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
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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.

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