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The distance, in feet, a stone drops in tt seconds is given by d(t)=16t2d(t)=16t2. The depth of a hole is to be approximated by dropping a rock and listening for it to hit the bottom. Assume that time measurement is accurate to 2 / 10th of a second. Use a linear approximation to estimate an upper bound for the propagated error if the measured time is

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Define the distance through the equation

[tex]x=16t^2[/tex]

The distance against the time is given by

[tex]\Delta x = 32 t \Delta t[/tex]

[tex]\Delta t = \frac{2}{10}s[/tex]

When t=2s,

[tex]\Delta x = 32(2)\frac{2}{10}[/tex]

[tex]\Delta x = 12.8[/tex]

There is a error in 12.8feet.

b) When t=5sec

[tex]\Delta x= 32t\Delta t[/tex]

[tex]\Delta x= 32(5)*\frac{2}{10}[/tex]

[tex]\Delta x = 32[/tex]

There is a error in 32 feet.

The distance a stone drops and hits the bottom, during this the propagation of the error is from 12.8 to 32.

What is the distance?

It is a measure of length that is traveled by an object. It is a scalar quantity.

Given

The distance, in feet, a stone drops in t seconds is given by

[tex]\rm x=16t2[/tex]

The depth of a hole is to be approximated by dropping a rock and listening for it to hit the bottom.

The distance against the time is given by

[tex]\rm \Delta x = 32t \Delta t[/tex]

And

[tex]\rm \Delta t = \dfrac{2}{10}= 0.2[/tex]

When t = 2 seconds, then

[tex]\rm \Delta x = 32 * 2 * 0.2\\\\\Delta x = 12.8[/tex]

There is an error in 12.8 ft.

When t = 5 seconds, then

[tex]\rm \Delta x = 32 * 5 * 0.2\\\\\Delta x = 32[/tex]

There is an error in 32 ft.

More about the distance link is given below.

https://brainly.com/question/989117

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