1.

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

Answer»

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.


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