A) Only numerically correct
B) Only dimensionally correct
C) Both numerically and dimensionally correct
D) Neither numerically nor dimensionally correct
Correct Answer: C
Solution :
We can derive this equation from equations of motion so it is numerically correct. \[{{S}_{t}}\]= distance travelled in tth second = \[\frac{\text{Distance}}{\text{time}}=[L{{T}^{-1}}]\] \[u\] = velocity = \[[L{{T}^{-1}}]\]and \[\frac{1}{2}a(2t-1)=[L{{T}^{-1}}]\] As dimensions of each term in the given equation are same, hence equation is dimensionally correct also.
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