2023TrialShort answer

Q14 (c)

4 marks

Question

The only force acting on a particle moving in a straight line is a resistance $m\lambda(c+v)$ acting in the same line. The mass of the particle is $m$ , its velocity is $v$ , and $\lambda$ and $c$ are positive constants. The particle starts to move with velocity $u>0$ and comes to rest after $T$ seconds. After half the time has elapsed, the particle's velocity is a quarter of its initial velocity.

Show that

c=\frac{u}{8}.

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Solution

\begin{aligned} &\text{when }t=0,\quad v=u\\ &\text{when }t=T,\quad v=0\\ &\text{when }t=\frac{T}{2},\quad v=\frac{u}{4} \end{aligned}

ma=-m\lambda(c+v)

\frac{dv}{dt}=-\lambda(c+v)

\frac{dv}{c+v}=-\lambda\,dt

\ln(c+v)-\ln(c+u)=-\lambda t

\ln\left(\frac{c+v}{c+u}\right)=-\lambda t

When $t=\frac{T}{2}$ , $v=\frac{u}{4}$ :

-\lambda\frac{T}{2}=\ln\left(\frac{c+\frac{u}{4}}{c+u}\right)

When $t=T$ , $v=0$ :

-\lambda T=\ln\left(\frac{c+0}{c+u}\right)

\ln\left(\frac{c+\frac{u}{4}}{c+u}\right)^2 =\ln\left(\frac{c}{c+u}\right)

\left(\frac{c+\frac{u}{4}}{c+u}\right)^2=\frac{c}{c+u}

\left(c+\frac{u}{4}\right)^2=c(c+u)

c^2+\frac{cu}{2}+\frac{u^2}{16}=c^2+cu

\frac{u^2}{16}=\frac{cu}{2}\quad\therefore\quad c=\frac{u}{8}

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Mechanics

Resisted motionDerive and solve motion models with resistance on horizontal and vertical paths, including terminal velocity.

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