The speed of a train is reduced from 60km/h at the same time as it travels a distance of 450m. If the retardation is uniform, find how much further (approx.) before coming to rest
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According to question
Vi=16.6ms-1
Vf=0 ms-1
( As it is not given hence we will consider it as zero )
S=450m
So according to the equation of motion
2aS=vf^2 -vi^2
So by putting in the values
a=0.28ms-2= 28m(approximately)
S=(vf^2-vi^2)/2a
Ab values place karo tou answer ayay ga
S2=480m aur uper walay distance ko S1 let karlo
∆S=S2-S1
∆S=480-450
∆S=30m
= 30m
Vi=16.6ms-1
Vf=0 ms-1
( As it is not given hence we will consider it as zero )
S=450m
So according to the equation of motion
2aS=vf^2 -vi^2
So by putting in the values
a=0.28ms-2= 28m(approximately)
S=(vf^2-vi^2)/2a
Ab values place karo tou answer ayay ga
S2=480m aur uper walay distance ko S1 let karlo
∆S=S2-S1
∆S=480-450
∆S=30m
= 30m
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