Here is a *v vs. t*, or *velocity vs. time*,
graph.

*Velocity* (v) is *vertical*.

*Time* (t) is *horizontal*.

Initially, when *t = 0 s*, the velocity of the object is
*v = 0 m/s*.

From then on as time passes the object moves with a greater and greater positive
velocity.

Let's look at two points on this graph....

Examine the *first* point, *(t*_{1}*,
v*_{1}).

When *t*_{1}*
= 2 s*, then
*v*_{1}* =
3 m/s*.

Examine the *second* point, *(t*_{2}*,
v*_{2}).

When *t*_{2}*
= 6 s*, then
*v*_{2}* =
9 m/s*.

Let's find the *slope* of this graph using those two
points...

Using those two points, here is the *rise* and the
*run* of the *slope* of this
*v vs. t* graph.

Here, the rise is the *difference of the velocity*
coordinates, or
*v*_{2}* -
v*_{1}, as in:

rise = v_{2} - v_{1}

rise = 9 m/s - 3 m/s

rise = 6 m/s

Here, the *run* is the *difference of the
time* coordinates, or
*t*_{2}* -
t*_{1}, as in:

run = t_{2} - t_{1}

run = 6 s - 2 s

rise = 4 s

The *slope* of this graph is **
***a change in velocity divided by a change in time*, as in:

slope = rise / run

slope = 6 m/s / 4 s

slope = 1.5 m/s/s (or m/s^{2})

This *slope is the acceleration* of the object, since
*acceleration* is *defined*
as the *change in velocity divided by the change in time*.

So, acceleration = 1.5 m/s/s.

## The *slope of the v vs. t graph* is the *acceleration* of the
object.