Breaks applied to a car produces retardation of 5m/s. If the bike take 5m to stop find the distance it covers during this time.
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a=-5m/ss
s=?
t=5s
a=(v-u)÷t
-5=o-u÷5
-u=-25m/s
s=ut+1÷2(att)
s=-25×5+(-5×5×5)÷2
s=-125-62.5
s=187.5m
s=?
t=5s
a=(v-u)÷t
-5=o-u÷5
-u=-25m/s
s=ut+1÷2(att)
s=-25×5+(-5×5×5)÷2
s=-125-62.5
s=187.5m
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