The exit of a school is through a corridor of width b = 3.5 m. After the school, on an average, N = 42 students pass through the corridor in 1 min and population density within the length of the corridor occupied becomes n = 0.4 students per square meter. Find the average speed of the students within the corridor.

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Solution

Let $x$ be distance of student from exit who passes in $1$ minute.
$x \times 3.5 \times 0.4 = 42$ [ $x \times b \times$ density $=$ Total student passing]
$\Rightarrow x = \frac{42 \times 100}{4 \times 35} = 30$
$∴$ $A$ student moves $30$ m in $1$ min
$30$ m $=$ $60$ sec.
$\Rightarrow 0.5$ m/s $=$ Speed of students within corridor.

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Let x be distance of student from exit who passes in 1 minute.x×3.5×0.4=42 [x×b× density = Total student passing]⇒x=42×1004×35=30∴ A student moves 30 m in 1 min 30 m = 60 sec.⇒0.5 m/s = Speed of students within corridor.