A body moves on a coordinate line such that it has a position $s=f(t)=t^{2}-4 t+3$ on the interval $0 \leq t \leq 5$, with $s$ in meters and $t$ in seconds.
a. Find the body's displacement and average velocity for the given time interval.
b. Find the body's speed and acceleration at the endpoints of the interval.
c. When, if ever, during the interval does the body change direction?

Step 1 ：At time \(t = k\), the body is at \(f(k) = k^{2}-4 k+3\).

Step 2 ：At time \(t = k + 1\), the body is at \(f(k+1) = (k+1)^{2}-4 (k+1)+3\).

Step 3 ：The displacement of the body is \(f(k+1) - f(k) = ((k+1)^{2}-4 (k+1)+3) - (k^{2}-4 k+3) = 2k - 3\).

Step 4 ：The average velocity of the body is the displacement divided by the time interval, which is \(\frac{2k - 3}{1} = 2k - 3\).

Step 5 ：At the endpoints of the interval, \(t = 0\) and \(t = 5\), the speed of the body is the absolute value of the derivative of the position function, which is \(|f'(t)| = |2t - 4|\).

Step 6 ：At \(t = 0\), the speed is \(|f'(0)| = |2*0 - 4| = 4\), and at \(t = 5\), the speed is \(|f'(5)| = |2*5 - 4| = 6\).

Step 7 ：The acceleration of the body is the derivative of the velocity, which is \(f''(t) = 2\). So, the acceleration is constant and equal to 2 at both endpoints.

Step 8 ：The body changes direction when the velocity changes sign, which happens when \(f'(t) = 0\). Solving the equation \(2t - 4 = 0\), we find that the body changes direction at \(t = 2\).

Step 9 ：So, the body's displacement and average velocity for the given time interval are \(2k - 3\) and \(2k - 3\) respectively, the body's speed and acceleration at the endpoints of the interval are 4 and 6, and 2 respectively, and the body changes direction at \(t = 2\).

Step 10 ：Final Answer: \(\boxed{Displacement = 2k - 3, Average Velocity = 2k - 3, Speed at t = 0 is 4, Speed at t = 5 is 6, Acceleration = 2, Change of direction at t = 2}\)