Rate (as used in the question) implies the change of distance over time.
The expression for rates is: [tex]\mathbf{s = \frac dn}[/tex]
The parameter is given as:
[tex]\mathbf{d = ns}[/tex]
Where:
d represents distance and n represents time
The complete question requires that we get an expression for rates
This means that
We make s the subject in [tex]\mathbf{d = ns}[/tex]
Divide both sides by n
[tex]\mathbf{\frac dn = \frac{ns}{n}}[/tex]
Cancel out common factors
[tex]\mathbf{\frac dn = s}[/tex]
Make s the subject
[tex]\mathbf{s = \frac dn}[/tex]
Hence, the expression for rates is:
[tex]\mathbf{s = \frac dn}[/tex]
Read more about rate, distance and time at:
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