Alan

In that case the angle (theta, say) subtended at the centre of the circle is found from:  theta/360 = 14/(2*pi*r)  (with theta in degrees)

The sector area is then found from:    area/(pi*r^2) = theta/360  or  area/(pi*r^2) = 14/(2*pi*r)  or area = 7r  

You'll need to supply more information, such as radius.

Let f be the fraction and d be the denominator.  Then  f = (d-1)/d  or f = 1 - 1/d

Let g = 1/d, then f = 1 - g.  This is a linear equation in the variables f and g.

Have a look at:  https://en.wikipedia.org/wiki/E_(mathematical_constant) 

You don't say what you are trying to calculate, so I've assumed you want to find the approaching velocities.

Also you can't have a mass in Newtons, so I've converted your 6500 pounds (mass) to kg.  (there is a difference between lbf (force) and lb (mass)).

On second thoughts I should probably have assumed kinetic friction.  However, static friction is usually higher than kinetic friction, so perhaps I've used the right numerical value after all!

Use a combination of graphical and numerical methods:

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"Actually though it looks like Alan has made a mistake in his arithmetic right at the end."

I did indeed make a mistake in the arithmetic of the last line.  It should have been:

                                         nCr(97,5)/nPr(97,5) = 0.00833...   (= 1/5!)

In general:   nCr(n,k)/nPr(n,k) = n!/[(n-k)!k!]*(n-k)!/n!  = 1/k!   (as given directly by Nauseated).

Consider a simpler problem first

Melody has calculated the tangential velocity (that is, the tangent to a circle of radius sqrt(10)).  The component of of this velocity along the (straight) shoreline is obtained by multiplying Melody's velocity by 3/sqrt(10).

y = 10 + 6(x - 1)       (assuming the distance is greater than 1 km)

Just replace x by 15 to work out the total fare.