The position of a particle moving on x-axis is given by x(t) = A sin t + B cos²t + Ct² + D, where t is time. The dimension of $\frac{\mathrm{ABC}}{\mathrm{D}}$ is-

We are given the position function of a particle moving along the x-axis:

$$x(t) = A \sin t + B \cos^2 t + C t^2 + D$$

where $x$ represents position, and $t$ is time.

We need to determine the dimension of $\frac{ABC}{D}$.

Step 1: Identify Dimensions of Each Term

Since all terms in the given equation represent position ($x$), they must have the same dimensional formula as length $[L]$.

1st Term: $A \sin t$

• $\sin t$ is dimensionless.
• So, $A$ must have the same dimension as $x$, which is $L$.

$$[A] = [L]$$

2nd Term: $B \cos^2 t$

• $\cos^2 t$ is dimensionless.
• So, $B$ must also have the same dimension as $x$, which is $L$.

$$[B] = [L]$$

3rd Term: $C t^2$

• $t^2$ has the dimension $T^2$.
• Since $C t^2$ must have the dimension of length $[L]$, we get:

$$[C] = \frac{[L]}{[T]^2} = [L T^{-2}]$$

4th Term: $D$

• Since $D$ represents a position constant, it has the same dimension as $x$, which is $L$.

$$[D] = [L]$$

Step 2: Compute $\frac{ABC}{D}$

$$\frac{ABC}{D} = \frac{([L])([L])([L T^{-2}])}{[L]}$$

Canceling [L] in the numerator and denominator:

$$[L \cdot L \cdot L T^{-2}] / [L] = L^2 T^{-2}$$

Thus, the dimension of $\frac{ABC}{D}$ is $\left[L^2{T}^{-2}\right]$