Car behind the truck
Regardation of truck = 20/5 = 4ms-2
Let the truck be at a distance x from the car when breaks are applied
Distance of truck from A at t > 0.5 s is x + 20t - 2t2
Distance of car from A is 10 + 20(t - 0.5) -(10/3)(t - 0.5)2
If the two meet
Second method: This method does not require the use of calculus. If the car is behind the truck
Vcar = 20 – (20/3)(t – 0.5) for t > 0.5 s as car declerate only after 0.5 s.
Vtruck = 20 – 4t
Find t from equating the two or from velocity vs time graph. This yields t = 5/4 s.
In this time truck would travel truck,
Struck= 20(5/4) – (1/2)(4)(5/4)2 = 21.875m
and car would travel, Scar = 20(0.5) + 20(5/4 – 0.5) – (1/2)(20/3) x (5/4 - 0.5)2 = 23.125m
Thus Scar – Struck = 1.25m.
If the car maintains this distance initially, its speed after 1.25s will he always less than that of truck and hence collision never occurs.