# Kinematics in 1D, stumped

1. Sep 2, 2009

### Satyr

1. The problem statement, all variables and given/known data
The left ventricle of the heart accelerates blood from rest to a velocity of +26 cm/s. (a) if the displacement of the blood during acceleration is +2.0 cm, determine its acceleration (in cm/s^2). (b) How much time does blood take to reach its final velocity

2. Relevant equations
V=Vi + at
D=RT
a=v/t

3. The attempt at a solution

I cannot figure out what I am doing wrong with this problem...I have logically approached this from two directions and am getting answers that aren't close to what the book is suggesting. Here's my work:

So I know initial velocity is 0 and final velocity is 26 cm/s. I know the displacement (delta x) is 2 and am looking for acceleration...a=v/t; t= D/R = (2cm/26cm/s) = .0769 seconds to travel those 2 centimeters. Since I now have t (just solved for it) and was given the final velocity to begin with, I can plug it in to a=v/t to yield 26/.0769 = 338 m/s^2. This approach was wrong.

By using the equation x=.5(Vo+V)t, I was able to get to x=1/2(Vo+V)(V-Vo/a). By rearranging the values and substituting in I got to 26^2/4 which equals 169 m/s^2. This is not correct either.

Aside from how to do this problem, my question is how can two different approaches that seem perfectly correct to me liberate two different answers?

2. Sep 2, 2009

### Jebus_Chris

Maybe you know this formula but forgot it?
$$v^2 = v_o^2 + 2a\Delta x$$ (just solve for a)
If you don't then you would have to use the average velocity. This would be used in your first approach.
$$V_a_v_g=\frac{v+v_o}{2}$$

3. Sep 2, 2009

### rl.bhat

In the first case if you indicate R as the velocity, t = D/R is wrong. Because the velocity is not uniform.
The second method is correct and the value of a is also correct.

4. Sep 2, 2009

### Satyr

Thank you for clearing up my error in the first approach, I can see it now. As for your suggestion, I have tried that as well and it yields the same answer as my second approach. I feel like a huge fool right now. I found my mistake and, come to find out, my second approach was correct after all.

You're right. Thanks!