A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a highway, the car is behind the truck both moving at 72 km/h. The truck gives 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 response time is 0.5 s.

Explanation

Given, speed of car = speed of truck = 72 km/h
`= 72 xx (5)/(18) m//s = 20 m//s`
Now, `v = u + a _(t)t `
`0= 20 + a _(t) xx 5`
`a _(t) = - 4 m //s ^(2)`
For car, `v = u + a _(c)t `
`0= 20 + a _(c) xx 3`
`a _(c) = - (20)/(3) m//s ^(2)`
Let car be at a distance x from truck, when truck gives 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 l,
`v _(t) = 20 - 4t`
To avoid the collision,
`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`
`therefore t = (2.5)/(2) = 5/4 s`
Distance travelled by the truck,
`x _(t) = u_(t) t + 1/2 a _(t) t ^(2) = 20 xx 5/4 +1/2 xx (-4) xx ((5)/(4)) ^(2) =21.875 m`
Distance travelled by car = {Distance travelled by car in `0.5 s` (with constnat speed )} + {Distance travelled by car in `(t - 0.5) s` (with retardation)}
`x _(c) = (20 xx 0.5) + 20 ((5)/(4) -0.5) - (1)/(2) ((20)/(3)) ((5)/(4) - 0.4) ^(2) = 23.125 m`
`thereforex =x _(c) - x _(t)=23.125 -21.875 =1.250m`
Therefore, to avoid the collision, the car must maintain a distance more than `1.250m` from the truck.