This is a straight line, since it is in the form

*y* = *mx *+* c*

See more on Straight Line.

Since we've recognized it is a straight line, we only need to plot 2 points and join them. But we find 3 points, just to make sure we have the correct line.

t |
0 | 1 | 2 |

v |
9 | -0.8 | -10.6 |

Graph of `v` against `t` - straight line.

Our graph starts at `t = 0` (since negative time values have no meaning in this example).

For the first `0.918\ "s"`, the ball is going up (positive velocity - that is, the blue line is above the *t*-axis), but slowing down.

Thereafter, the ball is coming down towards the ground and getting faster (the portion where the blue line is below the *t*-axis).

The ball hits the ground at approx `t = 2.04\ "s"` (we can
see this from Example 1). The **velocity** when the
ball hits the ground from the graph we just drew is about `-11\
"m/s"`. The graph stops at this point.

Our graph assumes the ball lands in sand and doesn't bounce.

Normally, as we have done here, we take velocity in the **up**
direction to be positive.

Easy to understand math videos:

MathTutorDVD.com