Here, ` u = 72 km//h = 20 m//s , Given, the car is behind the truck.
`Retardation of truck ` = (200/5 = 4 m//s^2`
Retardation of car ` =- (20)/3 m//s^2`
Let car be at a distanc e(x) from truck and (t) be the time taken to cover this distance.
As car decelerates only after ` 0.5 s`, so using relation, ` v=u = at , we have
` V_(car) = 20 - ( (20) /3) (t- 0.5)` and `V_(truck) = 20 -4 t`
The car will just touch to the truck if
or ` 20 - 4 t = 20 - (20)/3 ( t- 0.5) or 4 t = (20)/3 (t- 0.5)`
or ` t = 5/3 (t- 0.5 ) or ` 3 t= 5 t- 2.4 or t =2.5 //2 = (5//4) s`
Distance travelled by truck in this time (t),
` S_(truck) =50 (5/4) - 1/2 (4) ( 5/4) ^(2) = 21. 875 m`
Distanc etravelled by car in time (t).
` S_(car) = (20 xx 0.5 ) + 20 (5/4 - 0 .5 ) - 1/2 ( 200 /3) (5/4 - 0.5 )^2 = 23.125 m`
:. ` S_(car) - S_(trick) = 23. 125 - 21. 875 = 1.250 m.