another rate word problem

[wrightka]

A camper leaves the campsite walking due east at 3.5 mph. Another camper leaves the campsite at the same time but travels due west. In two hours the campers will be 15 miles apart. What is the walking rate of the second camper?
I have no clue except that rate = distance/time.

 

[Deleted member 4993]

A camper leaves the campsite walking due east at 3.5 mph. Another camper leaves the campsite at the same time but travels due west. In two hours the campers will be 15 miles apart. What is the walking rate of the second camper?
I have no clue except that rate = distance/time.

Another clue would be to draw a sketch:


B..................................C..............................A

C is the campsite.

A & B are the positions of the campers after 2 hours.

Let the speed of B be V[sub:16fcirnv]b[/sub:16fcirnv] ---> rate of walking of the second camper.

In 2 hours B walks the distance = CB = V[sub:16fcirnv]b[/sub:16fcirnv] * 2 miles

In 2 hours A walks the distance = CA = 3.5 * 2 miles

Now what ......

 

[wrightka]

My guess is to set the equations equal to each other and solve for Vb?

2hVb=(3.5mph)2h
2Vb=7mph^2
Vb =3.5mph

My big problem is setting up the equation right. You have been a lot of help to me so far. Where can I get more practice on math word problems?
Thanks,
wrightka

 

[Deleted member 4993]

Check out:

http://www.purplemath.com/modules/index.htm

 

[Deleted member 4993]

A camper leaves the campsite walking due east at 3.5 mph. Another camper leaves the campsite at the same time but travels due west. In two hours the campers will be 15 miles apart. What is the walking rate of the second camper?
I have no clue except that rate = distance/time.

Another clue would be to draw a sketch:


B..................................C..............................A

C is the campsite.

A & B are the positions of the campers after 2 hours.

Let the speed of B be V[sub:lr7phn5z]b[/sub:lr7phn5z] ---> rate of walking of the second camper.

In 2 hours B walks the distance = CB = V[sub:lr7phn5z]b[/sub:lr7phn5z] * 2 miles .......................(1)

In 2 hours A walks the distance = CA = 3.5 * 2 miles......................................(2)

Now what ......

My guess is to set the equations equal to each other and solve for Vb? <<< That is incorrect! Why would those two terms be equal?
2hVb=(3.5mph)2h
2Vb=7mph^2
Vb =3.5mph

No that is incorrect
wrightka

BC + CA = BA ............................ (3) from the picture

combining (1), (2) and (3) - we get:

V[sub:lr7phn5z]b[/sub:lr7phn5z] * 2 + 3.5 * 2 = 15

2V[sub:lr7phn5z]b[/sub:lr7phn5z] + 7 = 15

2V[sub:lr7phn5z]b[/sub:lr7phn5z] + 7 - 7= 15 - 7 .....................isolate V[sub:lr7phn5z]b[/sub:lr7phn5z]

2V[sub:lr7phn5z]b[/sub:lr7phn5z] = 8

Now continue...