SOLUTION: I've been working on word problems, but this one has me stumped. A car travels 120 miles. A second car; traveling 10 mph faster than the first car, makes the same trip in one hou

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: I've been working on word problems, but this one has me stumped. A car travels 120 miles. A second car; traveling 10 mph faster than the first car, makes the same trip in one hou      Log On


   

Question 5211: I've been working on word problems, but this one has me stumped.
A car travels 120 miles. A second car; traveling 10 mph faster than the first car, makes the same trip in one hour less time. Find the speed of each car.
Found 2 solutions by xcentaur, glabow:
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
distance travlled by first car=120 miles
speed of first car=x mph
time taken by first car=d/s=(120/x)hours


speed of second car=(x+10)mph
distance travelled=120 miles
time taken=d/s=[120/(x+10)]hours


Given,time taken by second car is one hour less than first car.
then,


120/x-120/(x+10)=1
[120(x+10)-120x]/x(x+10)=1
[120x-120x+1200]/x(x+10)=1
1200/x(x+10)=1
1200=x(x+10)
x^2+10x-1200=0
x^2+30x-40x-1200=0
x(x+30)-40(x+30)=0
(x+30)(x-40)=0
x=-30 or +40


Speed cannot be -ve.
hence x=40 mph


Thus speed of first car=40 mph,second car=50 mph

Hope this helps,
good luck.


Answer by glabow(165) About Me  (Show Source):
You can put this solution on YOUR website!
Always be careful and define the variables you are using.
Let x = the speed of the first car
Then x+10 = the speed of the second car
Let t1 = the time the first car takes to travel 120 miles
Then t2 = the time the second car takes to travel 120 miles
We know that t1 = t2 + 1 (why?)
The equation for time of travel is
t = distance / speed
For the two cars we have the following equations
t1 = 120 / x
t2 = 120 (x+10)
Using t1 = t2 +1 we combine the two equations to
+120+%2F+%28x%2B10%29+%2B+1+=+120+%2F+x+
This is solved by the following steps
+1+=+%28120+%2F+x%29+-+%28+120+%2F+%28x%2B10%29+%29 [rearranging terms]

[multiply first term on right by (x+10) and second term on right by
x to get a common denominator]
+1+=+%28120x+%2B+1200+-+120x%29+%2F+x%28x%2B10%29+ [simplify and do subtraction]
+x%28x%2B10%29+=+1200+ [simplify and rearrange]
+x%5E2+%2B+10x+-+1200+=+0+ [simplify and rearrange]
This is most easily solved by factoring. The factors are numbers that multiply to -1200 and add up to 10. Numbers that satisfy these criteria are 40 and -30.
So the equation becomes
+%28x%2B40%29+%28x-30%29+=+0+
which is true for x = -40 and x = 30. (why?)
You cannot have a speed of -40. So x = 30.
The speed of the first car is 30 miles/hr.
The speed of the second car is 40 miles/hr.
****Checking gives: the first car travels for 120/30 = 4 hours.
the second car travels for 120/40 = 3 hours.