Distance Rate Time problem: Bicyclists started at noon....

[ultrasonicsite]

I seem to have a problem progressing completely through Distance/Rate/Time problems without the aid of a teacher. I do not know what questions to ask myself that will allow me to continue towards a final answer.

Example(Taken directely from the current chapter I'm doing):

Bicyclists Brent and Jane started at noon from points 60km apart and rode toward each other, meeting at 1:30pm. Bren'ts speed was 4km/h greater than Jane's speed. Find their speeds.

So the first thing I would do is gather all the information I can.

-They started at 12:00pm
-They started 60km apart
-They rode towards each other meeting at 1:30pm
-Brent's speed is 4km/h faster than Jane's speed.
-I am trying to find both their speeds.

So:

let j = Jane's speed
let j+4 = Brent's speed

Now, both their distances traveled must equal 60km. The half-way point is 30km. Though they are going at different speeds, so they will not stop exactely in the middle.

-------------------------------------
-------Rate*Time=Distance--
-------------------------------------
Brent | 4+j | 1.5 | 1.5(4+j)
-------------------------------------
Jane | j | 1.5| 1.5j
-------------------------------------

-Stuck

 

[skeeter]

add the distances together ... what is the sum?

1.5(4+j) + 1.5j = ?

... then solve for j

 

[galactus]

As you know, distance = rate*time.

\(\displaystyle \L\\Jane=r\underbrace{(\frac{3}{2})}_{\text{time}}\)

\(\displaystyle \L\\Brent=\underbrace{(r+4)}_{\text{rate}}\underbrace{(\frac{3}{2})}_{\text{time}}\)

Total distance = 60

Add together and solve for r.

 

[ultrasonicsite]

Heres what I did. I created an equation (which is what I'm kind of struggling to get the hang of)

60=1.5(4+j)+1.5j
60=6+1.5j+1.5j
60-6=3j
54=3j

Divide both sides by 3 to have j alone

18=j
j=18

Brent = j+4
Brent = 18+4
Brent=22

Brent's speed = 22
Jane's speed = 18

Double check===

Brent's Distance = 1.5(4+j) = 1.5(4+18) = 33

Jane's Distance = 1.5j = 1.5(18) = 27

33 + 27 = 60

Distance = 60; True

 

[ultrasonicsite]

Let me try to figure a the next one out, if I have any problems I'll post =)

And thank you for your help.

 

[Denis]

Re: Distance Rate Time problem: Bicyclists started at noon..

So:
let j = Jane's speed
let j+4 = Brent's speed

You can now keep it real simple:
distance of 60 is travelled at speed of 2j+4 in 3/2 hours;
since time * speed = distance:
3/2(2j + 4) = 60
6j + 12 = 120
6j = 108
j = 18