# Solving an A Level kinematics problem using suvat formulas [closed]

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.

## closed as off-topic by user296602, A.P., jdods, kimchi lover, DidNov 19 '17 at 22:05

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• Please familiarize yourself with MathJax. – user499203 Nov 19 '17 at 19:44
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• When is this due? – Did Nov 19 '17 at 22:05

## 1 Answer

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 $x$.