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
1003loge{50011300}
B
10006loge{50011300}
C
1006loge{11300250}
D
10006loge{11300500}

Solution

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

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