An aeroplane lands at 432 km/h and stops after covering a runway of 4km.
Calculate (i) acceleration (ii) time in which it comes to rest.
Answers
Answered by
41
u=432km/h,v=0,s=4km
a=?
v^2-u^2=2as
(0)^2-(432)^2=2×4×a
-186624=8×a
a=-186624/8
a=-23328km/h^2
t=?
v=u+at
0=432+(-23328)t
-432=-(23328)t
t=-23328/-432
t=54sec
a=?
v^2-u^2=2as
(0)^2-(432)^2=2×4×a
-186624=8×a
a=-186624/8
a=-23328km/h^2
t=?
v=u+at
0=432+(-23328)t
-432=-(23328)t
t=-23328/-432
t=54sec
amaandola:
Good answer but one mistake of sign.
ok
which sign
When other units are in km/ h how you get time in seconds.
i already know just mistaken before ok
Answered by
55
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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