An aeroplane lands at 432 kilometer per hour and stops after covering a runaway of 4 km calculate acceleration and time in which it comes to rest
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acceleration is 23323 km/h^2
time is -432/23323
time is -432/23323
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Given:
Initial velocity=u=432 km/h =432×5/18=120m/s
Final velocity = V= 0 m/s
Time=?
DISTANCE covered= s=4km=4×1000m
a=?
From third equation of motion,
V^2 - u^2 = 2as
0^2 - (120*120)=2*ax4000
a= - 120x120/2*4000
= -1.8 m/s2
Therefore retardation is 1.8m/s2
From FIRST equation of motion :
V=u+ at
t= V-u/a
=0-120/-1.8
=120/1.8
=66.6 secs.
Initial velocity=u=432 km/h =432×5/18=120m/s
Final velocity = V= 0 m/s
Time=?
DISTANCE covered= s=4km=4×1000m
a=?
From third equation of motion,
V^2 - u^2 = 2as
0^2 - (120*120)=2*ax4000
a= - 120x120/2*4000
= -1.8 m/s2
Therefore retardation is 1.8m/s2
From FIRST equation of motion :
V=u+ at
t= V-u/a
=0-120/-1.8
=120/1.8
=66.6 secs.
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