A motor car moving at a speed of ` 72 km //h` can not come to a stop in less than ` 3.0 s` while for a truch this time interval is ` 5.0s` On a highway the car is behind the truck both moving at `72 km //h` The truck geives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does bot bump onto (collide with) the truck. Human responde time id ` 0.5 s`.

Explanation

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.