Question

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

Answer 1

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

Answer 2

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