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Okay folks, I'm keen to get a response to this brain teaser!!!

Let's say you're looking at a conveyor belt... The width = 2m, and the length = 10m.

Okay... This conveyor belt is moving from right to left at 1m/sec , meanwhile, you're scanning across the conveyor belt from right to left at 2m/sec. (so you're scanning just faster than the conveyor belt is moving).

What is the total area of the conveyor belt that you have scanned.

(Takes 5 seconds to scan from right to left).

OKAY... Next.. let's give you all two more scenarios...

What is the area if:

A) Conveyor belt is moving right to left at 1m/sec, But you are scanning LEFT to RIGHT at 2m/sec?

B) Conveyor belt is moving right to left at 2m/sec, but you are scanning RIGHT to LEFT at 1m/sec (moving slower than the machine)?

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closed as off-topic by Sconibulus, Rubio, JMP, Beastly Gerbil, Ian MacDonald Apr 3 '17 at 20:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Sconibulus, Rubio, JMP, Beastly Gerbil, Ian MacDonald
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ $2*(10-5)$, $2*(10+5)$, $2*(10+20)$. $\endgroup$ – Ian MacDonald Apr 3 '17 at 20:13
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Do you know the answer or are you asking for help?

Assuming the upper half of the belt's total length is fully visible and scannable (i.e. no effect from rounded ends, 5m are visible at any one time) and the width has no effect in the scanning (i.e. the entire width is scanned instantaneously) the answers are as follows:

first question, scanner moves R2L for 2.5s over a surface that moves at 1m/s relative to the scanner. Therefore the scanner covers 2.5m. Area scanned = 2.5m * 2m = 5m2

A) scanner moves L2R for 2.5s over a surface that moves 2at 3m/s relative to the scanner. Therefore the scanner covers 7.5m. Area scanned =7.5m * 2m = 15m2

B) scanner moves R2L for 5s over a surface that moves at 1m/s relative to the scanner. Therefore the scanner covers 5m. Area scanned = 5m*2m = 10m2.

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  • $\begingroup$ Thanks - Had a couple of possible answers, and wanted to confirm. $\endgroup$ – GrantRWHumphries Apr 3 '17 at 20:31
  • $\begingroup$ Sorry - wondering if you could clarify something - where did you get 2.5s? If the length is 10m, it should take 5 seconds to scan the entire length at 2m/s . So... I would have thought something like.. (2m/s - 1m/s) * 2m * 5s = 10m2 (essentially the same as 1/2 the total area because the conveyor belt is going 1/2 the speed) $\endgroup$ – GrantRWHumphries Apr 3 '17 at 21:28

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