Question 136647
Jacqui commutes 30 mi to her job each day. She finds that if she drives 10 mi/h faster, it takes her 6 min less to get to work. Find her new speed
:
Let s = new speed
then
(s-10) = original speed
:
Convert 6 min to hrs: 6/60 = .1 hrs
:
Write a time equation: Time = dist/speed
:
Fast trip + 6 min = slow trip
{{{30/s}}} + .1 = {{{30/((s-10))}}}
Multiply equation by s(s-10) to get rid of the denominators:
s(s-10)*{{{30/s}}} + .1(s(s-10) = s(s-10)*{{{30/((s-10))}}}
Results
30(s-10) + .1s^2 - 1s = 30s
:
30s - 300 + .1s^2 - s - 30s = 0
:
.1s^2 - s - 300 = 0; a quadratic equation
:
s^2 - 10s - 3000 = 0; mult by 10 to get rid of the decimal
Factors to:
(s-60)(s+50) = 0
:
S = 60 mph is the new speed
:
:
Check solution by finding the times of each trip:
30/50 = .6 hrs or 36 min
30/60 = .5 hrs or 30 min
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difference time = 6 min


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