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Chapter 2: Problem 77

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.

Short Answer

Expert verified
The acceleration of the blood is approximately 510.8 cm/s虏.

Step by step solution

01

Identify the Known Values

We are given the initial velocity, \( u = 0 \, \mathrm{cm/s} \), the final velocity, \( v = 425 \, \mathrm{cm/s} \), and the time for one heartbeat is \( \frac{60}{72} \) seconds, since there are 72 beats per minute. Calculate the time for one beat: \( t = \frac{60}{72} = 0.833 \) seconds.
02

Use the Formula for Acceleration

The formula for acceleration \( a \) is given by \( a = \frac{v - u}{t} \), where \( v \) is the final velocity, \( u \) is the initial velocity, and \( t \) is the time over which the change occurs. Substitute the known values into this formula.
03

Perform the Calculation

Substitute the values: \( a = \frac{425 - 0}{0.833} \). Compute the value: \( a = \frac{425}{0.833} \approx 510.8 \, \mathrm{cm/s^2} \).
04

Interpret the Result

The calculated acceleration is \( 510.8 \, \mathrm{cm/s^2} \). This represents how quickly the velocity of the blood increases as the heart beats.

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Key Concepts

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

Velocity
Velocity is a fundamental concept in physics that refers to the speed of an object in a specific direction. It is a vector quantity, which means it has both magnitude and direction. This is different from speed, which only has magnitude. In our exercise, the velocity of the blood is the rate at which it moves through the blood vessels.
  • Initial Velocity ( \( u \)): The starting speed of the object. For our problem, it is given as 0 cm/s because the blood starts from rest.
  • Final Velocity ( \( v \)): The speed of the object at the end of a period. Here, it reaches 425 cm/s.
Understanding velocity helps us describe how fast the blood is traveling within the body, which is vital for calculating acceleration.
Time Calculation
Time calculation is crucial in determining the duration over which changes occur. To understand this concept in the exercise, we need to know how long one heartbeat lasts.
To find this:
  • Consider the heart beats 72 times per minute.
  • Divide 60 seconds by 72 beats to get the time per beat: \( t = \frac{60}{72} = 0.833 \) seconds.
This time value is vital because it is used in the acceleration formula to determine the change in velocity over time, providing a clearer picture of how rapid these physiological processes are.
Unit Conversion
Unit conversion is a basic skill that ensures consistency and accuracy in calculations involving different measurement units. In physics, converting units is often necessary to apply equations correctly.
While our exercise is already in a consistent unit of cm/s, consider these possible conversions that might be necessary in different problems:
  • Length: Meters (m) to Centimeters (cm) by multiplying by 100.
  • Time: Minutes to Seconds by multiplying by 60.
  • Acceleration: \( ext{m/s}^2 \) to \( ext{cm/s}^2 \) by multiplying by 100.
In our exercise, understanding the unit conversion ensures we apply the acceleration formula correctly. Missteps in unit conversion can lead to incorrect results, so always make sure to use consistent units throughout your calculations.

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Most popular questions from this chapter

At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constant acceleration of 2.50 \(\mathrm{m} / \mathrm{s}^{2} .\) At the same instant, a truck, traveling with a constant speed of \(15.0 \mathrm{m} / \mathrm{s},\) overtakes and passes the automobile. (a) How far beyond its starting point does the automobile overtake the truck? (b) How fast is the automobile traveling when it overtakes the truck?

The "reaction time" of the average automobile driver is about 0.7 s. (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12.0 \(\mathrm{ft} / \mathrm{s}^{2}\) , compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of 15.0 \(\mathrm{mi} / \mathrm{h}(\mathrm{in}\) a school zone) and (b) from an initial velocity of 55.0 \(\mathrm{mi} / \mathrm{h}\)

A jetliner has a cruising air speed of 600 \(\mathrm{mph}\) relative to the air. How long does it take this plane to fly round-trip from San Francisco to Chicago, an east-west flight of \(2,000\) mi each way, (a) if there is no wind blowing and (b) if the wind is blowing at 150 \(\mathrm{mph}\) from the west to the east?

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of \(H,\) how high (in terms of \(H )\) will the faster stone go? Assume free fall.

The freeway blues! When you normally drive the freeway between Sacramento and San Francisco at an average speed of \(105 \mathrm{km} / \mathrm{hr}(65 \mathrm{mph}),\) the trip takes 1.0 \(\mathrm{hr}\) and 20 min. On a Friday afternoon, however, heavy traffic slows you down to an average of 70 \(\mathrm{km} / \mathrm{hr}(43 \mathrm{mph})\) for the same distance. How much longer does the trip take on Friday than on the other days?
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