Uniform acceleration (easy question)

Question

Two stunt drivers drive their cars along a straight horizontal road. The first car is travelling at $30$ m/s and is followed by the second car, $16$ m behind it, both cars are travelling with equal speeds. At an instant the driver of the first car applies the brakes decelerating at $3$ m/s/s. Two seconds later , the second car brakes and decelerates at $4$ m/s/s. The time it takes the cars to collide?. $$s1: 30t-1.5t^2+16$$ $$s2:30(t+2)-2(t+2)^2$$ $s1=s2 -b$ formula and $t=3.66$ rounded to 2 decimal places Just not sure of my answer can someone plz check my answer, Thank u.

Answer 1

Equation let $d_1, d_2$ be the distance the cars 1 and 2 pass before the collision. $d_1=30t-3t^2/2$, $d_2=30t-4(t-2)^2/2$ We know that $d2=d1+16$ So we get the following equation $$ 30t-4(t-2)^2/2=16+30t-3t^2/2 $$ So regrouping: $$ -2t^2-8+8t=16-3t^2/2 \to ~ t^2+48-16t=0 $$ The solution $t=4$ and $12$, but the feasible solution obviously is the first one. So $t=4s$ Remark : if u want to know the time since the second car starts the deacceleration then $t=t-2=2$

The simplest check. The fist car starts to slow down. For 2 sec it passes $60-3*2^2/2=54 m$ The second car passes $60m$.For the next 2 sec the first car passes $24*2-3 2^2/2=42m$. The second car passes $30*2-4*2^2/2=52m$. So the first car passes $96$ and the second car:$112=96+16$.