An aeroplane lands at 432 km/hr and stops after covering a runaway of 4 km. calculate the acceleration and the time in which it comes to rest.
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u=432km/hr
v=0
s=4km
t=?
a=?
_______________
v²=u²+2as
0²=(432)²+2a(4)
0=186624+8a
-8a=186624
a=-(186624/8)
a=-23328km/hr²
_______________
Now changing all km/h & km in m/s & m
u=432km/hr=(432×5/18)m/s= 120m/s
v=0 m/s
s=4km= (4×1000)m/s= 4000m/s
a= -23328km/hr²=(-23328×5/18)m/s²
= -6480m/s²
____________
v=u+at
0=120+(-6480)(t)
0=120-6480t
6480t = 120
t= 120/6480
t = 0.0185 seconds
@:-)
u=432km/hr
v=0
s=4km
t=?
a=?
_______________
v²=u²+2as
0²=(432)²+2a(4)
0=186624+8a
-8a=186624
a=-(186624/8)
a=-23328km/hr²
_______________
Now changing all km/h & km in m/s & m
u=432km/hr=(432×5/18)m/s= 120m/s
v=0 m/s
s=4km= (4×1000)m/s= 4000m/s
a= -23328km/hr²=(-23328×5/18)m/s²
= -6480m/s²
____________
v=u+at
0=120+(-6480)(t)
0=120-6480t
6480t = 120
t= 120/6480
t = 0.0185 seconds
@:-)
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