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What is the Instantaneous Velocity?

In this article, you will learn what we mean by instantaneous velocity when describing the motion of a particle.

Definition and formula for instantaneous velocity

The instantaneous velocity v that a particle has at an instant t is equal to the value that the average velocity, calculated over an interval of time Δt which includes t, approaches as Δt approaches 0. So, the instantaneous velocity v is expressed as:

vlim Δx
Δt0Δt

The interval of time Δt starts at some instant t1 and ends at some subsequent instant t2, so we can write:

Δt = t2t1

And since the instant t is within this interval of time, it follows that:

t1tt2

To better see what we're talking about, let's represent this on a position vs time graph of a particle:

Position vs time graph where the interval of time from t1 to t2, which includes t, is indicated to get smaller.xtOΔx1t2ttΔtxx12

The arrows next to t1 and t2 are there to indicate that we are considering the interval of time Δt to get smaller and smaller.

Now, since t1 ≤ t ≤ t2, let's consider t1 = t. When we do this, as Δt approaches 0, t2 gets closer and closer to t1:

Position vs time graph in which t1 is equal to t.xtOΔx1t2ttΔt=xx12

Since Δt = t2 − t1 and t1 = t, we can write:

Δt = t2t
t2 = t + Δt

This means that the extremes of Δt are t and t + Δt:

Position vs time graph in which the interval of time Δt starts at t and ends at t + Δt.tΔt+xtOΔxtttΔtΔt+xx

Since the instantaneous velocity at t is equal to whatever the average velocity for Δt approaches as Δt approaches 0:

vlim Δx
Δt0Δt

and we know that the change in position Δx is

Δx = xt+Δtxt

it follows that we can write the instantaneous velocity v as

vlim xt+Δtxt
Δt0Δt

Let's now go through an example to demonstrate how to use this formula.

Example: Instantaneous velocity of an object

Let's consider an object whose position in meters at an instant t, specified in seconds, is given by t2:

x = t2

So, at 1 s the position is 1 m, at 2 s the position is 4 m, at 3 s the position is 9 m, etc.

Our goal here is to find what the instantaneous velocity of the object is at any instant t.

As we've seen, the instantaneous velocity v at an instant t is given by:

vlim xt+Δtxt
Δt0Δt

Since in this case the position of the object at any instant t is given by t2, the positions at the instants t and t + Δt will be

xt = t2
xt+Δt = (t + Δt)2

Thus, the instantaneous velocity v becomes:

vlim (t + Δt)2t2
Δt0Δt

Let's work out the numerator:

vlim t2 + 2tΔt + Δt2t2
Δt0Δt
vlim 2tΔt + Δt2
Δt0Δt

Next, we can divide the numerator by Δt:

vlim 2t + Δt
Δt0

As Δt approaches 0, 2t + Δt approaches 2t, so we can write:

v = 2t

Therefore, the instantaneous velocity of the object at any instant t is given by 2t. So, at 1 s the instantaneous velocity is 2 m/s, at 2 s the instantaneous velocity is 4 m/s, at 3 s the instantaneous velocity is 6 m/s, etc.

Instantaneous velocity and derivatives

The limit

lim xt+Δtxt
Δt0Δt

is called the derivative of x with respect to t, and is indicated as

dx
dt

So, we can write

vlim xt+Δtxt = dx
Δt0Δtdt

Therefore, we say that the instantaneous velocity is the derivative of position with respect to time.

Position vs time graphs and instantaneous velocity

Instantaneous velocity as the slope of a tangent line

Let's return to the position vs time graph we've seen before and let's draw a secant line passing through the points t and t + Δt:

Position vs time graph with the secant line that passes through the points t and t + Δt on the graph.tΔt+xtOΔxtttΔtΔt+xx

As explained in the average velocity article, the slope of this secant line is equal to the average velocity for Δt because Δx/Δt is at the same time the average velocity and the slope of the secant line.

Since the instantaneous velocity at t is what the average velocity for Δt approaches as Δt approaches 0, it follows that the instantaneous velocity at t is what the slope of the secant line approaches as Δt approaches 0.

As Δt gets smaller and smaller, the secant line gets closer and closer to the line tangent to the graph at the point t:

Position vs time graph with the secant line approaching the tangent line at the point t as Δt approaches 0.tΔt+xtOΔxtttΔtΔt+xx

So, as Δt approaches 0, the slope of the secant line approaches the slope of the line tangent to the graph at the point t.

Therefore, the instantaneous velocity at t is equal to the slope of the line tangent to the graph at the point t.

If we indicate the slope of the tangent line with mT, we can write

vdx = mT
dt

Looking at the tangent line angle to determine the sign of instantaneous velocity

Let's call θ the angle that the line tangent to the graph at the point t makes with the positive t-axis:

Position vs time graph with the angle θ that the line tangent to the graph at the point t makes with the positive t-axis.xOtθt

We can look at the sign of the angle θ to determine whether the instantaneous velocity is positive, negative, or zero:

  1. if θ > 0 mT > 0 v > 0
  2. if θ < 0 mT < 0 v < 0
  3. if θ = 0 mT = 0 v = 0

Thus, we can easily determine when the velocity is positive, negative, and zero, simply by looking at the angle θ at different points on a position vs time graph:

Position-time graph with an indication of where the instantaneous velocity is positive, negative, and zero.xOv0>v0<v0=t

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