Question #56c3d

1 Answer
Jun 6, 2017

Answer:

#t=(ks)/e#

Explanation:

This will be a variant on #y=kx#

The one thing that will be in both is the need for a constant of variation #k#

What is happening? as the number of exits increases it should be the case that the people exit faster. Thus #t# will reduce

Let the total time of completed exist be #t#
Let the count of exits be #e#
Let the count of people (for this consideration) be the constant #s#

Then as #E# increases we need #t# to reduce

Logically we would have #t->s/e#

However this would need to incorporate some way of allowing for adjustment. This is where #k# comes in. Write as:

#t=kxxs/e" "->" "t=(ks)/e#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note that the choice of the variant #e# is not one I would have picked. In mathematics #e# is a recognised constant that is reserved.

It is used a lot in log to base #e" "->log_e" "->" "ln#