Question

An Automobile travelling with seeed of ( 60 mathrm{km} / mathrm{m} ) an brake
twice as fast in ( mathrm{RO} mathrm{km} / mathrm{m} ) the Stopping distenk will be
[
80 mathrm{m}
]
Given the t option
intialspeed ( u=60 mathrm{km} / mathrm{h} )
[
begin{array}{l}
v=0
s=20 m=20 eta 1000 m=0.02 mathrm{km}
end{array}
]
Acudding to kenemotic equebons of mobons
Given thef
[
begin{array}{l}
v^{2}-u^{2}=2 a s
0^{2}-60^{2}=2(a)(0.02)
-3600=0.04 a
a=-frac{3600}{0.04}
=frac{-3600}{100}=frac{3600 times 100}{4}
end{array}
]
speed becomes fortwice
[
-90000 mathrm{km} / mathrm{n}^{2}
]
there
[
c=120 mathrm{km} / mathrm{hr}
]
speed
[
begin{array}{l}
v=0
a=-90000 mathrm{km} 14^{2}
end{array}
]
Welove to find the ( theta ) stopping distance ( (s)=? )

# An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e. 120 km/h, the stopping distance will be [AIEEE-2004) (A) 20 m (B) 40 m (D) 80 m om

Solution