Question

check the dimensional correctness of v=v0+at​

Answer 1

Answer: It is dimensionally correct

Explanation:

An equation in which each term has equal dimensions is said to be dimensionally correct. Every equation used in science should be dimensionally accurate. The only time you run into something that isn't there is an error in the equation. So dimensional analysis is a valuable tool to help you spot an equation you made a mistake in algebra.

L H S :

dimensions in ( final velocity ) = [$LT^- 1$ ]

as $v = \frac{displacement}{time}$

RHS:

$v_{0}$ ( initial velocity ) = [$LT^-1$ ]

at = $[ LT^ - 2 ] [ T ]$

as $acceleration = \frac{change in velocity}{time}$

so,

at =$[ LT^ - 1 ]$

so, $v_{0}$$+ at =$ [$L T^ - 1$] according to the principles of homogeneity

L H S = R H S = [ $L T ^ -1$ ]

Thus,

It is the first equation of motion, so it is dimensionally correct.

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Answer 2

Answer: The given equation is dimensionally correct.

Dimensionally accurate equations are those whose terms all have the same dimensions. Dimensionally correct equations should be employed in all scientific applications. The only time you encounter anything that isn't there is when the equation contains an error. Dimensional analysis is a useful method for identifying algebraic errors in equations.

The given equation is:

                         $v = v_0+at$

We know that,

                     the velocity = Displacement/Time

Hence,

           LHS = $[LT^-1]$

And

          RHS = $[LT^-1]+[LT^-2][T]$

On solving we get,

          RHS = $[LT^-1]$

Here clearly,

          LHS = RHS = $[LT^-1]$    

Thus,

        The given equation is dimensionally correct.

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