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

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2 months ago

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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

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-02-28T19:47:38+03:00

AnnZ [12.3K]2 months ago

5 0

The distance can be defined by the equation

$x=16t^2$

The relationship of distance over time is expressed as

$\Delta x = 32 t \Delta t$

$\Delta t = \frac{2}{10}s$

When t=2s,

$\Delta x = 32(2)\frac{2}{10}$

$\Delta x = 12.8$

An error of 12.8 feet arises.

b) Now considering t=5 seconds

$\Delta x= 32t\Delta t$

$\Delta x= 32(5)*\frac{2}{10}$

$\Delta x = 32$

An error of 32 feet occurs.