Solving an A Level kinematics problem using suvat formulas

Question

A car comes to a stop from a speed of $30m/s$ in a distance of $804m$. The driver brakes so as to produce a deceleration of $\frac12m/s^2$ to begin with and then brakes harder to produce a deceleration of $\frac32m/s^2$. Find the speed of the car at the instant when the deceleration is increased and the total time the car takes to stop.

Answer 1

There are two phases to the motion, lets list the stats for each.

Phase 1: $u=30,v=x,t=t_1,s=s_1,a=-0.5$.

Phase 2: $u=x,v=0,t=t_2,s=s_2,a=-1.5$. And $s_1+s_2=804$.

Now use $v^2=u^2+2as$ for both phases \begin{eqnarray*} x^2=30^2-s_1 \\ 0=x^2-3s_2 \end{eqnarray*} Now multiply the first equation by $3$ and utilise $s_1+s_2=804$ to obtain a value for $x$.