The figure is a somewhat idealized graph of the velocity of blood in the ascending aorta during one beat of the heart. Approximately how far, in cm, does the blood move during one beat?

6.2 cm

Explanation:

Here we need to integrate the given graph to find the total distance moved by the blood.

We have two lines, one with the points:

(0.1s, 0m/s) and (0.15s, 0.8 m/s)

remember that:

A linear relationship can be written as:

y = a*x + b

where a is the slope and b is the y-axis intercept.

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

Then the slope for this line is:

a = (0.8m/s - 0m/s)/(0.15s - 0.1s) = 1.6 m/s^2

y = (1.6m/s^2)*x + b

To find the value of b we can just replace the values of one of the points, like (0.1s, 0m/s)

0m/s = (1.6m/s^2)*0.1s + b

0m/s = 0.16m/s + b

b = -0.16m/s

Then the equation is:

y = (1.6m/s^2)*x -0.16m/s

Now we integrate this between x = 0.1s and x = 0.15s

distance = (1/2)*(1.6m/s^2)*((0.15s)^2 - (0.1s)^2) - (0.16m/s)*(0.15s - 0.1s)

distance = 0.002m

Now we need to see the distance moved in the other part of the graph.

This line has the points:

(0.15s, 0.8 m/s) and (0.3s, 0m/s)

Then the slope of this line is:

a = (0m/s - 0.8m/s)/(0.3s - 0.15s) = -5.33m/s^2

y = (-5.33m/s^2)*x + b

Now we replace the values of one of the points, like (0.3s, 0m/s)

0m/s = (-5.33m/s^2)*0.3s + b

b = 5.33m/s^2*0.3s = 1.6m/s

Then the equation for this line is:

y = (-5.33m/s^2)*x + 1.6m/s

Now we integrate this between x = 0.15s  and x = 0.3s

distance = (1/2)*(-5.33m/s)*( (0.3s)^2 - (0.15s)^2) + (1.6m/s)*(0.3s - 0.15s)

distance = 0.06m

Then the total distance moved is:

D = 0.06m + 0.002m = 0.062m

And we want this in cm, remember that 1m = 100cm

D = 0.062m = 0.062*100cm = 6.2 cm