# College Physics 2.26 – Blood acceleration

## Solution:

### Part A

The sketch should contain the starting point and the final point. This will be done by connecting a straight line from the starting point to the final point. The sketch is shown below.

### Part B

The list of known variables are:

Initial velocity: $\text{v}_0=0\:\text{m/s}$
Final Velocity: $\text{v}=30.0\:\text{cm/s}$
Distance Traveled: $\text{x}-\text{x}_0=1.80\:\text{cm}$

### Part C

The best equation to solve for this is $\:\:\Delta \text{x}=\text{v}_{\text{ave}}\text{t}$ where $\text{v}_{\text{ave}}$ is the average velocity, and t is time. That is $\Delta \text{x}=\text{v}_{\text{ave}}\text{t}$ $\displaystyle \:\text{t}=\frac{\Delta \text{x}}{\text{v}_{\text{ave}}}$ $\displaystyle \:\text{t}=\frac{1.80\:\text{cm}}{\frac{\left(0\:\text{cm/s}+30\:\text{cm/s}\right)}{2}}$ $\:\:\:\text{t}=0.12\:\text{s}$

### Part D

Since the computed value of the time for acceleration of blood out of the ventricle is only 0.12 seconds (only a fraction of a second), the answer seems reasonable. This is due to the fact that an entire heartbeat cycle takes about one second. So, the answer is yes, the answer is reasonable.