Question

FIGURE EX2.7 showed the velocity graph of blood in the aorta. What is the blood’s acceleration during each phase of the motion, speeding up and slowing down?

Short answer

The acceleration of blood during the speeding up phase is $16m{s}^{-2}$.

The acceleration of blood during the speeding down phase is$-5.3m{s}^{-2}$.

Step-by-step solution

Given information

The velocity-time graph of blood in the aorta is shown below

Find the acceleration of blood in the aorta during the speeding up phase.

Average acceleration is given by the slope of the velocity-versus-time graph.

For the first phase, consider two points $\left(t_1,v_1\right)=\left(0.1,0\right)$and $\left(t_2,v_2\right)=\left(0.15,0.8\right)$.

The acceleration of the blood is

$$\begin{gathered} a=\frac{v_2-v_1}{t_2-t_1} \\ =\frac{0.8-0}{0.15-0.1} \\ =16\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore, the acceleration of blood during the speeding up phase is$16\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Find the acceleration of blood in the aorta during the speeding down phase.

For the second phase, consider two points $\left(t_2,v_2\right)=\left(0.15,0.8\right)$and $\left(t_3,v_3\right)=\left(0.3,0\right)$.

The acceleration of the blood is

$$\begin{gathered} a=\frac{v_3-v_2}{t_3-t_2} \\ =\frac{0-0.8}{0.3-0.15} \\ =-5.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore, the acceleration of blood during the speeding down phase is$-5.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.