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We are given data for the distance traveled, \( s(t) \), over time, and we need to estimate both the speed and acceleration at \( t = 10 \) and at other specified times using different methods:

### Part (i) Two-point central difference formula for speed at \( t = 10 \):
The two-point central difference formula for speed \( v(t) \) is given by:
\[
v(t) \approx \frac{s(t+h) - s(t-h)}{2h}
\]
For \( t = 10 \) and using \( h = 1 \) (since time intervals are spaced by 1 second):
\[
v(10) \approx \frac{s(11) - s(9)}{2} = \frac{30.302 - 21.460}{2} = \frac{8.842}{2} = 4.421 \, \text{m/s}
\]

### Part (ii) Three-point central difference formula for acceleration at \( t = 10 \):
The three-point central difference formula for acceleration \( a(t) \) is:
\[
a(t) \approx \frac{s(t+h) - 2s(t) + s(t-h)}{h^2}
\]
For \( t = 10 \) and \( h = 1 \):
\[
a(10) \approx \frac{s(11) - 2s(10) + s(9)}{1^2} = \frac{30.302 - 2(25.752) + 21.460}{1} = 30.302 - 51.504 + 21.460 = 0.258 \, \text{m/s}^2
\]

### Part (iii) MATLAB code using spline interpolation:
We are tasked with estimating the speed and acceleration at \( t = 8.5 \), \( t = 10.5 \), and \( t = 11.2 \) using spline interpolation in MATLAB. Below is a MATLAB script that accomplishes this:

```matlab
% Data points for time and distance
t = [8, 9, 10, 11, 12];
s = [17.453, 21.460, 25.752, 30.302, 35.084];

% Perform spline interpolation
spline_fit = spline(t, s);

% Derivatives of the spline for speed and acceleration
spline_speed = fnder(spline_fit, 1);  % First derivative (speed)
spline_acceleration = fnder(spline_fit, 2);  % Second derivative (acceleration)

% Estimations
t_values = [8.5, 10.5, 11.2];  % Times at which we need speed and acceleration

% Estimate speed and acceleration at given times
speed_estimates = ppval(spline_speed, t_values);  % Speed estimates
acceleration_estimates = ppval(spline_acceleration, t_values);  % Acceleration estimates

% Display results
disp('Speed estimates at t = 8.5, 10.5, 11.2:')
disp(speed_estimates)
disp('Acceleration estimates at t = 8.5, 10.5, 11.2:')
disp(acceleration_estimates)
```

This code will output the estimates for speed and acceleration at the specified times using spline interpolation.

Would you like further details on any of these methods? 

### Here are 5 related questions to explore further:
1. How does the accuracy of central difference formulas change with smaller time intervals?
2. What is the significance of using spline interpolation over linear interpolation?
3. How can the second derivative (acceleration) be interpreted in a physical context?
4. Can we apply higher-order numerical methods (like Runge-Kutta) to this problem, and how?
5. What are other methods to calculate derivatives from discrete data points?

### Tip:
When dealing with real-world data, numerical differentiation can be sensitive to noise. Always try to smooth the data or use methods that reduce error, like spline interpolation.

The distance traveled by an object is given at different time intervals, and the speed and acceleration are to be estimated at specific points using central difference formulas and MATLAB spline interpolation.

Explore the numerical estimation of speed and acceleration at specific times using two-point and three-point central difference formulas, along with MATLAB spline interpolation. Includes detailed data points and MATLAB code for t = 8.5, 10.5, and 11.2.

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