Water flows through a pipe with a velocity given by \overrightarrow{v}-(\overline{t}^{+x+y}) \hat{j} {m} \/ {s}, where \hat{j} is the unit vector in the y direction, t(>0) is in seconds, and x and y are in meters. The magnitude of total acceleration at the point (x, y)=(1,1) at t=2 s is _____________{m} {/} {s}^{2}

GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

Question : 35 of 65

Marks: +1, -0

Water flows through a pipe with a velocity given by

{v}↖{→}-(\ov{t}^{+x+y}) {j}↖{∧} {m} \/ {s}

, where

{j}↖{∧}

is the unit vector in the y direction,

t(>0)

is in seconds, and

x

and

y

are in meters. The magnitude of total acceleration at the point

(x, y)=(1,1)

t=2 s

is _____________

{m} {/} {s}^{2}

Solution:

{a}↖{→}=a_{x} {ı}↖{∧}+a_{y} {j}↖{∧}+a_{z} {k}↖{∧}

{V}↖{→}=f(x, y, z, t)

Here,

u=V_{x}=0 ; v=V_{y}=\;{4}/{t}+x+y ; w=v_{z}=0

a_{x}=u \;{∂ u}/{∂ x}+v \;{∂ u}/{∂ y}+w \;{∂ u}/{∂ z}+\;{∂ u}/{∂ t}=0

a_{y}=u \;{∂ v}/{∂ x}+v \;{∂ v}/{∂ y}+w \;{∂ v}/{∂ z}+\;{∂ v}/{∂ t}=\;{4}/{t}+x+y-\;{4}/{t^{2}}

a_{z}=u \;{∂ w}/{∂ x}+v \;{∂ w}/{∂ y}+w \;{∂ w}/{∂ z}+\;{∂ w}/{∂ t}=0

So,

{a}↖{→}=a_{x} {i}↖{∧}+a_{y} {}↖{∧}+a_{z} {k}↖{∧}=0+\;{4}/{t}+x+y-\;{4}/{t^{2}}+0

(x, y)=(1,1)

t=2 s

{a}↖{→}=0+\;{4}/{2}+1+1-\;{4}/{4}=2+1+1-1=3

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