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 - 2t^{2}

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

V_{car} = 20 – (20/3)(t – 0.5) for t > 0.5 s as car declerate only after 0.5 s.

V_{truck} = 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,

S_{truck}= 20(5/4) – (1/2)(4)(5/4)^{2} = 21.875m

and car would travel, S_{car} = 20(0.5) + 20(5/4 – 0.5) – (1/2)(20/3) x (5/4 - 0.5)^{2} = 23.125m

Thus S_{car} – S_{truck} = 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.