physics
posted by v .
A motor car moving at a speed of 72 km/h can not come to a stop in less than 3 s while for a truck this time interval is 5.0s . 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 the truck . Human response time is 0.5s

deceleration rate of the truck is
at = 20 m/s / 5s = 4 m/s²
deceleration rate of the car is
at = 20 m/s / 3s = 6⅔ m/s²
As the car can stop much quicker than the truck, then to avoid collision the car and truck velocities will be identical and non zero while the distance between them is near zero with a braking time (t) for the truck and (t  0.5) for the car
velocity of the truck
vt = 20  4t
velocity of the car
vc = 20  6⅔(t  0.5)
vc = 20  6⅔t + 3⅓
vc = 23⅓  6⅔t
to find the time when both of these are true, set them equal
23⅓  6⅔t = 20  4t
3⅓ = 2⅔t
t = 1.25 s
In 1.25 seconds the car will have traveled a distance before braking occurs and a distance after braking starts
dc = v₀t₀ + v₀t₁ + ½at₁²
dc = 20(0.5) + 20(0.75)  ½(6⅔)(0.75)²
dc = 23.125 m
in 1.25 seconds, the truck will have traveled
dt = 20(1.25)  ½(4)(1.25)²
dt = 21.875 m
so the minimum follow distance is
23.125  21.875 = 1.25 m <=== ANSWER
At 1.25 seconds, both vehicles will be traveling 15 m/s
20  4(1.25) = 15
20  6⅔(1.25  0.5) = 15
For times greater than 1.25 seconds, the car is traveling slower than the truck and the gap between the car and truck increases.
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