Question

check the correctness of the equations equal to x= Vot+1/2 at square​

Answer 1

Explanation:

Check the correctness of s = UT +1/2at2

s is distance so it's dimensions become L.

Nd other side we have ut+1\2at^2.

as 1\2 is a constant it will have dimensions and apply the dimensions to other quantities.

on solving u will get L there also i,e ur LHS = RHS.

thus the equation is dimensionally consistent.

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Answer 2

Explanation:

Your question:-

• Check the correctness of the given equation

Your solution:-

Given equation- x= Vot+1/2at^2

Dimensions of each terms:-

x=[M^0 L^1 T ^0]

Vo=[M^0 L^1 T^-1]

a=[M^0 L^1 T^-2]

T=[M^0 L^0 T^1]

Implementing given dimension in the equation:-

x= Vot+1/2at^2

[L]=[LT^-1]×[T]+1/2[LT^-2] [T^2]

→ {Anything to the power (1)=given value)} [Note]

After multiplying all dimensions we got:-

[L]=[L]+[L]

Hence LHS=RHS. So the given equation is dimensionally correct.