error approximation

Question

The distance, in feet, a stone drops in  seconds is given by . 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 3 / 10th of a second. Use a linear approximation to estimate an upper bound for the propagated error if the measured time is:

a) 3 seconds. 

Answer 1

According the physics, the position y(t) of an object after t seconds is...

y(t) = y₀ + v₀t - (1/2)gt²

Therefore the error in the displacement y(t) called E(y(t)) can be modeled as... Error is modeled as a strictly positive quantity...

E(y(t)) = E(y₀) + E(v₀)*t + v₀*E(t) + (1/2)g*2t*E(t)

Since we presume the initial velocity of the object is 0 and the initial position can be modeled as the 0 of the axis...

E(y(t)) = 0 + 0*t + 0*(3/10) + (1/2)g * 2t * (3/10)

There the equation of the error in the position of the stone is seen as ...

E(y(t)) = +(3/10)gt

Using g ~ 10 --> E(y(t)) = 3t

So with t = 3 seconds the error in the position (upper bound) is 9 feet