To determine The magnitude of the velocity and acceleration of the particle at point (1 m, 1 m) when t = 2 s . Explanation Given: Velocity components of fluid particles are u = ( 8 t 2 ) m / s and v = ( 7 y + 3 x ) m / s . Calculation: Determine the velocity component of flow along the x -direction, u , at (1 m, 1 m) and when t = 2 s . u = 8 t 2 u = 8 × ( 2 s ) 2 = 32 m / s Determine the velocity component of flow along the y -direction, v , at (1 m, 1 m) and when t = 2 s . v = ( 7 y + 3 x ) v = [ ( 7 × 1 m ) + ( 3 × 1 m ) ] = 10 m / s Determine the direction of the velocity field with respect to the horizontal. tan θ v = v u θ v = tan − 1 ( v u ) θ v = tan − 1 ( 10 m / s 32 m / s ) = tan − 1 ( 0.3125 ) = 17.4 ° Determine the magnitude of the velocity, V , of a particle located at (1 m, 1 m) and when t = 2 s . V = u 2 + v 2 V = ( 32 m / s ) 2 + ( 10 m / s ) 2 = 1 , 024 + 100 = 33.5 m / s Thus, the magnitude of the velocity of the particle at point (1 m, 1 m) when t = 2 s is 33.5 m / s . Write the generalized two dimensional Eulerian expression for acceleration ( a ) . a = ∂ V ∂ t + u ∂ V ∂ x + v ∂ V ∂ y Determine the scalar component of the acceleration along the x -direction ( a x ) . a x = ∂ u ∂ t + u ∂ u ∂ x + v ∂ u ∂ y a x = [ ∂ ∂ t ( 8 t 2 ) ] + [ ( 8 t 2 ) × ∂ ∂ x ( 8 t 2 ) ] + [ ( 7 y + 3 x ) × ∂ ∂ y ( 8 t 2 ) ] a x = ( 8 × 2 t ) + [ ( 8 t 2 ) × 0 ] + [ ( 7 y + 3 x ) × 0 ] a x = 16 t a x = ( 16 × 2 s ) a x = 32 m / s 2 Determine the scalar component of the acceleration along the y -direction ( a y )

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ISBN: 9780134626055

Author: HIBBELER

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 3, Problem 39P

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The magnitude of the velocity and acceleration of the particle at point (1 m, 1 m) when t=2 s.

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Chapter 3, Problem 38P

Chapter 3, Problem 40P

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