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. How long does the acceleration take?

Answer 1

Answer:

Acceleration will take for 0.12 seconds.

Explanation:

Given:

Initial velocity of the blood, $v_i$ = 0 m/s

Final velocity of the blood, $v_f$ = 30 cm/s = 0.3 m/s

Distance to be considered, $x$ = 1.80 cm = 0.018 m

We have to find the time taken to covered the said distance.

Let the time taken be "t" seconds.

Equation of motion:

⇒ $v_f=v_i+at$ ...equation (i)

⇒ $x=v_it+\frac{at^2}{2}$  ...equation (ii)

Now,

Using equation (i) we can say that:

⇒ $v_f=v_i+at$

⇒ $0.3=0+at$

⇒ $0.3=at$            ...equation (iii)

Plugging x = 0.018 m in equation (ii).

⇒ $x=v_it+\frac{at^2}{2}$

⇒ $0.018=0+\frac{(at)(t)}{2}$   ...breaking at^2 as (at)(t)

⇒ $0.018=\frac{(at)(t)}{2}$

⇒ $0.018=\frac{0.3\times (t)}{2}$      ...plugging at=0.3 from equation (iii)

⇒ $0.018\times 2=\frac{0.3\times (t)}{2}\times 2$

⇒ $0.036=0.3\times (t)$

⇒ $\frac{0.036}{0.3} =\frac{0.3\times t}{0.3}$

⇒ $0.12\ s = t$

The acceleration will take for 0.12 seconds.