A car weighting 600kg travels with 72km/h stops at a distance 50m de-acceleration uniformly , the force exerted by the braks will be ?
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mass of car=600 kg
initial velocity u = 72km/h
distance s= 50m
final velocity v=0 (car stop)
By equation of motion
Vsq. =Usq. +2as
0sq.=72sq.+2×a×50
a=5184/-100
a=-51.84m/s^2
after this
F=600×(-51.84)=-31104
thus,force exerted by break will be 31104N.
initial velocity u = 72km/h
distance s= 50m
final velocity v=0 (car stop)
By equation of motion
Vsq. =Usq. +2as
0sq.=72sq.+2×a×50
a=5184/-100
a=-51.84m/s^2
after this
F=600×(-51.84)=-31104
thus,force exerted by break will be 31104N.
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