Question

an aeroplane lands at the speed of 180 km per hour and stops after covering a Runway of 1 km calculate the retardation and the time in which the plane come to rest . with solution

Answer 1

first convert all unit in SI unit
180km/h = 180×5/18=50 m/s
1km = 1000m

solution:-. v^2=u^2 + 2as
0 = 50^2 + 2a×1000
2500/2000=a
1.25m/s =a

now time:-
v=u+at
0= 50 + 1.25t
50/1.25 =t
t=40s

Answer 2

The required formulae are i) v² = u² - 2aS and ii) v = u - at

where, v = final velocity of the aeroplane, u = initial velocity of the aeroplane, a = retardation, S = distance, t = time.

Given, u = 180 km/hr ; v = 0 km/hr ; S = 1 km

Putting the values in formula (i);

0² = 180² - 2.a.1

or, 2.a = 32400

or, a = 16200

Thus the retardation of the aeroplane is 16200 km/hr².

Putting the values in formula (ii);

0 = 180 - 16200.t

or, t = 180/16200

or, t = 1/90

Thus, the time required for the aeroplane to stop is 1/90 hr.

1/90 hr = (1 x 3600)/90 sec = 40 sec.