Define velocity and speed.

II. Short Answer Questions · Q8

Q. Define velocity and speed.

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Concept understanding — Instantaneous Velocity

Instantaneous Velocity: From "How Fast" to "How Fast Right Now"

You already know average velocity. If a car travels 120 km in 2 hours, its average velocity is 60 km/h. That tells you the overall rate, but it hides everything that happened in between — the traffic jams, the sudden bursts of speed, the moments the car was completely stopped.

Now imagine you want to know the car's velocity at exactly 10:15 AM, not averaged over an hour or a minute. That's instantaneous velocity — the velocity at a single instant of time.

The Intuition: Zooming In

Think of a speedometer needle. When you drive, the needle doesn't stay fixed at 60 km/h. It jumps up when you accelerate, drops when you brake. At any given moment, the needle points to a specific number. That number is your instantaneous speed (velocity, if direction matters).

But here's the puzzle: at a single instant, the car hasn't moved any distance. How can you have a speed if $\Delta t = 0$ ? You can't divide by zero.

The trick is to shrink the time interval smaller and smaller, and see what the average velocity approaches.

The Precise Definition

Let $s(t)$ be the position of an object at time $t$ . The average velocity over a time interval $[t, t+h]$ is:

v_{\text{avg}} = \frac{s(t+h) - s(t)}{h}

Now, let $h$ get closer and closer to 0 (but never equal to 0). If the average velocity settles down to a single number as $h \to 0$ , that number is the instantaneous velocity at time $t$ :

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

This limit is exactly the derivative of position with respect to time. In calculus notation: $v(t) = s'(t)$ .

A Concrete Example

Suppose a ball is dropped from rest, and its height (in meters) after $t$ seconds is $s(t) = 4.9t^2$ (ignoring air resistance).

Average velocity from $t=2$ to $t=2.1$ seconds:

v_{\text{avg}} = \frac{4.9(2.1)^2 - 4.9(2)^2}{0.1} = \frac{4.9(4.41 - 4)}{0.1} = \frac{4.9 \times 0.41}{0.1} = 20.09 \text{ m/s}

Average velocity from $t=2$ to $t=2.01$ :

v_{\text{avg}} = \frac{4.9(2.01)^2 - 4.9(2)^2}{0.01} = \frac{4.9(4.0401 - 4)}{0.01} = 19.649 \text{ m/s}

Average velocity from $t=2$ to $t=2.001$ :

v_{\text{avg}} = \frac{4.9(2.001)^2 - 4.9(2)^2}{0.001} = 19.6049 \text{ m/s}

The numbers are converging to 19.6 m/s. That's the instantaneous velocity at $t=2$ seconds.

Using the derivative: $v(t) = 9.8t$ , so $v(2) = 19.6$ m/s. Matches perfectly.

Key Takeaways for Exams

Concept	Meaning	Formula
Average velocity	Total displacement ÷ total time	$\frac{\Delta s}{\Delta t}$
Instantaneous velocity	Velocity at a single moment	$\lim_{h \to 0} \frac{s(t+h)-s(t)}{h}$

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Q.Define velocity and speed.