Question

Ive tried numerous methods, but I cant seem to get this right. Please help, and explain step by step.

An object's position is given by x=bt+ct^3, where b=1.50m/s and c=0.640m/s3. To study the limiting process leading to the instantaneous velocity, calculate the object's average velocity over time intervals from 1.00 s to 3.00 s.

Also, from 1.95 s to 2.05 s.

Answer 1

well, just for fun, let's see hat the form of the limit is
v = dx/dt = b + 3 c t^2
= 1.5 + 1.92 t^2
=========================
now
from 1 to 3
x(3)= 1.5*3 +.64*27=4.5+17.28
= 21.78
x(1) = 1.5+.64 = 2.14
so
x(3)-x(1) = 19.64
delta t = 3 - 1 = 2 seconds
so
v during interval = 19.64/2
= 9.82 m/s
now check that with calculus v in the middle of that range
v(2) =1.5+ 1.92 (4) = 9.18 m/s
that makes sense, the parabola is headed up so the instant speed i the middle should be a bit less than average.
Do you see what is going on? try some other intervals.

Answer 2

Also, from 1.95 s to 2.05 s

this is set up so you can compare your calculated speed with the real speed at 2 seconds from calculus. By coincidence we already did that last part, v(2) = 9.18 m/s and your average from 1.95 to 2.05 should be very close to that, like really, really close ;)

Answer 3

Thank you!