The equation for the change of position of a train starting at x=0 m is given by x=(1)/(2)at^2+bt^3 . The dimensions of b are: a) T^-3 b) LT^-3 C LT^-2 D LT^-1 E L^-1T^-1

The correct answer is:<br /><br />b) $LT^{-3}$<br /><br />Explanation:<br />The equation for the change of position of the train is given by $x=\frac{1}{2}at^{2}+bt^{3}$, where $x$ represents the position, $a$ is a constant, $t$ is the time, and $b$ is the coefficient we need to find the dimensions for.<br /><br />To find the dimensions of $b$, we need to analyze the dimensions of each term in the equation.<br /><br />The dimensions of $x$ are $L$ (length), and the dimensions of $t$ are $T$ (time).<br /><br />The dimensions of the first term, $\frac{1}{2}at^{2}$, are $L$ (since $\frac{1}{2}$ is dimensionless and $a$ has dimensions of $LT^{-2}$).<br /><br />The dimensions of the second term, $bt^{3}$, are also $L$ (since $b$ has dimensions of $LT^{-3}$ and $t^{3}$ has dimensions of $T^{-3}$).<br /><br />Since the dimensions of both terms are the same ($L$), we can equate the dimensions of the coefficients:<br /><br />$\frac{1}{2}at^{2} \sim bt^{3}$<br /><br />$a \sim b t$<br /><br />$a = b t^4$<br /><br />Therefore, the dimensions of $b$ are $LT^{-3}$.