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Delicious77 [7]
3 years ago
5

Write a differential equation that fits the physical description. The velocityvelocity of a particle movingof a particle moving

along a straight linealong a straight line at time t is proportional to the fourthfourth power of its position xposition x.
Mathematics
1 answer:
Lady_Fox [76]3 years ago
8 0

Answer:

The differential equation is

dx/dt - Kx^4 = 0

Step-by-step explanation:

Let V represent the velocity of the particle moving along a straight line at time t.

We have the position to be x.

Then we have that

V is proportional to x^4

=> V = Kx^4

Where K is constant of proportionality.

Velocity is the derivative of the position vector with respect to time t, so we can write

V = dx/dt

And then

dx/dt = Kx^4

So that

dx/dt - Kx^4 = 0

This is the differential equation

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Examplelt: A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall
harina [27]

Answer :

  • 4167 bricks.

Explanation :

Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid.

Here,

{\qquad \dashrightarrow{ \sf{Length=10 \: m=1000 \: cm}}}

\qquad \dashrightarrow{ \sf{Thickness=24 \: cm}}

\qquad \dashrightarrow{ \sf{Height=4 m=400 \: cm}}

Therefore,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: the  \: wall = length \times breadth \times height}}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: the  \: wall = 1000 \times 24 \times 400 \:  {cm}^{3} }}}

Now, each brick is a cuboid with Length = 24 cm, Breadth = 12 cm and height = 8 cm.

So,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: each  \: brick = length \times breadth \times height}}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: each  \: brick = 24 \times 12 \times 8 \:  {cm}^{3} }}}

So,

{\qquad \dashrightarrow{ \bf{Volume \:  of  \: bricks  \: required =  \dfrac{volume \: of \: the \: wall}{volume \: of \: each \: brick} }}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: bricks  \: required =  \dfrac{1000 \times 24 \times 400}{24 \times 13 \times 8} }}}

{\qquad \dashrightarrow{ \sf{Volume \:  of  \: bricks  \: required =   \bf \: 4166.6} }}

Therefore,

  • <u>The wall requires 4167 bricks. </u>

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Since the limit is less than 1, the series converges.

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