Instantaneous Velocity

The problem with average velocity

Average velocity tells you the overall result over a time interval. But it says nothing about what the body was doing at any particular moment within that interval.

Consider a ball thrown upward. At the top of its path, it is momentarily at rest — its velocity is zero at that instant. But the average velocity over the entire flight (up and back down) is also zero. These are two very different zeros — one is an instantaneous fact, the other is an overall average.

To talk about velocity at a specific instant, we need instantaneous velocity.

Building the idea

Take a small time interval $\Delta t$ around the instant we care about. Measure the displacement $\Delta x$ during that interval. The average velocity over $\Delta t$ is $\frac{\Delta x}{\Delta t}$.

Now make $\Delta t$ smaller and smaller — shrink it toward zero. The average velocity over the shrinking interval approaches a definite value. That limiting value is the instantaneous velocity:

$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$

In the language of calculus, this limit is the derivative of position with respect to time:

$\boxed{v = \frac{dx}{dt}}$

What the derivative means geometrically

On a position-time graph ($x$ vs $t$), the average velocity over an interval is the slope of the secant line joining two points on the curve.

As the interval shrinks, the secant line rotates and approaches the tangent line at the point of interest.

The instantaneous velocity at any moment is the slope of the tangent to the position-time curve at that moment.

• Steep upward slope → large positive velocity
• Gentle upward slope → small positive velocity
• Horizontal tangent → zero velocity (momentarily at rest)
• Downward slope → negative velocity (moving in negative direction)

Finding instantaneous velocity from a position function

If $x(t)$ is given as a function of time, differentiate it:

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

Example: $x = 3t^2 + 2t + 1$ (in metres, $t$ in seconds)

$v = \frac{dx}{dt} = 6t + 2 \text{ m/s}$

At $t = 0$: $v = 2$ m/s

At $t = 3$ s: $v = 6(3) + 2 = 20$ m/s

The velocity is different at every instant — the body is accelerating.

Relation to speed

Instantaneous speed is the magnitude of instantaneous velocity:

$\text{speed} = |v| = \left|\frac{dx}{dt}\right|$

Speed is always non-negative. Velocity can be negative (when moving in the negative direction).

Special cases

The third row recovers the first equation of motion — consistent with everything we have derived.