Question

The landing speed of a particular type of aircraft is $216$ $kmph$ and from the moment its wheels touch the ground, its motion is resisted by a force $F$ given by $F=(500+3$$v^2$) $N/tonne$ of aircraft. The minimum length of runway required by this aircraft to ensure a safe-landing is (in $m$)

• A. $\frac{100}{3}lo{g}_e\left\{\frac{500}{11300}\right\}$
• B. $\frac{1000}{6}lo{g}_e\left\{\frac{500}{11300}\right\}$
• C. $\frac{100}{6}lo{g}_e\left\{\frac{11300}{250}\right\}$
• D. $\frac{1000}{6}lo{g}_e\left\{\frac{11300}{500}\right\}$

Answer 1

Answer: (D)

$\frac{1000}{6}lo{g}_e\left\{\frac{11300}{500}\right\}$

Since the force resists the motion, it is negative
$\Rightarrow Fperunitmass=\frac{(500+3v^2)}{1000}N/kg$
$\Rightarrow F=-m\frac{(500+3v^2)}{1000}N=m.a=m\frac{dv}{dt}$
Where m = mass of aircraft (Applying Newton's Second Law)
$\Rightarrow 500+3v^2=-1000v\frac{dv}{dx}$
$\Rightarrow \int dx=-1000\int \frac{v}{500+3v^2}dv$
$\Rightarrow x=-\frac{1000}{6}\int \frac{6v}{500+3v^2}+dv$
$=-\frac{1000}{6}lo{g}_e(500+3v^2)+C$
Where C is constant of integration.
At the moment the wheels touch the ground,
$x=0$ and $v=216km/hr=$$\frac{216}{3.6}$ $=60m/s$
$\Rightarrow 0=-\frac{1000}{6}lo{g}_e(500+3(60{)}^2)+C$
$\Rightarrow C=+\frac{1000}{6}lo{g}_e\left(11300\right)$
Distance travelled along the runway, $x=\frac{1000}{6}lo{g}_e\left(11300\right)-\frac{1000}{6}lo{g}_e(500+3v^2)$
$=\frac{1000}{6}lo{g}_e\left\{\frac{11300}{500+3v^2}\right\}$
When aircraft comes to rest, $v=0$
$\Rightarrow x=\frac{1000}{6}lo{g}_e\left\{\frac{11300}{500}\right\}$