A graph of Runner Velocity vs. Time has a horizontal axis labeled Time (s) and a vertical axis labeled velocity $(\mathrm{m} / \mathrm{s})$. Straight lines connect adjacent points, which have approximate values as follows.
\[
(0,0),(2,6),(4,12),(6,12),(8,12) \text {, and }(10,12)
\]
(a) What is his average velocity (in $\mathrm{m} / \mathrm{s}$ ) for the first $4 \mathrm{~s}$ ?
$\mathrm{m} / \mathrm{s}$
(b) What is his instantaneous velocity (in $\mathrm{m} / \mathrm{s}$ ) at $t=6 \mathrm{~s}$ ?
$\mathrm{m} / \mathrm{s}$
(c) What is his average acceleration (in $\mathrm{m} / \mathrm{s}^{2}$ ) between 0 and $4 \mathrm{~s}$ ? $\mathrm{m} / \mathrm{s}^{2}$
(d) What is his time (in s) for the race?

Step 1 ：Given the points on the graph, we can see that the runner's velocity is increasing linearly from 0 to 12 m/s over the first 4 seconds.

Step 2 ：The average velocity is calculated by the total displacement divided by the total time.

Step 3 ：The total displacement can be calculated as the area under the velocity-time graph, which forms a trapezoid in this case.

Step 4 ：The formula for the area of a trapezoid is \((\text{base1} + \text{base2}) / 2 * \text{height}\). Here, base1 is the initial velocity (0 m/s), base2 is the final velocity (12 m/s), and the height is the time interval (4 s).

Step 5 ：Calculate the total displacement: \(\text{displacement} = (0 + 12) / 2 * 4 = 24.0\) m

Step 6 ：Calculate the average velocity: \(\text{average_velocity} = \text{displacement} / \text{time} = 24.0 / 4 = 6.0\) m/s

Step 7 ：Final Answer: The average velocity for the first 4 seconds is \(\boxed{6 \, \mathrm{m/s}}\).