91Ó°ÊÓ

(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.