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

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

thank you

To find the values of b0, b1, and b2 that best fit the data, we need to solve a system of equations using the given data points.

We can substitute the values of ti and si into the equation s(t) = b0 + b1t + b2t^2 to create three equations:

- For t = 0, s(0) = b0 + b1(0) + b2(0)^2 = b0.
- For t = 1, s(1) = b0 + b1(1) + b2(1)^2 = b0 + b1 + b2.
- For t = 2, s(2) = b0 + b1(2) + b2(2)^2 = b0 + 2b1 + 4b2.

Using the observed values of s(t), we can substitute them into the equations:

- For t = 0, s(0) = b0 = 0.
- For t = 1, s(1) = b0 + b1 + b2 = 0.6.
- For t = 2, s(2) = b0 + 2b1 + 4b2 = 4.2.

Now, we have a system of three equations:

b0 = 0,
b0 + b1 + b2 = 0.6,
b0 + 2b1 + 4b2 = 4.2.

Substituting b0 = 0 into the second equation, we get:

0 + b1 + b2 = 0.6.

Simplifying the third equation, we have:

0 + 2b1 + 4b2 = 4.2.

Now, we have the following equations:

b1 + b2 = 0.6,
2b1 + 4b2 = 4.2.

We can solve this system of linear equations to find the values of b1 and b2:

Multiplying the first equation by 2, we get:

2b1 + 2b2 = 1.2.

Subtracting this new equation from the second equation, we can eliminate b1:

2b1 + 4b2 - (2b1 + 2b2) = 4.2 - 1.2,
2b2 - 2b2 = 3.

Simplifying, we have:

0 = 3.

Since this equation is not true, there is no solution to the system.

Therefore, it is not possible to find exact b0, b1, and b2 values that best fit the given data.

To determine the values of b0, b1, and b2 that best fit the observed data, we will use the method of least squares. This method minimizes the sum of the squared differences between the predicted values and the observed values.

First, let's write down the equations using the given data:

For ti = 0: s(0) = b0 + b1(0) + b2(0)^2 = 0
For ti = 1: s(1) = b0 + b1(1) + b2(1)^2 = 0.6
For ti = 2: s(2) = b0 + b1(2) + b2(2)^2 = 4.2
For ti = 3: s(3) = b0 + b1(3) + b2(3)^2 = 9.4

We now have a system of four equations with three unknowns (b0, b1, b2). To solve this system, we can rearrange the equations and write them in matrix form:

| 1 0 0 | | b0 | | 0 |
| 1 1 1 | x | b1 | = | 0.6 |
| 1 2 4 | | b2 | | 4.2 |
| 1 3 9 | | 9.4 |

We can solve this system of linear equations using various methods such as Gaussian elimination or matrix inversion.

Once we solve the system, we will obtain the values of b0, b1, and b2, which will give us the desired quadratic function s(t) = b0 + b1t + b2t^2 that best fits the given data.