A horizontal pipe has a cross-sectional area of $A_1 = 48 \text{ cm}^2$ at point x. The pipe narrows to a cross-sectional area of $A_2 = 12 \text{ cm}^2$ at point y. Water with a density of $\rho = 1000 \text{ kg/m}^3$ flows through the pipe. The velocity of the water at point y is $V_2 = 24 \text{ m/s}$. We need to find the velocity of the water at point x, $V_1$.

1. Problem Description

A horizontal pipe has a cross-sectional area of $A_1 = 48 \text{ cm}^2$ at point x. The pipe narrows to a cross-sectional area of $A_2 = 12 \text{ cm}^2$ at point y. Water with a density of $\rho = 1000 \text{ kg/m}^3$ flows through the pipe. The velocity of the water at point y is $V_2 = 24 \text{ m/s}$. We need to find the velocity of the water at point x, $V_1$.

2. Solution Steps

We can use the equation of continuity, which states that for an incompressible fluid, the mass flow rate is constant throughout the pipe. This can be expressed as:

$A_1 V_1 = A_2 V_2$

We are given $A_1 = 48 \text{ cm}^2$, $A_2 = 12 \text{ cm}^2$, and $V_2 = 24 \text{ m/s}$. We need to find $V_1$.

Rearranging the equation to solve for $V_1$:

$V_1 = \frac{A_2 V_2}{A_1}$

Plugging in the given values:

$V_1 = \frac{(12 \text{ cm}^2)(24 \text{ m/s})}{48 \text{ cm}^2}$

$V_1 = \frac{12}{48} \times 24 \text{ m/s}$

$V_1 = \frac{1}{4} \times 24 \text{ m/s}$

$V_1 = 6 \text{ m/s}$

3. Final Answer

The velocity of the water at point x is $6 \text{ m/s}$.