Let $\Delta t$ be the duration of one time interval and $\Delta x_n$ be the displacement during the $n$-th interval.

[Q1] Uniformly Accelerated Motion & [Q3] Reasoning (Part 1)

For uniformly accelerated motion with acceleration $a$, the displacement in the $n$-th interval is given by $\Delta x_n = v_{n-1}\Delta t + \frac{1}{2}a(\Delta t)^2$, where $v_{n-1}$ is the velocity at the start of that interval.

The velocity at the start of the next interval, $(n+1)$, is $v_n = v_{n-1} + a\Delta t$. The displacement in this interval is:

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

The difference between consecutive displacements is a key indicator:

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

This derivation shows that for uniformly accelerated motion, the difference in displacements between any two consecutive, equal time intervals must be constant.

Let's check the given data: $\Delta x_1 = 3$ m, $\Delta x_2 = 9$ m, $\Delta x_3 = 15$ m, $\Delta x_4 = 21$ m, $\Delta x_5 = 27$ m. The differences are:

$$ \Delta x_2 - \Delta x_1 = 9 \text{ m} - 3 \text{ m} = 6 \text{ m} $$ $$ \Delta x_3 - \Delta x_2 = 15 \text{ m} - 9 \text{ m} = 6 \text{ m} $$ $$ \Delta x_4 - \Delta x_3 = 21 \text{ m} - 15 \text{ m} = 6 \text{ m} $$ $$ \Delta x_5 - \Delta x_4 = 27 \text{ m} - 21 \text{ m} = 6 \text{ m} $$

Since the difference is constant, the motion is uniformly accelerated.

[Q2] Zero Initial Velocity & [Q3] Reasoning (Part 2)

Let $v_0$ be the initial velocity at the start of the first interval ($t=0$). The displacement in the $n$-th interval, $\Delta x_n$, can be expressed as the difference between the total displacement at time $t_n = n\Delta t$ and $t_{n-1} = (n-1)\Delta t$.

$$ x(t) = v_0 t + \frac{1}{2}at^2 $$ $$ \Delta x_n = x(t_n) - x(t_{n-1}) = \left[v_0(n\Delta t) + \frac{1}{2}a(n\Delta t)^2\right] - \left[v_0((n-1)\Delta t) + \frac{1}{2}a((n-1)\Delta t)^2\right] $$ $$ \Delta x_n = v_0\Delta t + \frac{1}{2}a(\Delta t)^2[n^2 - (n-1)^2] = v_0\Delta t + \frac{1}{2}a(\Delta t)^2(2n-1) $$

If the initial velocity is zero ($v_0=0$), the expression simplifies to:

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

This implies that for motion starting from rest, the displacements in consecutive equal time intervals are proportional to consecutive odd integers:

$$ \Delta x_1 : \Delta x_2 : \Delta x_3 : \dots = 1:3:5:\dots $$

Let's check the ratio of the given displacements:

$$ 3 : 9 : 15 : 21 : 27 $$

Dividing all terms by the first term, 3, we get the simplified ratio:

$$ 1 : 3 : 5 : 7 : 9 $$

This ratio matches the theoretical prediction for uniformly accelerated motion with zero initial velocity.