Let's carefully analyze and solve this problem step-by-step. --- ## Given Data and what is asked: ### Data: - Velocity of gas along center streamline: \[ u = 10x^2 + 200t + 6 \quad \text{(m/s)} \] where: - \( x \) is in meters - \( t \) is in seconds - Time just before leaving nozzle: \[ t = .01 \, \text{sec} \] - Position of interest: \[ x = A \] (where \( A \) is the nozzle exit point) - Distance from the pipe wall to the centerline: \[ .6 \, \text{m} \] ### To find: - The **local acceleration** \( a_{local} \) at the exit point just before the particle leaves (at \( t=.01\, \text{s} \) and \( x=A \)) - The **convective acceleration** \( a_{conv} \) at the same point - The **total acceleration** \( a_{total} \) --- ## Step 1: Find the velocity \( u \) at \( t=.01\,\text{s} \) and \( x=A \) Since the velocity is given as a function of \( x \) and \( t \), at the point of interest: \[ u = 10A^2 + 200(.01) + 6 \] Calculate: \[ u = 10A^2 + 2 + 6 = 10A^2 + 8 \] --- ## Step 2: Find the **local acceleration** \( a_{local} \) **Local acceleration** is the partial derivative of the velocity with respect to time at a fixed position: \[ a_{local} = \frac{\partial u}{\partial t} \] Given: \[ u = 10x^2 + 200t + 6 \] The partial derivative with respect to \( t \): \[ a_{local} = \frac{\partial}{\partial t} (10x^2 + 200t + 6) \] Since \( 10x^2 \) and 6 are independent of \( t \): \[ a_{local} = 200 \] **Answer for local acceleration:** \[ a_{local} = 200\, \text{m/s}^2 \] --- ## Step 3: Find the **convective acceleration** \( a_{conv} \) **Convective acceleration** reflects the change in velocity as the particle moves along the streamline, given by: \[ a_{conv} = u \frac{\partial u}{\partial x} \] First, compute \( \frac{\partial u}{\partial x} \): \[ u = 10x^2 + 200t + 6 \] Derivative with respect to \( x \): \[ \frac{\partial u}{\partial x} = 20x \] At \( x = A \): \[ \frac{\partial u}{\partial x} = 20A \] Now, the convective acceleration: \[ a_{conv} = u \times 20A \] Recall that: \[ u = 10A^2 + 8 \] So: \[ a_{conv} = (10A^2 + 8) \times 20A \] Simplify: \[ a_{conv} = 20A \times 10A^2 + 20A \times 8 \] \[ a_{conv} = 200A^3 + 160A \] --- ## Step 4: Find the **total acceleration** \( a_{total} \) The total acceleration combines local and convective accelerations: \[ a_{total} = a_{local} + a_{conv} \] Substitute the values: \[ a_{total} = 200 + (200A^3 + 160A) \] \[ a_{total} = 200A^3 + 160A + 200 \] --- ## **Final Answer:** \[ \boxed{ \text{Local acceleration} = 200\, \text{m/s}^2 } \] \[ \boxed{ \text{Convective acceleration} = 200A^3 + 160A\, \text{m/s}^2 } \] \[ \boxed{ \text{Total acceleration} = 200A^3 + 160A + 200\, \text{m/s}^2 } \] --- **Note:** To get a numerical value for total acceleration, you need the specific value of \( A \).