Give-the-velocity-field-v-6-2xy-t-2-i-xy-2-10t-j-25k-what-is-the-acceleration-of-the-particle-at-3-0-2-at-time-t-1-

Question Number 10528 by Saham last updated on 16/Feb/17

Give the velocity field

v = (6 + 2 xy + t^{2}) i - ({xy}^{2} + 10 t) j + 25 k

what is the acceleration of the particle at (3, 0, 2)

at time t = 1 .

Answered by robocop last updated on 16/Feb/17

v = 18 + 6 x y + 3 + 50

a = v^{'}

a = 6 (x + y)

Commented by Saham last updated on 16/Feb/17

Thanks sir .

Commented by FilupS last updated on 18/Feb/17

incorrect . you are refering to

linear velocity

i . e . v \in R

This question is about a particle in a field

i . e . v \in R^{3}

Answered by ajfour last updated on 16/Feb/17

a_{x} = 2 ({y v}_{x} + {x v}_{y} + t); a t t h a t t i m e a n d p l a c e i s = 2 [0 + 3 (- 10) + 1] = - 58

a_{y} = - (y^{2} v_{x} + 2 {x y v}_{y} + 10); t h e n a n d t h e r e w i l l b e = - [0 + 0 + 10] = - 10

a_{z} = 0

S o \bar{a} t h e n a n d t h e r e (3, 0, 2, 1) w o u l d b e - 58 \hat{i} - 10 \hat{j} .

Commented by Saham last updated on 16/Feb/17

Thanks sir .

Answered by mrW1 last updated on 16/Feb/17

v_{x} = 6 + 2 x y + t^{2}

v_{y} = - ({x y}^{2} + 10 t)

v_{z} = 25

a_{x} = \frac{{d v}_{x}}{d t} = 2 x \frac{d y}{d t} + 2 y \frac{d x}{d t} + 2 t = 2 ({x v}_{y} + {y v}_{x} + t)

a_{y} = \frac{{d v}_{y}}{d t} = - (x y \frac{d y}{d t} + y^{2} \frac{d x}{d t} + 10) = - ({x y v}_{y} + y^{2} v_{x} + 10)

a_{z} = 0

a t (3, 0, 2) a n d t = 1 :

v_{x} = 6 + 2 \times 3 \times 0 + 1^{2} = 7

v_{y} = - (3 \times 0^{2} + 10 \times 1) = - 10

v_{z} = 25

a_{x} = 2 [3 \times (- 10) + 0 \times 7 + 1] = - 58

a_{y} = - [3 \times 0 \times (- 10) + 0^{2} \times 7 + 10] = - 10

a_{z} = 0

\Rightarrow a = - 58 i - 10 j + 0 k

Commented by Saham last updated on 16/Feb/17

Thanks sir .

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