[Q1] Find the average velocity during the time interval Δt that begins at time t.

The average velocity is defined as the total displacement divided by the time interval:

$$ \bar{v} = \frac{\Delta s}{\Delta t} $$

The time interval begins at time $t$ and ends at $t + \Delta t$. The positions at these times are given by the equation $s = 3t^2$:

$$ s(t) = 3t^2 $$ $$ s(t+\Delta t) = 3(t+\Delta t)^2 $$

The displacement $\Delta s$ during this interval is the difference between the final and initial positions:

$$ \Delta s = s(t+\Delta t) - s(t) = 3(t+\Delta t)^2 - 3t^2 $$

Expanding the expression:

$$ \Delta s = 3(t^2 + 2t\Delta t + (\Delta t)^2) - 3t^2 = 6t\Delta t + 3(\Delta t)^2 $$

Substituting this into the definition of average velocity:

$$ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{6t\Delta t + 3(\Delta t)^2}{\Delta t} $$ $$ \bar{v} = 6t + 3\Delta t $$

[Q2] Find the magnitude of the instantaneous velocity at time t.

The problem states the motion is uniformly accelerated. We compare the given position equation, $s = 3t^2$, with the general kinematic equation for uniformly accelerated motion, $s = s_0 + v_0 t + \frac{1}{2} a t^2$. This comparison yields: Initial position $s_0 = 0$ Initial velocity $v_0 = 0$

$\frac{1}{2}a = 3 \implies a = 6$ m/s²

The instantaneous velocity for uniformly accelerated motion is given by $v(t) = v_0 + at$. Substituting the values we found:

$$ v(t) = 0 + (6)t = 6t $$

Since time $t \ge 0$, the magnitude of the instantaneous velocity is equal to its value.

Alternative (Calculus-based) Method: The instantaneous velocity is the limit of the average velocity (from Q1) as the time interval $\Delta t$ approaches zero:

$$ v(t) = \lim_{\Delta t \to 0} \bar{v} = \lim_{\Delta t \to 0} (6t + 3\Delta t) = 6t $$

This is equivalent to the time derivative of the position function:

$$ v(t) = \frac{ds}{dt} = \frac{d}{dt}(3t^2) = 6t $$