Find Speed

[nycmathguy]

Doomtown is 200 miles due West of Sagebrush and Joshua is due West Doomtown. At 9 in the morning Mr. Archer leaves Sagebrush for Joshua. At 1 in the afternoon Mr. Samson leaves Doomtown for Joshua. If Mr. Samson travels at an average speed of 20 mph faster than Mr. Archer and they each reach Joshua at 4 in the afternoon, how fast is each traveling?

Note: I think there is missing information here. What do you say? I can't sense of this word problem.

Thank you.

P. S. Back to precalculus this Friday, Saturday and Sunday.

 

[MathLover1]

Josh-----------------d--------------Doom----------------200------------Sage

A's travel time: 9am to 4pm = 7 hr
S's travel time: 1pm to 4pm = 3 hr

A's speed = s
S's speed = (s+20)

Let d = distance from Doom to Josh; (Distance = time * speed)

d = 3(s+20); (S's speed times 3 hr)
d = 3s + 60; (use for substitution)

(d + 200) = 7s; (A's speed times 7 hrs)

In the above equation, Replace d with (3s+60) and find s:

(3s + 60 + 200 = 7s
3s + 260 = 7s
260 = 7s - 3s
260 = 4s
s = 260/4
s = 65 mph is A's speed
Then
65 + 20 = 85 mph is S's speed


Check solution by finding d
d = 3s + 60
d = 3(65) + 60
d = 195 + 60
d = 255 mi

255/85 = 3 hrs S's time
(200+255)/65 = 7 hrs, A's time


for this kind of problems, you can use this for set up

 

[nycmathguy]

Josh-----------------d--------------Doom----------------200------------Sage

A's travel time: 9am to 4pm = 7 hr
S's travel time: 1pm to 4pm = 3 hr

A's speed = s
S's speed = (s+20)

Let d = distance from Doom to Josh; (Distance = time * speed)

d = 3(s+20); (S's speed times 3 hr)
d = 3s + 60; (use for substitution)

(d + 200) = 7s; (A's speed times 7 hrs)

In the above equation, Replace d with (3s+60) and find s:

(3s + 60 + 200 = 7s
3s + 260 = 7s
260 = 7s - 3s
260 = 4s
s = 260/4
s = 65 mph is A's speed
Then
65 + 20 = 85 mph is S's speed


Check solution by finding d
d = 3s + 60
d = 3(65) + 60
d = 195 + 60
d = 255 mi

255/85 = 3 hrs S's time
(200+255)/65 = 7 hrs, A's time


for this kind of problems, you can use this for set up

View attachment 1922

Let me be honest with you. I don't recall ever solving this type of involved distance problem back in my school days. Of course, I was a remedial student and thus placed in remedial classes throughout my high school years. I never took Algebra 1 and 2 in high school.

 

[Country Boy]

Do you know that speed is measured in "miles per hour", or "kilometers per hour" or, in the lab, "meters per second"? Then you know that speed is "distance divided by time".

"Doomtown is 200 miles due West of Sagebrush and Joshua is due West of Doomtown. At 9 in the morning Mr. Archer leaves Sagebrush for Joshua. At 1 in the afternoon Mr. Samson leaves Doomtown for Joshua. If Mr. Samson travels at an average speed of 20 mph faster than Mr. Archer and they each reach Joshua at 4 in the afternoon, how fast is each traveling?

There are two "unknowns" here since we are told that "Joshua is due West of Doomtown" but are not told how far. Let "v" be Mr. Archer's speed in miles per hour. Mr Samson's speed is v+ 20 miles per houf. Let "x" be the distance, in miles, from Joshua to Doomtown.

Mr. Archer, traveling from Sagebrush to Joshua, travels 200+ x miles at speed v. He leaves at "9 in the morning" and arrives at "4 in the afternoon" so he travels for 3+ 4= 7 hours. Since speed is "distance divided by time", v= (200+ x)/7.

Mr. Samson, traveling from Doomtown to Joshua, travels x miles at v+ 20 miles per hour. He leaves at "1 in the afternoon" and also arrives at "4 in the afternoon" so he travels for 3 hours. Since speed is "distance divided by time", v+ 20= x/3. Subtracting 20 from both sides, v= x/3- 20.

Then v= (200+ x)/7= x/3- 20. Solve that equation for v.

Since I don't like fractions the first thing I would do is multiply both sides by 3(7)= 21. That gives 3(200+ x) = 7x+ 420.

Remember that you must give TWO answers, the speed each was traveling, v and v+ 20. Solving that equation gives you x and then v= (200+ x)/7.

 

[nycmathguy]

Do you know that speed is measured in "miles per hour", or "kilometers per hour" or, in the lab, "meters per second"? Then you know that speed is "distance divided by time".

"Doomtown is 200 miles due West of Sagebrush and Joshua is due West of Doomtown. At 9 in the morning Mr. Archer leaves Sagebrush for Joshua. At 1 in the afternoon Mr. Samson leaves Doomtown for Joshua. If Mr. Samson travels at an average speed of 20 mph faster than Mr. Archer and they each reach Joshua at 4 in the afternoon, how fast is each traveling?

There are two "unknowns" here since we are told that "Joshua is due West of Doomtown" but are not told how far. Let "v" be Mr. Archer's speed in miles per hour. Mr Samson's speed is v+ 20 miles per houf. Let "x" be the distance, in miles, from Joshua to Doomtown.

Mr. Archer, traveling from Sagebrush to Joshua, travels 200+ x miles at speed v. He leaves at "9 in the morning" and arrives at "4 in the afternoon" so he travels for 3+ 4= 7 hours. Since speed is "distance divided by time", v= (200+ x)/7.

Mr. Samson, traveling from Doomtown to Joshua, travels x miles at v+ 20 miles per hour. He leaves at "1 in the afternoon" and also arrives at "4 in the afternoon" so he travels for 3 hours. Since speed is "distance divided by time", v+ 20= x/3. Subtracting 20 from both sides, v= x/3- 20.

Then v= (200+ x)/7= x/3- 20. Solve that equation for v.

Since I don't like fractions the first thing I would do is multiply both sides by 3(7)= 21. That gives 3(200+ x) = 7x+ 420.

Remember that you must give TWO answers, the speed each was traveling, v and v+ 20. Solving that equation gives you x and then v= (200+ x)/7.

Cool.