Write down $\displaystyle\int{\frac{1}{{{a}^{2}}-{{v}^{2}}}\,\mathrm{d}v}$, where $a$ is a positive

Question

Write down $\displaystyle\int{\frac{1}{{{a}^{2}}-{{v}^{2}}}\,\mathrm{d}v}$, where $a$ is a positive constant. [1] In the motion of an object through a certain medium, the medium furnishes a resisting force proportional to the square of the velocity of the moving object. Suppose that a body falls vertically through the medium, the model used to describe the velocity, $v$ms$^{-1}$ of the body at time t seconds after release from rest is given by the differential equation $\frac{{{a}^{2}}}{10}\frac{\mathrm{d}v}{\mathrm{d}t}={{a}^{2}}-{{v}^{2}}$, where $a$ is a positive constant. Show that $v=a\left( \frac{{{e}^{\frac{20t}{a}}}-1}{{{e}^{\frac{20t}{a}}}+1} \right)$. [8] The rate of change of the displacement, $x$ metres, of the body from the point of release is the velocity of the body. Given that $a=2$, find the value of $x$ when $t=1$, giving your answer correct to $3$ decimal places. [3]

Accepted Answer

(i) $\int\frac{1}{a^2-v^2}\,\mathrm{d}v=\frac{1}{2a}\ln\left|\frac{a+v}{a-v}\right|+C$. (ii)(a) $v=a\left(\frac{e^{20t/a}-1}{e^{20t/a}+1}\right)$. (ii)(b) When $a=2$ and $t=1$, $x=\frac{2}{5}\ln(\cosh5)\approx1.723\text{ m}$.

2019 RI P1 Q9

2019 RI P1 Q9

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