Question

The exit of a school is through a corridor of width $b=3.5m$ . 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.

Answer 1

Hint : The average speed can be calculated by finding the average length crossed by the student. The product of the population density and the area of the corridor, the number of students in the corridor can be found

Complete Step By Step Answer:
Let us note down the given data;
Width of the corridor $b=3.5m$
Length of the corridor $l=?$
Area of the corridor $A=l\times b$
$\therefore A=l\times 3.5m$
Number of students passing through the corridor in $1\;min$ is $N=40$
As we need to find the speed, we should convert the rate for $1\;min$ to seconds
$\therefore N=\dfrac{42}{1\min }=\dfrac{42}{60\sec }$
$\therefore N=\dfrac{7}{10\sec }$
Hence, the number of students passing through the corridor in $1\;sec$ is $N=0.7$
Population density of students occupying the corridor $\therefore n=0.4$
Now, we know that population density of the students can be defined as the number of students present in the corridor per unit area.
So if we multiply the population density with the total area of the corridor, we can get the total number of students present in the corridor.
$\therefore$ The total number of students present in the cupboard $S=n\times A$
Substituting the given values,
$\therefore S=0.4\times l\times 3.5$
$\therefore S=1.4\times l$
This is the average number of students present and passing through the corridor.
This students move out of the corridor at a rate of $N=0.7$ per second
Hence, we can equate the both the equations,
$\therefore 1.4\times l=0.7$
$\therefore l=0.5$
Hence, a distance of $0.5\;m$ is covered per second
Thus, the average velocity of the students is $0.5m{{s}^{-1}}$ .

Note :
Here, we need to note that the width of the corridor can be explained as the vastness of the corridor or how wide the corridor is. To find the average speed of the students, we need to calculate the length of the corridor which is the distance covered by the students.