Mechanics 1: Linear Kinematics and Calculus - 3D Math Primer for Graphics and Game Development - page 500

v(2.0) = −2(2.0) + 6 = 2.0 v(2.4) = −2(2.4) + 6 = 1.2 v(2.5) = −2(2.5) + 6 = 1.0

v(2.6) = −2(2.6) + 6 = 0.8 v(3.0) = −2(3.0) + 6 = 0.0

Figure 11.7
The hare's velocity and corresponding tangent line at selected times

So the instantaneous velocity of the hare at t = 2.5 was precisely 1 furlong
per minute, just as our earlier arguments predicted. But now we can say it
with confidence.
Figure 11.7 shows this point and several others along the interval we've
been studying. For each point, we have calculated the instantaneous veloc-
ity at that point according to Equation (11.5) and have drawn the tangent
line with the same slope.
It's very instructive to compare the graphs of position and velocity side
by side. Figure 11.8 compares the position and velocity of our fabled racers.
There are several interesting observations to be made about Figure 11.8.
• When the position graph is a horizontal line, there is zero velocity,

and the velocity graph traces the v = 0 horizontal axis (for example,
during the hare's nap).

• When the position is increasing, the velocity is positive, and when the

position is decreasing (the hare is moving the wrong way) the velocity
is negative.