SOLUTION: Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, she realizes that she must increase her speed by 15 mph in order to get the

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, she realizes that she must increase her speed by 15 mph in order to get the      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 143527This question is from textbook Introductory Algebra
: Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, she realizes that she must increase her speed by 15 mph in order to get there on time. What was her speed for the first 30 miles?
I am not sure which motion formula to use. Could you assist me with this problem? This question is from textbook Introductory Algebra

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, she realizes that she must increase her speed by 15 mph in order to get there on time. What was her speed for the first 30 miles?
:
Let s = speed for 1st 30 mi
then
(s+15) = speed for last 20 mi
:
Write a time equation: Time = Dist/speed
:
1st 30mi time + Last 20mi time = 1 hr
30%2Fs + 20%2F%28%28s%2B15%29%29 = 1
:
Multiply equation by s(s+15):
s(s+15)*30%2Fs + s(s+15)*20%2F%28%28s%2B15%29%29 = s(s+15)(1)
:
Cancel out the denominators:
30(s+15) + 20s = s(s+15)
;
30s + 450 + 20s = s^2 + 15s
:
Arrange as quadratic equation:
s^2 + 15s - 50s - 450 = 0
:
s^2 - 35s - 450 = 0
:
Factor to:
(s - 45)(s + 10) = 0
:
Positive solution:
s = 45 mph on the 1st 30 mi
;
:
Check solution by finding the times at each speed, (60 mph on the last 20 mi)
30/45 = .67
20/60 = .33
------------
total = 1.0