Discussion
The displacement of a particle is given by \(x = at + bt^{\frac{3}{2}}\) where t has dimensions, T and a and b are constants. The dimensions of \(\frac{b}{a}\) are:
Answer: E
The dimensions of \(\frac{b}{a}\) are \(T^{-1/2}\).
Each term on the right must have the dimension of length, like x.
\[ \begin{aligned} [a]\,T &= L \Rightarrow [a] = LT^{-1} \\ [b]\,T^{3/2} &= L \Rightarrow [b] = LT^{-3/2} \\ \left[\tfrac{b}{a}\right] &= \frac{LT^{-3/2}}{LT^{-1}} = T^{-1/2} \end{aligned} \]
Dividing the other way, a by b, gives \(T^{1/2}\), the usual slip. The integer powers \(T^{-2}\), \(T^{-1}\) and \(T\) would need an integer exponent on t in the second term.
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