How to Solve a Kinematics Problem Involving Acceleration and Time

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Homework Help Overview

The discussion revolves around a kinematics problem involving the acceleration of blood in the heart, where the left ventricle accelerates blood from rest to a velocity of +26 cm/s over a displacement of +2.0 cm. Participants are tasked with determining the acceleration and the time taken to reach the final velocity.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to solve for acceleration using two different approaches but encounters discrepancies in the results. They express confusion over why two logically sound methods yield different answers.
  • Some participants suggest using the equation v² = v₀² + 2aΔx to find acceleration, while others critique the use of t = D/R due to the non-uniform nature of velocity during acceleration.
  • There is a discussion about the average velocity and its relevance to the problem, with some participants questioning the assumptions made in the calculations.

Discussion Status

Participants are actively engaging with the problem, providing feedback on each other's approaches. Some guidance has been offered regarding the correct application of kinematic equations, and there is acknowledgment of errors in the original poster's reasoning. The conversation reflects a collaborative effort to clarify misunderstandings without reaching a definitive conclusion.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available or the methods they can use. There is an emphasis on understanding the underlying principles of kinematics rather than simply arriving at a solution.

Satyr
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Homework Statement


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


Homework Equations


V=Vi + at
D=RT
a=v/t


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 into 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?

Thanks for your help.
 
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Satyr said:
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 into a=v/t to yield 26/.0769 = 338 m/s^2. This approach was wrong. The reason why that is wrong is because of t = D/R. You have R equal to 26. R is equal to 0 at the start and equal to 26 at the end

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.There is nothing to rearrange in this equation. You know all the variables except t
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}
 
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.
 
Jebus_Chris said:
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 avTHerage velocity. This would be used in your first approach.
V_a_v_g=\frac{v+v_o}{2}

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.

rl.bhat said:
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.

You're right. Thanks!
 

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