How to Solve a Kinematics Problem Involving Acceleration and Time

In summary, the conversation discusses a problem involving determining the acceleration of blood in the left ventricle of the heart, as well as the time it takes for the blood to reach its final velocity. Two different approaches are used, one using the equation v^2 = v_o^2 + 2a\Delta x and the other using the average velocity. The second approach is found to be correct, with the mistake in the first approach being the incorrect use of R as the velocity.
  • #1
Satyr
12
0

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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  • #2
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?
[tex] v^2 = v_o^2 + 2a\Delta x[/tex] (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.
[tex] V_a_v_g=\frac{v+v_o}{2}[/tex]
 
  • #3
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
Jebus_Chris said:
Maybe you know this formula but forgot it?
[tex] v^2 = v_o^2 + 2a\Delta x[/tex] (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.
[tex] V_a_v_g=\frac{v+v_o}{2}[/tex]

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!
 

1. What is Kinematics in 1D?

Kinematics in 1D is a branch of classical mechanics that deals with the motion of objects in one-dimensional space. It involves studying the position, velocity, and acceleration of objects without considering the forces that cause the motion.

2. How is Kinematics in 1D different from Kinematics in 2D or 3D?

Kinematics in 1D only considers motion in a single dimension, whereas Kinematics in 2D and 3D involves motion in two or three dimensions, respectively. This means that in 1D, only one coordinate is needed to describe the position of an object, while in 2D and 3D, two or three coordinates are required, respectively.

3. What are the basic equations used in Kinematics in 1D?

The basic equations used in Kinematics in 1D are the equations of motion, which include position, velocity, acceleration, and time. These equations include: x = x0 + v0t + 1/2at2, v = v0 + at, and v2 = v02 + 2a(x-x0), where x is the position, v is the velocity, a is the acceleration, t is time, and x0 and v0 are the initial position and velocity, respectively.

4. How is acceleration related to velocity and position in Kinematics in 1D?

In Kinematics in 1D, acceleration is the rate of change of velocity over time. This means that the acceleration of an object is directly related to its change in velocity and inversely related to the time it takes for that change to occur. Additionally, acceleration is also related to the position of an object through the equation a = dv/dt, where a is acceleration, v is velocity, and t is time.

5. What are some real-life applications of Kinematics in 1D?

Kinematics in 1D has many real-life applications, such as predicting the motion of objects in sports, analyzing the movement of vehicles on a straight road, and understanding the motion of projectiles. It is also used in engineering and physics to design and analyze machines and structures, such as bridges and roller coasters.

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