Experimental data indicate that in a region downstream of a given louvered supply vent the velocity of the emitted air is defined by τ= 0.18τo/x where τ and x are expressed in m/s and meters, respectively, and τo is the initial discharge velocity of the air. for τo 3.6 m/s. determine:

a. the acceleration of the air at x= 2m.
b. the time required for the air to flow from x=1 to x= 3m.

a) a =  - 0.0524 m/s²

b) t = 6.17 s

Explanation:

Given

τ = 0.18*τ₀ / x

τ₀ = 3.6 m/s

Determine:

a) The acceleration of the air at x= 2m.

Knowing

a = dτ / dt

Multiplying both the numerator and denominator by dx

a = (dτ / dx) (dx / dt)

Substituting τ for dx / dt

a = τ*(dτ / dx)

⇒ a = τ*(-0.18*τ₀ / x²) = (0.18*τ₀ / x)(-0.18*τ₀ / x²)

⇒ a = - 0.18²*τ₀² / x³ = - 0.18²*(3.6)² / x³

⇒ a = - 0.4199 / x³

If x = 2m

⇒ a = - 0.4199 / (2)³

⇒ a =  - 0.0524 m/s²

b) The time required for the air to flow from x=1 to x= 3m.

If

τ = dx / dt = 0.18*τ₀ / x

⇒   dt = (x / 0.18*τ₀) dx

⇒   t = (1 / 0.18*τ₀) ∫x dx

⇒   t = (1 / 0.18*τ₀)*(1 / 2)*x²

then

t = (1 / (0.18*3.6))*(1 / 2)*((3)² - (1)²)

⇒   t = 6.17 s

can you make it more obvious. i can't see it properly