since s(0) = 0, b0=0 and
s = t(b1+b2t)
since s(1) = 0.6,
b1+b2 = 0.6
since s(2) = 4.2,
2(b1+2b2) = 4.2
and we have the system
b1+b2 = .6
b1+2b2 = 2.1
b2 = 1.5
b1 = -.9
s(t) = 1.5t^2 - 0.9t
This correctly figures for t=0,1,2, but starts being off at t=3.
You can make a parabola fit any 3 points, but not necessarily 4 given points.
If we don't require that b0=0, then we have
s(t) = 1/5 (4t^2+6t-7)
which is correct for t=1,2,3 but off at t=0
A truck rolls down a hill subject to acceleration due to gravity and resistance (including
air resistance, rolling friction, etc) and the distance si (metres) travelled after ti seconds
is observed. Classical mechanics predicts that the distance travelled by the truck should
be a quadratic function of time:
s(t) = b0 + b1t + b2t2.
Given the observed time (ti) and corresponding distance (si) data below, what are the
values of b0, b1, b2 which best fit this data?
ti 0 1 2 3
si 0 0.6 4.2 9.4
2 answers
thank you