Question

Show that the following are dimensionally correct.
a) a=v-u÷t

Answer 1

Using LHS:

The dimensional formula will be,

${M}^{0} {L}^{1} {T}^{ - 2}$

For RHS,

$\frac{v - u}{t} = \frac{v}{t} - \frac{u}{t}$

Showing both the terms have the same dimensional formula we can prove the terms on LHS and RHS are equal, hence it's dimensionally correct.

For the first term on RHS,
$\frac{ {L}^{1} {T}^{ - 1} }{ {T}^{1} } \\ = {M}^{0}{L}^{1} {T}^{ - 2}$


Similarly for second term,

$\frac{ {L}^{1} {T}^{ - 1} }{ {T}^{1} } \\ = {M}^{0}{L}^{1} {T}^{ - 2}$


Since both terms' DF match on the Dimensional formula on RHM.


Hence proved.

Answer 2

Yes, the equation is dimensionally correct

Explanation:

• Dimension correction of any equation can be proved by homogeneity of equation. This can be done by equating left hand dimension and right hand dimension. Dimensional formula of each quantity is substituted in the equation then equated

Given, a = (v-u)/t

where a= acceleration

v= final velocity

u = initial velocity

t = time taken

Acceleration is rate of change of velocity, that is fraction of velocity and time taken

L.H.D

=> a = $\frac{change in velocity}{time taken}$

=> a = $\frac{[LT^{-1}] }{[T]}$

=> a = [LT⁻²]                 ..........................(1)

R.H.D

=> $\frac{v-u}{t}$ = $\frac{final velocity-initial velocity}{time taken}$

=> $\frac{v-u}{t}$ = $\frac{[LT^{-1}] -[LT^{-1}] }{[T]}$

=>$\frac{v-u}{t}$ = $\frac{[LT^{-1}] }{[T]}$                  

  (same dimensional quantities can be added and subtracted)

=> $\frac{v-u}{t}$ = [LT⁻²]                 ............................(2)

From equation (1) and (2)

L.H.D = R.H.D

Hence the equation is dimensionally correct

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