Question

The left ventricle of the heart accelerates blood from rest to a velocity of +26 cm/s. (a) If the displacement of the blood during the acceleration is +2. 0 cm, determine its acceleration

( in cm/s2)\left( \text { in } \mathrm { cm } / \mathrm { s } ^ { 2 } \right)

(b) How much time does blood take to reach its final velocity?

Solutions

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Using the following equation of motion:

v2=v02+2axv^2=v_0^2+2ax

we can get the acceleration as:

a=v2v022x\begin{align}a=\dfrac{v^2-v_0^2}{2x}\end{align}

(a)\textbf{(a)} Assume we have left ventricle of the heart, which accelerates blood from rest v0=0v_0=0 to a speed of v=26 cms1v=26 \mathrm{~cm\cdot s^{-1}}, if the displacement of the blood is x=2x=2 cm, the acceleration is therefore:

a=(26 cms1)202(2 cm)=169 cms1\begin{align*}a&=\dfrac{(26 \mathrm{~cm\cdot s^{-1}})^2-0}{2(2 \mathrm{~cm})}\\ &=169 \mathrm{~cm\cdot s^{-1}} \end{align*}

a=169 cms1\boxed{a=169 \mathrm{~cm\cdot s^{-1}}}

(b)\textbf{(b)} Using the following equation of motion:

v=v0+atv=v_0+at

we can get the time as:

t=vv0a=(26 cms1)0(169 cms1)=0.154 s\begin{align*}t&=\dfrac{v-v_0}{a}\\ &=\dfrac{(26 \mathrm{~cm\cdot s^{-1}})-0}{(169 \mathrm{~cm\cdot s^{-1}})}\\ &=0.154 \mathrm{~s} \end{align*}

t=0.154 s\boxed{t=0.154 \mathrm{~s}}

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