Q. An aircraft approaches the threshold of a runway strip at a speed of 200 km/h. The pilot decelerates the aircraft at a rate of 1.697 m/s2 and takes 18 s to exit the runway strip. If the deceleration after exiting the runway is 1 m/s2, then the distance (in m, up to one decimal place) of the gate position from the location of exit on the runway is _
Ans: 310 – 314
Sol:
From the equations of motion:
v = u + at
v2 = u2 + 2as
s = ut + 0.5 × at2
Where, v = final velocity attained by body, u = initial velocity of the body, s = displacement of the body and t = time taken
Calculation: v = u + at = 55.55 + (-1.697) × 18 = 25.004 m/s ∴ 0 = 25.004 + (-1) × t t = 25.004 sec