Question

Is v=u+ax dimensionally correct?

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

The equation is : $\sf{v \: = \: u \: + \: at}$

Put u = 0

$\implies {\sf{MLT^{-1} \: = \: [MLT^{-2}] [T]}} \\ \\ \implies {\sf{[MLT^{-1} \: = \: [MLT^{-1}]}}$

Hence Proved

Its dimensions are equal

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Some Important Formulas

1) v = u + at

2) S = ut + ½at²

3) v^2 - u^2 = 2as

Answer 2

Answer:

NOOOO..

Explanation:

V = $LT^{-1}$   (FINAL VELO)

U= $LT^{-1}$    ( INIT VELO)

A = acceleration =  $LT^{-2}$

X= DISTANCE =  L

SO AX = $L^{2} T^{-2}$  

HENCE :- DIMMENSION OF

V=U

BUTTT:----

V$\neq$ AX