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Because of a traffic jam, Alana's 18-mile commute to work took 4 minutes longer than usual, and her average speed was decreased by 9 mph. How many minutes did it take her to get to work that day?

 Nov 21, 2018
 #1
avatar+107348 
+2

Let her normal time be  T

Let her normal rate be R

 

So

 

RT = D

RT = 18    ⇒   R  =   18/T     (1)

 

We need to convert 4 minutes to hours  =  4/60   = 1/15 hr

 

So....when she is slowed we have

 

(R - 9) ( T + 1/15)  = 18     (2)

 

Sub   (1)   into (2)   and we have

 

(18/T - 9) (T + 1/15) = 18      simplify

 

18 - 9T + (18)/ (15T) - 9/15 = 18     subtract 18 from both sides

 

 - 9T  + (6)/(5T) - 3/5 =  0       multiply through by -5T

 

45T^2 - 6 + 3T = 0

 

45T^2 + 3T - 6  = 0       divide through by 3

 

15T^2 + T-  2  =   0

 

(5T + 2) ( 3T - 1)  = 0

 

The first factor set to 0 and solved  for T produces a negative anser

 

The second factor set to 0 and solved for T produces   T  =1/3

 

This is in hrs

 

So   (1/3) hr   =  20 min   = her normal time to get to work

 

So....now...at the slowed rate   it takes her 4  minutes longer = 24min

And  24 minutes = 24/60 hrs  = 2/5 hrs

 

 

Check

R =  18 / (1/3) = 54mph

And   54mph * 1/3 hr = 18 miles

 

And at the slower rate

(54 - 9) ( 1/3 + 1/15)

(45) ( 18/ 45)   =  18 miles

 

 

cool cool cool

 Nov 22, 2018
 #2
avatar+20217 
0

ANother perspective;

the situation in the question is described by the following equation:

 

18/(x-9) = 18/x +4/60      where x = her usual speed in mph   (x-9) = new (slower speed) &  4 minutes = 4/60 hr

 

solve for x

18 = (18x-162)/x  +4/60x - 4/60(9)

18x = 18x-162  + 4/60 x^2 - 36/60 x

0 = .06667 x^2 -.6x - 162       Use Quadratic Formula to find x = 54 mph

Her new (slower) speed = 54-9 = 45 mph     so the trip will take her   18 m / 45m/hr = .6hr= 24 minutes

 Nov 22, 2018

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