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 not bump onto (collide with) the truck. Human responde time is` 0.5 s`.

Explanation

In this problem equations related to one dimensional motion will be applied for acceleration positive sign will be used and for retardation negative sign will be used.
Given, speed of car as well as truck `= 72 km//h`
`= 72 xx 5/18 m//s = 20 m//s`
Retarded motion for truck `v = u + a_(t)t`
`0 = 20 + a_(t) xx 5`
or `a_(t) = -4 m//s^(2)`
Retarded motion for the car `v = u + a_(C) t`
`0 = 20 + a_(C) xx 3`
`a_(C) = - (20)/(30) m//s^(2)`
Let car be at a distance x from truck, when truck give the signal and t be the time taken to cover this distance.
As human response time is `0.5s`, therefore time of retarded motion of car is `(t - 0.5)s`,
velocity of car after time t,
`v_(C) = u - at`
`= 20 - ((20)/(3)) (t - 0.5)`
Velocity of truck after time `t`,
`v_(t) = 20 - 4t`
To avoid the car bump onto the truck,
`v_(C) = v_(t)`
`20 - (20)/(3) (t - 0.5) = 20 - 4t`
`4t = (20)/(3) (t-0.5)`
`t = 5/3 (t - 0.5)`
`3t = 5t - 2.5`
`rArr t = (2.5)/(2) = 5/4 s`
Distance travelled by the truck in time `t `,
`s_(t) = u_(t)t + 1/(2) a_(t) t^(2)`
`= 20 xx 5/4 + 1/2 xx (-4) xx ((5)/(4))^(2)`
`= 21875 m`
Distance travelled by car in time `t =` Distance travelled by car in 0.5s (without retardation)
+ Distance travelled by car in `(t - 0.5)s` (with retardation)
`s_(C) = (20 xx 0.5) + 20(5/4 - 0.5) - 1/2((20)/(3)) (5/4 - 0.5)^(2)`
`= 23.125 m`
`:. s_(C) - s_(t) = 23.125 - 21.875 = 1.250 m`
Therefore, to avoid the bump onto the truck, the car must maintian a distance from the truck more than `1.250 m`.