Speeds of two identical cars are $u$ and $4u$ at the specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is
(1) $1 : 1$
(2) $1 : 4$
(3) $1 : 8$
(4) $1 : 16$

A car, moving with a speed of $50 \mathrm{~km} / \mathrm{hr}$, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of $100 \mathrm{~km} / \mathrm{hr}$, the minimum stopping distance is
(1) 12 m
(2) 18 m
(3) 24 m
(4) 6 m

An automobile travelling with speed of $60 \mathrm {~km} / \mathrm { h }$, can brake to stop within a distance of 20 cm . If the car is going twice as fast, i.e $120 \mathrm {~km} / \mathrm { h }$, the stopping distance will be
(1) 20 m
(2) 40 m
(3) 60 m
(4) 80 m

For a train engine moving with speed of $20 \mathrm {~ms} ^ { - 1 }$, the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt { x } \mathrm {~ms} ^ { - 1 }$. The value of $x$ is $\_\_\_\_$ . (Assuming same retardation is produced by brakes)