Q.The position of an object moving along x-axis is given by $x = a + bt^{2}$ where $a = 8.5\ \text{m}$, $b = 2.5\ \text{m s}^{-2}$ and $t$ is measured in seconds. What is its velocity at $t = 0\ \text{s}$ and $t = 2.0\ \text{s}$? What is the average velocity between $t = 2.0\ \text{s}$ and $t = 4.0\ \text{s}$?

Concept: Instantaneous Velocity — the derivative of position with respect to time.

1. Given $x = a + bt^{2}$, the instantaneous velocity is

   $v(t) = \frac{dx}{dt} = 2bt$.

2. At $t = 0$:

   $v(0) = 2b(0) = 0\ \text{m/s}$.

3. At $t = 2.0\ \text{s}$:

   $v(2) = 2(2.5)(2.0) = 10\ \text{m/s}$.

4. Average velocity between $t = 2.0\ \text{s}$ and $t = 4.0\ \text{s}$:

   $x(2) = 8.5 + 2.5(4) = 18.5\ \text{m}$,

   $x(4) = 8.5 + 2.5(16) = 48.5\ \text{m}$, …

The key idea here is that instantaneous velocity is the rate of change of position at a single instant — the derivative of $x$ with respect to $t$. Average velocity, on the other hand, is simply the total displacement divided by the total time interval. Both are fundamental in kinematics, and the distinction between them is a classic point of confusion.

Why does this approach work? Because the position function $x(t) = a + bt^{2}$ is a simple quadratic in time. The constant $a$ just sets the starting position; it doesn't affect velocity at all. The $bt^{2}$ term means the object speeds up as time increases — its velocity grows linearly with time. Taking the derivative gives us that linear relationship directly.

Let’s work through each part step by step.

1. Find the instantaneous velocity function. Velocity $v(t)$ is the first derivative of position with respect to time:

Since $a$ is constant, its derivative is zero. The derivative of $bt^{2}$ is $2bt$. So:

This tells us velocity increases linearly from zero, with slope $2b$.

2. Velocity at $t = 0\ \text{s}$. Substitute $t = 0$ into $v(t)$:

The object starts from rest — the derivative is zero at the vertex of the parabola.

3. Velocity at $t = 2.0\ \text{s}$. Given $b = 2.5\ \text{m s}^{-2}$:

So at $t = 2.0$ seconds, the object is moving at $10\ \text{m/s}$ along the $x$-axis.

4. Average velocity between $t = 2.0\ \text{s}$ and $t = 4.0\ \text{s}$. Average velocity is defined as: …

Concept: Instantaneous Velocity as a Limiting Process

Method: First-Principles Limit Definition (no differentiation shortcut)

Instead of quoting the differentiation rule $\frac{d}{dt}(bt^2) = 2bt$, this method builds instantaneous velocity directly from its definition — the limit of average velocity over a shrinking time interval — which is how the x-$t$ relation, not a memorized rule, tells you what's happening physically.

Steps

1. Set up the average velocity over a small interval $\Delta t$ starting at time $t$:

2. Expand $x(t+\Delta t)$ for $x = a + bt^2$:

3. Subtract $x(t) = a + bt^2$:

4. Divide by $\Delta t$:

5. Take the limit $\Delta t \to 0$ — this is the formal definition of instantaneous velocity, $v = \lim_{\Delta t \to 0} v_{\text{avg}}$. The term $b\,\Delta t \to 0$, leaving

This is exactly the derivative, but reached by shrinking a real physical interval to zero rather than by quoting a calculus rule — the same limiting idea used to justify every differentiation rule in the first place.

6. Evaluate at the two required instants, with $b = 2.5\ \text{m s}^{-2}$: $v(0) = 2(2.5)(0) = 0\ \text{m s}^{-1}, \qquad v(2.0) = 2(2.5)(2.0) = 10\ \text{m s}^{-1}$ …