1. It is presumed that 8m/s^2 and 4m/s^2 are in fact retardation of car after brakes are applied.

i) Consider the first case of retardation being 8m/s^2. If the car is traveling at 36 kph which means 10 m/s and the Reaction time is 0.25 sec, distance traveled during this time will be s = time x speed = 0.25(10) = 2.5 m

After brakes are applied its retardation is 8m/s^2. Distance traveled by the car before stopping would be given by v^2= u^2 - 2aS, v=0, u=10m/s, a= 8m/s^2

Hence 0= 100-2(8)S or, S= 6.25 m

Similarly, if u= 72kph, or 20m/s, the distance traveled before applying the brakes would be 0.25(20)= 5.00 m and the stopping distance would be = 25m.

and if u= 90kph, or 25 m/s, the distance traveled before applying the brakes would be 0.25(25)=6.25m and the stopping distance would be = 39.0625m

ii) Next consider the retardation being 4m/s^2. The distance traveled before applying the brakes would be same as above in each case and the stopping distance when speed is 36kph will be = 12.5 m.

“ “ 72kph will be= 50 m

“ “ 90kph will be= 78.125 m

The above data can be put in tabular form as follows:

Speed of car	Distance traveled during reaction time 0.25 s	Retardation	Stopping Distance	Total distance
32kph=10m/s	02.50 m	8m/s^2	06.25m	08.75m
72kph=20m/s	05.00 m	8m/s^2	25.00m	30.00m
90kph=25m/s	06.25 m	8m/s^2	39.0625m	45.3125m
32kph=10m/s	02.50 m	4m/s^2	12.50m	15.00m
72kph=20m/s	05.00 m	4m/s^2	50.00m	55.00m
90kph=25m/s	06.25 m	4m/s^2	78.125m	84.375m

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