SOLUTION: please help!! i need help on two simlar questions. q1) Doomtown is 200 miles due west of Sagebrush and Joshua is due west of Doomtown. At 9AM Mr. Archer leaves Sagebrush for Jos

Question 152412: please help!! i need help on two simlar questions.
q1) Doomtown is 200 miles due west of Sagebrush and Joshua is due west of Doomtown. At 9AM Mr. Archer leaves Sagebrush for Joshua. At 1PM Mr. Sassoon leaves downtown for Joshua. If Mr. Sassoon travels at an average speed of 20 mph faster than Mr.Archer and they each reach Joshua at 4 PM, how fast is each traveling?
q2) Timothy leaves home for Skedunk 400 miles away. After 2 hours, he has to reduce his speed by 20 mph due to rain. If he takes 1 hour for lunch and gas, and and reaches Skedunk 9 hours after he left home, what was his initial speed?
please help!!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
q1) Doomtown is 200 miles due west of Sagebrush and Joshua is due west of Doomtown. At 9AM Mr. Archer leaves Sagebrush for Joshua. At 1PM Mr. Sassoon leaves doomtown for Joshua. If Mr. Sassoon travels at an average speed of 20 mph faster than Mr.Archer and they each reach Joshua at 4 PM, how fast is each traveling?
A pictorial representation:
;
Josh*-------------Xmi--------------Doom*-----------200mi-------------Sage*
:
Let dist from Doom to Josh = x
Dist from Josh to Sage = x+200
:
Let r = Sass's speed
then
(r-20) = Arch's speed
:
From the problem we learn:
A's travel time = 7 hrs (9am to 4 pm)
S's travel time = 3 hrs (1pm to 4 pm)
:
Write a dist equation for each Dist = time * speed
:
3r = x (Sass's dist)
and
7(r-20) = x + 200 (Arch's dist)
7r - 140 = x + 200
7r - x = 200 + 140
:
Substitute 3r for x in the above equation:
7r - 3r = 340
4r = 340
r =
r = 85 mph is Sass's speed
then
85 - 20 = 65 mph is Arch's speed
:
:
We can find x (dist from Josh to Doom) to check our solutions
x = 3r
x = 3(85)
x = 255 mi
:
Sass's time: 255/85 = 3 hr
Arch's time: (255+200)/65 = 7 hrs
:
:
:
q2) Timothy leaves home for Skedunk 400 miles away. After 2 hours, he has to reduce his speed by 20 mph due to rain. If he takes 1 hour for lunch and gas, and and reaches Skedunk 9 hours after he left home, what was his initial speed?
:
By reading the problem we can learn that:
Actual travel time = 8 hr
If he travels at full speed for 2 hrs, he will be at reduced speed for 6 hr
:
Let s = his initial speed
then
(s-20) = his reduced speed
;
Write a simple distance equation: dist = time * speed
:
2s + 6(s-20) = 400
2s + 6s - 120 = 400
8s = 400 = 120
s =
s = 65 mph is his initial speed
:
:
Check solution by finding the dist of each:
2*65 + 6*45 = 400 confirms our solution

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