FIGURE EX2.8 showed the velocity graph of blood in the aorta. ...

Question

FIGURE EX2.8 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?

Accepted Answer

Hey, everyone in this problem, a lubricating oil is driven through a pipe using an electric pump. The speed of an oil particle is plotted in the 1st 10 seconds of operation. We're asked to determine the acceleration of an oil particle through every stage shown on the graph. We're given a graph with velocity in meters per second on the Y axis and time in seconds on the X axis stage one starts at time zero with a velocity of zero and continues to a time of two seconds and a velocity of seven seconds, seven m per second. Sorry stage two. OK. Starts at seven m per second at time two seconds. The velocity decreases to 4m/s at seven seconds. And in stage three, we have no change in velocity. The velocity remains at four m per second from seven seconds to seconds. We're given four answer choices. Each of them have a different Acceleration for the three stages. For stage one, we have differences of either 3.5 m per second squared or negative 3.5 m per second squared depending on the answer trace for the second stage. We have either positive 0.6 m per second squared or negative 0.6 m per second squared. And for this third stage, we have either zero m per second squared, four m per second squared or negative four m per second squared. Now, if this was a test question, it was multiple choice and you were trying to finish this as quickly as possible. What you could notice right away is that stage two, we have a negative slope. The velocity is decreasing. If the velocity is decreasing, then we must have a negative acceleration. And there's only one answer choice with the negative acceleration which is answer choice. A. So you could choose answer choice a right away. Now we're gonna go through and work this out mathematically. So we can see how it works so that you know how to do this. If it's not a multiple choice problem, let's recall the acceleration we're looking for is going to be equal to the derivative of the velocity. So D V divided by D T and when we have a velocity time curve, the derivative is just the slope. And so the acceleration is going to just be the slope of a V T graph. So what we wanna do to solve this problem if we wanna look at each stage and find the slope of that stage, Let's start with stage one. The acceleration is going to be equal to the slope like we just mentioned. And recall that the slope we have the rise over the red. Well, the rise is gonna be the change in the Y axis. So it's gonna be delta V X in this case divided by the run which is a change in the X axis. So delta T in this case, now for stage one, the final velocity is seven m per second, the initial velocity is zero m per second. And so the change is going to be seven m per second zero m/s. And with the time, stage two has a final time of two seconds or sorry. Stage one has a final time of two seconds and an initial time of zero seconds. And so we have two seconds minus zero seconds in the denominator. This gives us an acceleration of 3.5 m/s squared Moving now to stage two. And we're gonna do the exact same thing. It's the same process. Our acceleration is going to be equal to the slope is equal to the right, the change in the Y axis divided by the run, the change in the X axis. In this case, the final Velocity in stage two is m/s And the initial velocity is seven m/s. So the change or delts of VX is going to be four m/s -7 m/s and be careful that you're getting these in the right order. You always want to do the final minus the initial, OK. And that's gonna dictate the sign of the acceleration that you get. So it is important to have them in the right order. And we're gonna divide by the change in time. Stage two ends at seven seconds and it started at two seconds. So we have seven seconds minus two seconds. This gives us negative three divided by five m per second squared, Which is equal to negative 0.6 meters per second squared. And again, that's stage two, that's that negative acceleration that we expected when we first looked at this problem and we looked at the answer choices. So that's our acceleration for stage two. Now let's move to stage three In stage three, same process, the acceleration is equal to the slope which is equal to the change in the Y axis delta V X divided by the change in the X axis delta T Our graph shows that in stage three, the velocity doesn't change the initial velocity and the final velocity are both 4m/s. So acceleration is going to be four m per second minus four m per second divided by the change in time. We go from seven seconds up to 10 seconds. So we have 10 seconds minus seven seconds. In the denominator, we get a slope of zero m per second squared. So we have an acceleration of zero m per second squared. And that makes sense. This is a horizontal line. So we expect that it would have no slope. So we've found our three accelerations for the three stages. Now, we can compare them to our answer choices. We found that the first stage had an acceleration of 3.5 m/s squared. The second stage had an acceleration of negative 0.6 m per second squared. And the third stage had an acceleration of zero m per second squared. This corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

FIGURE EX2.8 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?

Key Concepts

Velocity

Acceleration

Slope of a Graph

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