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A jet plane has a takeoff speed of 75 m/s and can move along the runway at an average acceleration 1.3 m/s squared. If the length of the runway is 2.5 km, will the plane be able to use this runway safely?

Respuesta :

Donek
[tex]Given:\\a=1.3 \frac{m}{s^2} \\s=2.5km=2500m\\v_0= 75\frac{m}{s} \\\\Solution\\\\a= \frac{\Delta v}{t} \Rightarrow t= \frac{\Delta v}{a} \\\\v_k=0\Rightarrow \Delta v=v_0\Rightarrow t= \frac{v_0}{a} \\\\s= v_0t-\frac{at^2}{2} = \frac{v_0^2}{a} - \frac{v_0^2}{2a} = \frac{v_0^2}{2a} \\\\s= \frac{(75\frac{m}{s})^2}{2\cdot1.3 \frac{m}{s^2}} \approx 2163,5m\\\\\\proof\;that \;s= v_0t-\frac{at^2}{2} \\\\\\-a= \frac{dv}{dt} \Rightarrow dv=adt\\\\v= \int {-a} \, dt =-a\int \, dt =at+c\\\\v(t=0)=v_0\Rightarrow c=v_0[/tex]

[tex]v=-at+v_0\\\\v= \frac{ds}{dt} \Rightarrow ds=vdt\\\\v= \int {(-at+v_0)} \, dt =-a\int \,t dt +v_0\int \, dt=- \frac{at^2}{2} +v_0t+c_1\\\\s(t=0)=0\Rightarrow c_1=0\\\\s=v_0t - \frac{at^2}{2}[/tex]
aachen

Explanation:

Initial speed of the jet, u = 0

Final speed of the jet, v = 75 m/s

Average acceleration of the jet, [tex]a=1.3\ m/s^2[/tex]

Length of the runway, l = 2.5 km = 2500 m

Let d is the distance covered by the jet. Using the third equation of motion as :

[tex]d=\dfrac{v^2-u^2}{2a}[/tex]

[tex]d=\dfrac{75^2-0}{2\times 1.3}[/tex]

d = 2163.46 m

or

d = 2.16 km

As the value of d is less than the length of the runway, so the aircraft will be safe.

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