Answer
The acceleration during the speeding up phase is $16 ~m/s^2$
The acceleration during the slowing down phase is $-5.33 ~m/s^2$
Work Step by Step
The acceleration is equal to the slope of the velocity versus time graph.
We can find the slope from point A to point B when the blood is speeding up. We can use point A (0.1 s, 0 m/s) and point B (0.15 s, 0.8 m/s) to calculate the slope.
$a_y=\frac{\Delta v_y}{\Delta t}$
$a_y = \frac{0.8 ~m/s - 0}{0.15 ~s-0.1~s}$
$a_y = \frac{0.8 ~m/s}{0.05 ~s}$
$a_y = 16 ~m/s^2$
The acceleration during the speeding up phase is $16 ~m/s^2$
We can find the slope from point B to point C when the blood is slowing down. We can use point B (0.15 s, 0.8 m/s) and point C (0.3 s, 0 m/s) to calculate the slope.
$a_y=\frac{\Delta v}{\Delta t}$
$a_y = \frac{0-0.8 ~m/s}{0.3 ~s-0.15~s}$
$a_y = \frac{-0.8 ~m/s}{0.15 ~s}$
$a_y = -5.33 ~m/s^2$
The acceleration during the slowing down phase is $-5.33 ~m/s^2$
