car traveling speed 15m/s is bracked and it slows down with uniform retardation. it cover distance 88m as it's velocity reduces to 7m/s.if the car continue to slow down with same rate .after what further distance will it be brought to rest
pathakak9708:
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v=7m/s. u= 15m/s s= 88m. a=?
v^2=u^2+2as
7^2=15^2+2×a×88
49=225+176a
49-225=176a
-176=176a
a=-176÷176=-1m/s^2
deacceleration=1m/s^2
now still more distance it will travel is
v=0 (as the car is going stop)
u=7m/s. a=1 s=?
v^2=u^2+2as
0=7^2+2×-1×s
0=49-2s
2s=49
s=49÷2
s=24.5m
v^2=u^2+2as
7^2=15^2+2×a×88
49=225+176a
49-225=176a
-176=176a
a=-176÷176=-1m/s^2
deacceleration=1m/s^2
now still more distance it will travel is
v=0 (as the car is going stop)
u=7m/s. a=1 s=?
v^2=u^2+2as
0=7^2+2×-1×s
0=49-2s
2s=49
s=49÷2
s=24.5m
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Answer:
car traveling speed 15m/s is bracked and it slows down with uniform retardation. it cover distance 88m as it's velocity reduces to 7m/s.if the car continue to slow down with same rate .after what further distance will it be brought to rest
Explanation:
car traveling speed 15m/s is bracked and it slows down with uniform retardation. it cover distance 88m as it's velocity reduces to 7m/s.if the car continue to slow down with same rate .after what further distance will it be brought to rest
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