Question

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?

Short answer

(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.

Step-by-step solution

Sketch for the situation

(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)

Determination of known for problem (b)

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

Determination of duration of acceleration(c)

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 \(0.12\;{\rm{s}}\).

Comparison of heartbeat duration with duration of acceleration(d)

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.