Distance, Speed, and Time A boardwalk is parallel to and 210 ft inland from a straight shoreline. A sandy beach lies between the boardwalk and the shoreline. A man is standing on the boardwalk, exactly 750 ft across the sand from his beach umbrella, which is right at the shoreline. The man walks 4 ft/s on the boardwalk and 2 ft/s on the sand. How far should he walk on the boardwalk before veering off onto the sand if he wishes to reach his umbrella in exactly 4 min 45 s?

The problem's image

The solution given in the site:

The solution

My solution:

I first found out the distance from the man to the straight line which ends at the broadwalk like given in the solution and it turned out to be $720 ft.$ After that, I assumed the distance to be walked by the man in the broadwalk to be $x$, so the distance from the man to the end of the broadwalk is now $(720 - x)$. Total time we have: $285 seconds$.

And here's the relation I got:

$285 = \frac{x}{4} + \frac{\sqrt{(210)^2 + (720 - x)^2}}{2}$

Solving this, I get $x = 860 or 300$. Since the broadwalk itself if only $720 ft$, the answer must be $300 ft.$

Though, I have no idea why it's even allowed to go the way it's given in the solution or whether the answer is correct. Where am I committing the error and what is it?


$$\begin{align*} \frac x4 + \frac{\sqrt{210^2+(720-x)^2}}2 &= 285\\ 2\sqrt{210^2+(720-x)^2} &= 4\cdot285-x\\ 2^2\left[210^2+(720-x)^2\right] &= (1140-x)^2\\ 4\left(210^2+ 720^2-1440x+x^2\right) &= 1140^2-2280x + x^2\\ 3x^2 - 3480x+ 950400 &= 0\\ x^2 - 1160x + 316800 &= 0\\ (x - 580)^2 - 19600 &= 0\\ x &= 580\pm140\\ &= 720\text{ or }440 \end{align*}$$

And check that for both cases, $4\cdot285-x\ge 0$, so both answers satisfy the equation.

  • $\begingroup$ I forgot to divide by two in the last step-- 2nd term on the RHS after (plus or minus). Stupid me! $\endgroup$ – MathEnthusiast Apr 22 '16 at 13:42

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