Posted by v on Tuesday, September 18, 2012 at 9:36pm.
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.