After how long do these accelerating trucks meet ?

Let the time the two trucks meet after started from $A$ and $B$ be $t \text{ h}$ . Then the distances traveled by the two trucks are:

$\begin{cases} s_A = v_A t + \frac 12 a_A t^2 \\ s_B = v_B t + \frac 12 a_B t^2 \end{cases}$

And we note that when the two trucks meet, we have:

$\begin{aligned} s_A + s_B & = 100 \\ (v_A + v_B) t + \frac 12 (a_A+a_B) t^2 & = 100 \\ \implies 5t^2 + 50t - 100 & = 0 \\ t^2 + 10 t - 20 & = 0 \end{aligned}$

$\begin{aligned} \implies t & = \frac {-10+\sqrt{10^2-4(-20)}}2 & \small \blue{\text{Since }t > 0} \\ & = 3\sqrt 5 - 5 \approx \boxed{1.71} \text{ h} \end{aligned}$

If they meet at a distance x kms from A. The equations of motion under constant acceleration is given by s = ut + ½at² Hence applying it to the Truck at A and the Truck at B we get two equations x = 20t + 3t^2 -- (1) 100-x = 30t + 2t^2 (2)

Adding (1) and (2) we get 100 = 50t + 5*t^2 Solving the quadratic in t

5*t^2 + 50t - 100 = 0 , t = 1.708 and t = -11.708 , as t cannot be < 0 , t = 1.708 hours