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A moving dot P departs from O at an initial speed of 6 m/s and accelerates at 2m/s^2 in the east direction. 2 seconds after P's departure, dot Q departs from O to chase down P at a constant speed of k m/s in the east direction. What is the minimum speed, k, required for Q to eventually catch up to P?

The distance for P is given by 6t+t^2. For Q to catch up k(t-2)= 6t+t^2. To find the point do we have to compare the rate of change?If yes, why?

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3 Answers 3

You don't have to compare the rate of change. Once you have $k(t-2)=6t+t^2$ then $Q$ have catch up $P$. You want to minimize $k$ so the only things left to do is to minimize $\frac{6t+t^2}{t-2}$ for $t>2$.

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@RossMillikan I'm not sure what you mean. $k(t-2)=6t+t^2$ hence $k=\frac{6t+t^2}{t-2}$ no? –  wece Mar 18 '13 at 14:31

Let $k$ be the smallest speed that lets $Q$ catch up, and let $t$ be the time that she catches up. Then $$k(t-2)=6t+t^2.$$ Also, at the instant $t$ that $Q$ catches up, the velocities of $P$ and $Q$ must match. For if at time $t$ $Q$ is slower than $P$, she would not have caught up at time $t$. And if at time $t$ $Q$ is faster than $P$, then a cheaper speed than $k$ would have been sufficient. Thus $$k=6+2t.$$

Now use the two equations to eliminate $t$ and solve for $k$.

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That a nice way to see the problem :) and it should gives the same result as the minimisation. –  wece Mar 18 '13 at 14:36

The quadratic for the meeting time, $$t^2+(6-k)t+2k=0$$will have two positive roots, since k must be greater than 6 for Q to have any chance of catching P. The first positive root refer to the point where Q first catches and passes P, and the second to the point when the accelerating P passes back in front of Q. The minimum k comes when these two events are simultaneous, meaning the discriminant of this quadratic is zero; that is $$(6-k)^2-8k=0$$Solving this quadratic gives $k=2$, clearly impossible since $k>6$, and $k=18$

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