This solution explains the calculation of the average pick velocity during air-jet weft insertion, considering a two-phase motion.

Calculating Average Pick Velocity

The problem describes a pick's motion during air-jet insertion, divided into two distinct phases over the reed width ($W = 2$ m):

• Phase 1: Uniform acceleration ($a = 18$ m/s$^2$) from rest ($u = 0$ m/s) up to the midpoint of the reed width ($s_1 = W/2 = 1$ m).
• Phase 2: Constant velocity ($v_1$, the velocity achieved at the midpoint) for the remaining half of the reed width ($s_2 = W/2 = 1$ m).

We need to find the average velocity ($v_{avg}$) over the total distance $W$. The average velocity is calculated as Total Distance / Total Time.

Phase 1: Acceleration Calculation

First, find the velocity ($v_1$) at the midpoint using the kinematic equation:

$v_1^2 = u^2 + 2as_1$

Substituting the known values:

$v_1^2 = (0 \text{ m/s})^2 + 2 \times (18 \text{ m/s}^2) \times (1 \text{ m})$

$v_1^2 = 36 \text{ m}^2/\text{s}^2$

$v_1 = \sqrt{36} \text{ m/s} = 6 \text{ m/s}$

Next, calculate the time ($t_1$) taken for this acceleration phase:

$v_1 = u + at_1$

$6 \text{ m/s} = 0 \text{ m/s} + (18 \text{ m/s}^2) \times t_1$

$t_1 = \frac{6 \text{ m/s}}{18 \text{ m/s}^2} = \frac{1}{3} \text{ s}$

Phase 2: Constant Velocity Calculation

The velocity during this phase ($v_2$) is constant and equal to the velocity at the midpoint:

$v_2 = v_1 = 6 \text{ m/s}$

Calculate the time ($t_2$) taken for this constant velocity phase:

$t_2 = \frac{s_2}{v_2}$

$t_2 = \frac{1 \text{ m}}{6 \text{ m/s}} = \frac{1}{6} \text{ s}$

Total Time and Average Velocity

Calculate the total time ($T$) for the entire weft insertion process:

$T = t_1 + t_2$

$T = \frac{1}{3} \text{ s} + \frac{1}{6} \text{ s} = \frac{2}{6} \text{ s} + \frac{1}{6} \text{ s} = \frac{3}{6} \text{ s} = \frac{1}{2} \text{ s}$

Finally, calculate the average pick velocity ($v_{avg}$):

$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{W}{T}$

$v_{avg} = \frac{2 \text{ m}}{1/2 \text{ s}}$

$v_{avg} = 4 \text{ m/s}$

Final Answer

The average pick velocity in the entire duration of weft insertion is 4 m/s.