SOLUTION: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. Thank you in advance for the help!

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. Thank you in advance for the help!      Log On


   

Question 69992: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. Thank you in advance for the help!
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
---------------
1st trip DATA:
distance = 600 miles ; rate= x mph ; time= d/r = 600/x hrs.
-------------
Imaginary trip DATA:
distance = 600 miles ; rate = "x+20" mph ; time = d/r = 600/(x+20)
=========
EQUATION:
1st trip time = imaginary trip time + 1 hr.
600/x = 600/(x+20) + 1
Multiply thru by x(x+20)
600(x+20)= 600x +x(x+20)
12000 = x^2+20x
x^2+20x-12000=0
(x+120)(x-100)=0
x=100 mph (speed of his vehicle)
Cheers,
Stan H.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

Steve traveled 600 miles at a certain speed.
Had he gone 20mph faster, the trip would have 
taken 1 hour less. Find the speed of his vehicle. 
Thank you in advance for the help!

I'll do it a different way from Stanbon:

Let r be the actual rate and t be the actual time.
Then the rate for the hypothetical trip would be r+20 and
the time for the hypothetical trip would be t-1.

So put the information into a DRT chart:

                   DISTANCE   RATE    TIME
Actual trip      |   600   |   r   |   t
Hypothetical trip|   600   |  r+20 |  t-1

Now use DISTANCE = (RATE)(TIME)

     600 = rt
     600 = (r+20)(t-1)

FOIL out the right side of the second equation:

     600 = rt - r + 20t - 20

Multiply the 1st equation by -1 and add it to
the second equation

     600 =  rt - r + 20t - 20
    -600 = -rt
   ---------------------------
       0 =      -r + 20t - 20
       r = 20t - 20

Substitute in the first equation

     600 = rt
     600 = (20t - 20)t
     600 = 20tē - 20t
       0 = 20tē - 20t - 600
Divide every term through by 20
       0 = tē - t - 30
       0 = (t - 6)(t + 5)

t - 6 = 0     t + 5 = 0
    t = 6 hrs     t = -5 hrs

We discard the negative answer.

To find r, substitute in
            
     600 = rt
     600 = r(6)
     600 = 6r
     100 = r

So he traveled at 100 mi/hr.
His vehicle must have been a
small plane.

Edwin