For uniformly accelerated motion, the distance traveled, $d_n$, during the $n$-th consecutive time interval, $\Delta t$, is given by:

$$d_n = v_{n-1} \Delta t + \frac{1}{2}a(\Delta t)^2$$

where $v_{n-1}$ is the velocity at the beginning of the $n$-th interval.

The distance in the next interval, $d_{n+1}$, is:

$$d_{n+1} = v_n \Delta t + \frac{1}{2}a(\Delta t)^2$$

The velocity $v_n$ is related to $v_{n-1}$ by $v_n = v_{n-1} + a\Delta t$. Substituting this into the expression for $d_{n+1}$:

$$d_{n+1} = (v_{n-1} + a\Delta t)\Delta t + \frac{1}{2}a(\Delta t)^2 = \left(v_{n-1}\Delta t + \frac{1}{2}a(\Delta t)^2\right) + a(\Delta t)^2$$ $$d_{n+1} = d_n + a(\Delta t)^2$$

This shows that the difference in distance between consecutive intervals is constant: $\Delta d = d_{n+1} - d_n = a(\Delta t)^2$.

[Q1] Distance during the seventh 2-second interval

1. Calculate the acceleration ($a$) From the problem, the time interval is $\Delta t = 2$ s. The distances in the first two intervals are $d_1 = 24$ m and $d_2 = 20$ m. The change in distance is $\Delta d = d_2 - d_1 = 20 \text{ m} - 24 \text{ m} = -4$ m. Using the derived relation:

$$a = \frac{\Delta d}{(\Delta t)^2} = \frac{-4 \text{ m}}{(2 \text{ s})^2} = -1 \text{ m/s}^2$$

2. Calculate the initial velocity ($v_0$) Using the distance traveled in the first interval ($n=1$):

$$d_1 = v_0 \Delta t + \frac{1}{2}a (\Delta t)^2$$ $$24 \text{ m} = v_0(2 \text{ s}) + \frac{1}{2}(-1 \text{ m/s}^2)(2 \text{ s})^2$$ $$24 = 2v_0 - 2 \implies v_0 = 13 \text{ m/s}$$

3. Determine the time to stop ($t_{stop}$) The object stops when its final velocity is zero ($v_f=0$).

$$v_f = v_0 + at_{stop}$$ $$0 = 13 \text{ m/s} + (-1 \text{ m/s}^2)t_{stop} \implies t_{stop} = 13 \text{ s}$$

4. Analyze the motion in the seventh interval The seventh interval spans from $t_6 = 6\Delta t = 12$ s to $t_7 = 7\Delta t = 14$ s. The object comes to a stop at $t_{stop} = 13$ s, which is inside this interval. Therefore, the object moves forward, stops, and reverses direction. The total distance is the sum of the path lengths.

• Distance from $t=12$ s to $t=13$ s ($d_A$): First, find the velocity at the start of the interval, $t=12$ s:

  $$v(12) = v_0 + a(12) = 13 + (-1)(12) = 1 \text{ m/s}$$

  The distance traveled until it stops at $t=13$ s can be found using $v_f^2 = v_i^2 + 2ad$:

  $$0^2 = (1 \text{ m/s})^2 + 2(-1 \text{ m/s}^2)d_A \implies d_A = 0.5 \text{ m}$$

• Distance from $t=13$ s to $t=14$ s ($d_B$): The object starts from rest at $t=13$ s and moves for 1 s. The distance traveled is $d = |v_i t + \frac{1}{2}at^2|$:

  $$d_B = |(0)(1 \text{ s}) + \frac{1}{2}(-1 \text{ m/s}^2)(1 \text{ s})^2| = |-0.5 \text{ m}| = 0.5 \text{ m}$$

• Total distance in the seventh interval ($d_7$):

  $$d_7 = d_A + d_B = 0.5 \text{ m} + 0.5 \text{ m} = 1.0 \text{ m}$$