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Q26PE

Expert-verifiedFound in: Page 83

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?**

(a) The sketch of the situation is given in **figure (1).**

(b) The known variables for problem are **initial velocity**, **final velocity** of blood and **duration** for acceleration of blood.

(c) The unknown variables in problem are acceleration and duration of acceleration with value \({\bf{0}}{\bf{.12}}\;{\bf{s}}\).

(d) The answer is reasonable and comparable with heartbeat.

**(a)**

**Given Data:**

The initial speed of blood is \(u = 0\)

The final speed of blood is \(v = 30\;{\rm{cm}}/{\rm{s}}\)

The distance for blood is \(D = 1.80\;{\rm{cm}}\)

**The time for acceleration and deceleration of commuter train is found by using first equation of motion.**

The sketch for the situation is given in below figure:

Figure (1)

The known variables for blood are initial velocity \(u = 0\),, final velocity \(v = 30\;{\rm{cm}}/{\rm{s}}\) and distance covered by blood \(D = 1.80\;{\rm{cm}}\).

The unknowns for the problem are acceleration of blood and duration for acceleration of blood.

The acceleration of blood is given as:

\(\begin{array}{c}a = \frac{{v - u}}{t}\\a = \frac{{30\;{\rm{cm}}/{\rm{s}} - 0}}{t}\\a = \left( {\frac{{30\;{\rm{cm}}/{\rm{s}}}}{t}} \right)\end{array}\)

Here, \(d\) is the distance covered by car.

The duration for acceleration is calculated as:

\(D = ut + \frac{1}{2}a{t^2}\)

Substitute all the values in the above equation.

\(\begin{array}{c}1.80\;{\rm{cm}} = \left( 0 \right)t + \frac{1}{2}\left( {\frac{{30\;{\rm{cm}}/{\rm{s}}}}{t}} \right){t^2}\\t = 0.12\;{\rm{s}}\end{array}\)

Therefore, the duration for acceleration of blood is \(0.12\;{\rm{s}}\).

The duration of one heartbeat for human is \(0.83\;{\rm{s}}\) which is greater than the duration of acceleration of blood \(0.12\;{\rm{s}}\) so the answer is reasonable.

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