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, Did Nov 19 '17 at 22:05

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  • $\begingroup$ Please familiarize yourself with MathJax. $\endgroup$ – user499203 Nov 19 '17 at 19:44
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    $\begingroup$ Welcome to StackExchange. You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: what you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$ – A.P. Nov 19 '17 at 20:14
  • $\begingroup$ When is this due? $\endgroup$ – Did Nov 19 '17 at 22:05

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 $x$.


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