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average speed of an unknown distance


Best Answer jhawk, 21 August 2013 - 07:15 PM

Spoiler for I make it to be

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#1 BMAD

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Posted 21 August 2013 - 03:34 PM

I drove x miles at 55 mph. I then drove x + 20 miles at 40 mph. I drove for a total of 100 minutes. How far did I drive?

Edited by BMAD, 21 August 2013 - 03:35 PM.

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#2 jhawk

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Posted 21 August 2013 - 07:15 PM   Best Answer

Spoiler for I make it to be


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#3 Joyandwarmfuzzies

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Posted 21 August 2013 - 07:15 PM

Spoiler for D=R*T


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#4 dgreening

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Posted 21 August 2013 - 07:41 PM

X = miles at 55 mph

X+20 = miles driven at 40 mph

 

and

55 mi/ hr = 0.917  mi/ min

40 mi/ hr = 0.667 mi/ min

 

D = Total Distance = D1 (at 55 mph) + D2 (at 40 mph)

 

T = Total Time = T1 (at 55 mph) + D2 (at 40 mph)

 

D1 = T1 ( 0.917  mi/ min)

D2 = T2 ( 0.667 mi/ min)

 

Therefore 

D = D1 + D2 =   T1 ( 0.917  mi/ min) + T2 ( 0.667 mi/ min)

 

We also know that D1 +20 = D2

 

These can be rearranged into 2 simultaneous equations in the form

 

20


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#5 dgreening

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Posted 21 August 2013 - 08:00 PM

X = miles at 55 mph

X+20 = miles driven at 40 mph

 

and

55 mi/ hr = 0.917  mi/ min

40 mi/ hr = 0.667 mi/ min

 

D = Total Distance = D1 (at 55 mph) + D2 (at 40 mph)

 

T = Total Time = T1 (at 55 mph) + D2 (at 40 mph)

 

D1 = T1 ( 0.917  mi/ min)

D2 = T2 ( 0.667 mi/ min)

 

Therefore 

D = D1 + D2 =   T1 ( 0.917  mi/ min) + T2 ( 0.667 mi/ min)

 

We also know that D1 +20 = D2

 

These can be rearranged into 2 simultaneous equations in the form

 

20

I hit the wrong key

 

20    =   - T1 ( 0.917  mi/ min) + T2 ( 0.667 mi/ min)              [Eqn 1]

100   =        T1                                +T2                                            [Eqn 2]

 

Multiplying all terms in  Eqn 1 by {- 1/0.667} yeilds

 

-30  =   1.375 T1                           - T2

 

Adding this to Eqn2 results in

 

70 =  2.375 T1 Therefore 

T1 = 70 /2.375  = 29.47 min, so

D1 = 29.47 min * ( 0.917  mi/ min) = 27.02

D2 = D1 + 20 = 47.02

 

Therefore Total distance = 74.04 Miles


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