In this article, you will learn what we mean by instantaneous velocity when describing the motion of a particle.
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:
| v = | lim | Δx | |
| Δt→0 | Δt |
The interval of time Δt starts at some instant t1 and ends at some subsequent instant t2, so we can write:
And since the instant t is within this interval of time, it follows that:
To better see what we're talking about, let's represent this on a position vs time graph of a particle:
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:
Since Δt = t2 − t1 and t1 = t, we can write:
This means that the extremes of Δt are t and t + Δt:
Since the instantaneous velocity at t is equal to whatever the average velocity for Δt approaches as Δt approaches 0:
| v = | lim | Δx | |
| Δt→0 | Δt |
and we know that the change in position Δx is
it follows that we can write the instantaneous velocity v as
| v = | lim | xt+Δt − xt | |
| Δt→0 | Δt |
Let's now go through an example to demonstrate how to use this formula.
Let's consider an object whose position in meters at an instant t, specified in seconds, is given by 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:
| v = | lim | xt+Δt − xt | |
| Δt→0 | Δt |
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
Thus, the instantaneous velocity v becomes:
| v = | lim | (t + Δt)2 − t2 | |
| Δt→0 | Δt |
Let's work out the numerator:
| v = | lim | t2 + 2tΔt + Δt2 − t2 | |
| Δt→0 | Δt |
| v = | lim | 2tΔt + Δt2 | |
| Δt→0 | Δt |
Next, we can divide the numerator by Δt:
| v = | lim | 2t + Δt |
| Δt→0 |
As Δt approaches 0, 2t + Δt approaches 2t, so we can write:
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.
The limit
| lim | xt+Δt − xt | |
| Δt→0 | Δt |
is called the derivative of x with respect to t, and is indicated as
| dx |
| dt |
So, we can write
| v = | lim | xt+Δt − xt | = | dx | |
| Δt→0 | Δt | dt |
Therefore, we say that the instantaneous velocity is the derivative of position with respect to time.
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:
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:
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
| v = | dx | = mT |
| dt |
Let's call θ the angle that the line tangent to the graph at the point t makes with the positive t-axis:
We can look at the sign of the angle θ to determine whether the instantaneous velocity is positive, negative, or zero:
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:
