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2001 q10b

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horse9996:
For the second part, I don't understand why finding the maximum value of the time difference between the bus and Claire gives the answer, but I get the process of doing this

RuiAce:
Already addressed in the compilation. Although please clarify more of the problem if you need more detail than that, because understanding that question is the hardest bit of it.

horse9996:

--- Quote from: RuiAce on October 07, 2018, 05:30:47 pm ---Already addressed in the compilation. Although please clarify more of the problem if you need more detail than that, because understanding that question is the hardest bit of it.

--- End quote ---

After you get the two time equations, I don't understand why finding the maximum of the difference of them gets the answer

RuiAce:

--- Quote from: horse9996 on October 07, 2018, 05:41:04 pm ---After you get the two time equations, I don't understand why finding the maximum of the difference of them gets the answer

--- End quote ---
It is like stated. The difference of the times reflects the time elapsed since 8AM for when Claire gets out of the house.

If we want Claire to leave her house later, then we really want the time elapsed to be larger. Hence, since the difference of the times helps us measure exactly that, we would like to maximise it.

The reason why the time difference reflects the time elapsed since 8AM for Claire to leave the house, is because by itself \( \frac{5}{18} (8+\tan \theta) \) is just the time (elspsed since 8AM where) Claire must've arrived at the bus depot. The extra \( \frac{25}{24}\sec \theta\) reflects that she actually takes time herself to get to the bus depot itself, and doesn't just magically teleport there or something.

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