We can define the instantanous velocity as a limit of an average velocity,
as the time interval gets smaller and smaller. Let s (t) be the
position of an object at time t. The instantaneous velocity at t = a is defined as
1. Graph of height vs time
Initially, the applet shows a graph of height above the ground versus time, for
the object that we examined in the previous page. Note that this graph is
not a drawing of the path of the object, but is a graph
of height versus time. The object was actually tossed straight up and
fell straight back down.
2. Velocity at 1 second
Select the second example from the drop down menu. This just zooms in
on the interval between 0 and 3 seconds. Notice the green line, which
extends from the green dot at the point (2,142) to the red dot at the
point (1,90). What is the slope of this green line? It's just rise over
But this is also the average velocity over the interval
from 1 to 2 seconds. In other words, we can visualize the average
velocity over an interval as the slope of the secant line between the
endpoints of that interval. The slope of the green secant line is
displayed in a small box on the graph.
What we want to find out is the instantaneous velocity at t = 1
second. We can approach this just like on the previous page by making the
interval smaller. Click-drag the green dot closer to the red dot (zooming
with the mouse is turned off on this applet so you don't accidentally
zoom when you really meant to click on a dot). Notice that the slope is
increasing as the green dot approaches the red dot.
In fact, the slope of
the secant is approaching the slope of the red line, which is tangent to
the curve at the point (1,90). The slope of this tangent line is 68,
which is the instantaneous velocity at t = 1. We can think of
instantaneous velocity as the slope of the tangent line at a point on our
position curve, just like average velocity is the slope of the secant
While you are here..
... I have a small favor to ask. Over the years we have used advertising to support the site so it can remain free for everyone.
However, advertising revenue is falling and I have always hated the ads. So, would you go to Patreon and become a patron of the site?
When we reach the goal I will remove all advertising from the site.
It only takes a minute and any amount would be greatly appreciated.
Thank you for considering it! – John Page
Become a patron of the site at patreon.com/mathopenref
Other differentiation topics
Derived from the work of Thomas S. Downey under a Creative Commons Attribution 3.0 License.