Biology Forums - Study Force

For the position function s(t) = −16t^2+110t :

a) Complete the following table with the appropriate average velocities (show work and
give up to 3 decimal places)
Time Interval    [0.99, 1]           [1, 1.01]
Average
Velocity
b) Make a conjecture about the value of the instantaneous velocity at t = 1. _______

Substitute 0.99 and 1 into the function to get s:

\(s\left(0.99\right)=-16\left(0.99\right)^2+110\left(0.99\right)=93.2184\)
\(s\left(1\right)=-16\left(1\right)^2+110\left(1\right)=94\)

Average them out:

\(\frac{93.2184+94}{2}=93.6092\)

Do the same for the other interval:

\(s\left(1.01\right)=-16\left(1.01\right)^2+110\left(1.01\right)=94.7784\)
\(s\left(1\right)=-16\left(1\right)^2+110\left(1\right)=94\)

Average them out:

\(\frac{94.7784+94}{2}=94.3892\)


Umm should be the same as the average of the averages

i think it's not true

Actually, you're right. I'll revise my answer, let's use the formula:

\(Average\ Velocity=\frac{s\left(f\right)-s\left(i\right)}{f-i}\)

Where f = 1 and i = 0.99

Average Velocity = (94 - 93.2184) / (1 - 0.99) = 78.16

Do it again for the other interval:

Average Velocity = (94.7784 - 94) / (1.01 - 1) = 77.84


It will be @ 80 because the average of these two is 80.

My apologies for the incorrect understanding :-:) :-$