<PRE><B>Question 1034986</B>
A salesman drives from Ajax to Barrington, a distance of 114 mi, at a steady speed.
 He then increases his speed by 14 mi/h to drive the 141 mi from Barrington to Collins.
 If the second leg of his trip took 3 min more time than the first leg, how fast was he driving between Ajax and Barrington?
:
A-------114------B--------141---------C
:
Let s = his speed from A to B
then
(s+14) = his speed from B to C
:
Change 3 min to hrs: 3/60 reduces to 1/20
:
Write a time equation; time = dist/speed
AB time - BC time = 3 min
{{{141/((s+14))}}} - {{{114/s}}} = {{{1/20}}}
multiply equation by 20s(s+14), cancel the denominators and you have
20s(141) - 20(s+14)(114) = s(s+14)
20s(141) - 2280(s+14) = s(s+14)
2820s - 2280s - 31920 = s^2 + 14s
540s - 31920 = s^2 + 14s
A quadratic equation
0 = s^2 + 14s - 540s + 31920
s^2 - 526s + 31920 = 0
The quadratic formula; a=1; b=-426; c=31920; reveals two solutions
s = 456, not reasonable in a car
and
s = 70 mph is the speed from A to B
:
;
See if that checks out, find the actual time for each part (BC is 84mph)
141/84 = 1.67857 hrs
114/70 = 1.62857 hrs
--------------------
difference: .05 hrs;  which is 3 min.  (3/60 = .05)
:
Hey, did this make sense to you? I like to know if I am explaining this sufficiently.  Thanks. CK</PRE>