in second​

Answer 1

Topic :- Units and Dimensions

$\maltese\:\underline{\sf Correct \:Question :} \: \maltese$

For an equation 'x'

$\sf x = a + bt + ct^{2}$

what will be the dimension of b and c where x is in meter and t is in second.

$\maltese\:\underline{\sf AnsWer :} \: \maltese$

We are given that x is in meters so, x is nothing but length. Hence, the dimensions of the 'x' is ;

$:\implies \sf x = [L]$

And t is in second means Dimensions of t will be :

$:\implies \sf t = [T]$

But we are also given that :

$:\implies \sf x = a + bt + ct^2$

It can also be written as :

$:\implies \sf x = a$

$:\implies \sf x = bt$

$:\implies \sf x = ct^2$

Now, let's find the dimensions of a, b, and c :

$:\implies \sf x = a$

$:\implies \underline{ \boxed{ \sf a = [L]}}$

$:\implies \sf x = bt$

$:\implies \sf [L] = b\:[T]$

$:\implies \sf \dfrac{[L]}{[T]}= b$

$:\implies \sf b = [L][T^{ - 1} ]$

$:\implies \underline{ \boxed{ \sf b = [LT^{ - 1} ]}}$

$:\implies \sf x = ct^2$

$:\implies \sf [L] = c\:[T ^{2} ]$

$:\implies \sf \dfrac{[L]}{[T ^{2} ]}= c$

$:\implies \sf c = [L][T^{ - 2} ]$

$:\implies\underline{\boxed{ \sf c = [LT^{ - 2} ]}}$

$\bullet\: \underline{\sf Dimensions \: of \: b = [M^0 L^1 T^{-1}]}.$

$\bullet\: \underline{\sf Dimensions \: of \: c = [M^0 L^1 T^{-2}]}.$