Question

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?

Answer 1

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


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

Answer 2

Step-by-step explanation:

Initial speed of the jet, u = 0

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

Average acceleration of the jet,
`a=1.3\ m/s^2`

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 :

`d=(v^2-u^2)/(2a)`

`d=(75^2-0)/(2* 1.3)`

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.