Let t1 be the time deceleration is initially applied, then v=30-t1/2 is the speed as reduced by the deceleration. The average speed is (30+v)/2 so the distance travelled in time t1 is (30+v)t1/2. The car now decelerates at 3/2 m/s^2. So 0=v-3t2/2, or 2v=3t2, where t2 is the time it takes to come to a stop. The distance travelled in this time is v/2 (the average speed) multiplied by t2. We know the distance in total is 804m so (30+v)t1/2+vt2/2=804; (30+v)t1+vt2=1608; 30t1+v(t1+t2)=1608. t1+t2=t the time it takes to come to a stop after braking at a speed of 30 m/s.
Now we can do some calculations. 3t2/2=v=30-t1/2, so 3t2/2=30-t1/2; 3t2=60-t1; t2=20-t1/3, giving us a relationship between the deceleration times, t1 and t2. And v=30-t1/2, so we have relationship between v and t1. That means we can solve by substitution the distance equation: 30t1+v(t1+20-t1/3)=1608; 30t1+(30-t1/2)(20+2t1/3)=1608. If we multiply through by 2 it will get rid of a fraction: 60t1+(60-t1)(20+2t1/3)=3216; now multiply by 3: 180t1+(60-t1)(60+2t1)=9648. This gives us a quadratic in t1: 180t1+3600+60t1-2t1^2=9648; 2t1^2-240t1+6048=0; t1^2-120t1+3024=0=(t1-84)(t1-36), so t1=36 or 84. This makes t2=20-36/3=8 seconds, or 20-84/3=-6, which can be ignored. So t=36+8=44 seconds and v=30-18=12 m/s (decelerating over a distance of (30+v)t1/2=756m).