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Chapter 2: Problem 75

If the contraction of the left ventricle lasts \(250 \mathrm{~ms}\) and the speed of blood flow in the aorta (the large artery leaving the heart) is \(0.80 \mathrm{~m} / \mathrm{s}\) at the end of the contraction, what is the average acceleration of a red blood cell as it leaves the heart? A. \(310 \mathrm{~m} / \mathrm{s}^{2}\) B. \(31 \mathrm{~m} / \mathrm{s}^{2}\) C. \(3.2 \mathrm{~m} / \mathrm{s}^{2}\) D. \(0.32 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
C. \(3.2\,\mathrm{m/s^2}\)

Step by step solution

01

Understand the Problem

We need to find the average acceleration of a red blood cell while it leaves the heart. We know the contraction time is the duration of acceleration, and the final speed of blood flow, which becomes the cruising speed.
02

Identify Known Values

From the problem, we know the time duration is 250 ms, which is \(0.250\) seconds in standard units. The final velocity \(v_f\) is given as \(0.80\,\mathrm{m/s}\). The initial velocity \(v_i\) is zero as the blood starts from rest.
03

Use the Formula for Acceleration

The formula for average acceleration \(a\) is \(a = \frac{v_f - v_i}{t} \). Here, \(v_f = 0.80\,\mathrm{m/s}\), \(v_i = 0\,\mathrm{m/s}\), and \(t = 0.250\,\mathrm{seconds}\).
04

Substitute Known Values into the Formula

Plugging the known values into the formula gives us: \[ a = \frac{0.80 - 0}{0.250} \].
05

Calculate the Acceleration

Calculate the numerical value: \[ a = \frac{0.80}{0.250} = 3.2\,\mathrm{m/s^2}. \]
06

Find the Correct Answer

Matching the calculated value \(3.2\,\mathrm{m/s^2}\) with the options given, we find the correct answer is C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is a core component of physics that deals with the motion of objects without considering the forces that cause this motion. In our example, kinematics principles help us understand the motion of a red blood cell as it leaves the heart. By analyzing the velocity and acceleration over time, we can determine how the blood cell behaves in this system. Kinematics often employs equations to solve problems related to motion, and in this case, the relationship between velocity, time, and acceleration is key to finding the solution.
Velocity
Velocity is a vector quantity that describes the speed of an object in a specific direction. In this exercise, the velocity of the blood cell is given as its speed at the end of the contraction, specified as 0.80 meters per second (m/s).
Velocity is crucial because it allows us to determine how fast the blood moves through the aorta and in what direction. We differentiate velocity from speed, which is a scalar and does not include directional information.
  • Final velocity (\( v_f \)) is the speed of the object at the end of the period being observed.
  • Initial velocity (\( v_i \)) is the speed of the object at the start, which in our problem, is 0 since the blood starts from rest.

Understanding these values is essential to calculating other dynamic quantities like acceleration.
Time Conversion
Time conversion is an important step in physics problem-solving. Often, we need to convert time into standard units to ensure accurate calculations.
In this problem, the contraction duration is given in milliseconds (ms), but most physics equations use seconds (s). To convert milliseconds to seconds, you divide by 1000.
Here's how we converted the given time:
  • Given: \( 250 \mathrm{~ms} \)
  • Conversion: \( 250 \mathrm{~ms} = 0.250 \mathrm{~s} \)
This conversion allows us to apply the standard equations of motion without error.
Always make sure to check units before performing calculations.
Physics Problem-Solving
Physics problems require a systematic approach to solve. Here's a concise guide to tackling such challenges effectively:
  • **Understand the Problem**: Identify what's being asked and note the known quantities. For this problem, the goal is to find average acceleration.
  • **Identify Known Values**: Clearly recognize the given information, like velocities and time, and convert them to standard units if necessary.
  • **Choose the Right Formula**: Select appropriate equations or formulas. For finding acceleration, we use \( a = \frac{v_f - v_i}{t} \).
  • **Substitute and Calculate**: Insert known values into the formula. Here, calculate using \( v_f = 0.80 \mathrm{~m/s}, v_i = 0 \mathrm{~m/s}, \) and \( t = 0.250 \mathrm{~s} \).
  • **Verify the Solution**: Check your calculated value against provided options or interpret the result's meaning.

This structured method aids in solving or verifying steps in numerous physics problems, ensuring clarity and accuracy.

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Most popular questions from this chapter

A car is traveling in the negative \(x\) direction at \(25 \mathrm{~m} / \mathrm{s}\). After \(5 \mathrm{~s}\) passes, it is traveling at \(50 \mathrm{~m} / \mathrm{s},\) again in the negative \(x\) direction. (a) What is the magnitude of its average acceleration? (b) What is the direction of its average acceleration? (c) How far does the car travel during this period?

A healthy heart pumping at a rate of 72 beats per minute increases the speed of blood flow from 0 to \(425 \mathrm{~cm} / \mathrm{s}\) with each beat. Calculate the acceleration of the blood during this process.

A mouse travels along a straight line; its distance \(x\) from the origin at any time \(t\) is given by the equation \(x=\) \(\left(8.5 \mathrm{~cm} \cdot \mathrm{s}^{-1}\right) t-\left(2.5 \mathrm{~cm} \cdot \mathrm{s}^{-2}\right) t^{2} .\) Find the average velocity of the mouse in the interval from \(t=0\) to \(t=1.0 \mathrm{~s}\) and in the interval from \(t=0\) to \(t=4.0 \mathrm{~s}\)

A helicopter 8.50 m above the ground and descending at \(3.50 \mathrm{~m} / \mathrm{s}\) drops a package from rest (relative to the helicopter). Just as it hits the ground, find (a) the velocity of the package relative to the helicopter and (b) the velocity of the helicopter relative to the package. The package falls freely.

A hot-air balloonist, rising vertically with a constant speed of \(5.00 \mathrm{~m} / \mathrm{s},\) releases a sandbag at the instant the balloon is \(40.0 \mathrm{~m}\) above the ground. (See Figure \(2.54 .)\) After it is released, the sandbag encounters no appreciable air drag. (a) Compute the position and velocity of the sandbag at \(0.250 \mathrm{~s}\) and \(1.00 \mathrm{~s}\) after its release. (b) How many seconds after its release will the bag strike the ground? (c) How fast is it moving as it strikes the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch graphs of this bag's acceleration, velocity, and vertical position as functions of time.
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