Uniform Motion Problems (here i go again)

[Richay]

I need desperate help on this problem. I've tried this problem twice, and both times i was wrong (As my supervisor has said) And i have yet to figure it out. Can i get some help?

At exactly 5 p.m., Alex rode off from Asteroid. One hour later, Jessica rode off from Asteroid in the opposite direction at a speed 400 kph less than that of Alex. If they were 7900 km apart at 11 p.m., how fast did each travel?

Answer in the form: Alex, Jessica

I understand the steps to the process of solving this problem BESIDES one. They're going in opposite directions. And i'm unable to solve because I don't know what to do when they're both going in a different direction. I only understand this problem if they're going the same way. What do i do?

 

[TchrWill]

At exactly 5 p.m., Alex rode off from Asteroid. One hour later, Jessica rode off from Asteroid in the opposite direction at a speed 400 kph less than that of Alex. If they were 7900 km apart at 11 p.m., how fast did each travel?

1--In one hour, Alex travelss V(1) km in one direction.
2--Over the next 5 hours, Alex travels 5V additional km in the same direction while Jessica traveled 5(V - 400) km in the oppositie direction.
3--Therefore, 1V + 5V + 5(V - 400) = 7900 km.

I think you can take it from here.

 

[tkhunny]

I need desperate help on this problem.

I think you mean, "I desparately need help on this problem." Thanks for the laugh. :lol:

 

[soroban]

Hello, Richay!

Did you make a sketch?

At exactly 5 p.m., Alex rode off from Asteroid.
One hour later, Jessica rode off from Asteroid in the opposite direction at a speed 400 kph less than that of Alex.
If they were 7900 km apart at 11 p.m., how fast did each travel?

      A                                     J
      * ← - - - - - - - - - + - - - - - - → *
      :          6v         :   5(v-400)    :

Alex rode off at 5 p.m. at \(\displaystyle v\) kph.
By 11 p.m. (6 hours later), his distance was: \(\displaystyle 6v\) km.

Jessica rode off at 6 p.m. at \(\displaystyle v-400\) kph.
By 11 p.m. (5 hours later), her distance was: \(\displaystyle 5(v - 400)\) km.

The total distance between them was 7900 km.
There is our equation . . . \(\displaystyle 6v\,+\,5(v\,-\,400) \;= \;7900\)

 

[Richay]

Hello, Richay!

Did you make a sketch?

At exactly 5 p.m., Alex rode off from Asteroid.
One hour later, Jessica rode off from Asteroid in the opposite direction at a speed 400 kph less than that of Alex.
If they were 7900 km apart at 11 p.m., how fast did each travel?

      A                                     J
      * ← - - - - - - - - - + - - - - - - → *
      :          6v         :   5(v-400)    :

Alex rode off at 5 p.m. at \(\displaystyle v\) kph.
By 11 p.m. (6 hours later), his distance was: \(\displaystyle 6v\) km.

Jessica rode off at 6 p.m. at \(\displaystyle v-400\) kph.
By 11 p.m. (5 hours later), her distance was: \(\displaystyle 5(v - 400)\) km.

The total distance between them was 7900 km.
There is our equation . . . \(\displaystyle 6v\,+\,5(v\,-\,400) \;= \;7900\)

Hey, thanks for the help TchrWill and Soroban.

This is what I figured out.

6 x 900 = 5400 (Roger)

900 - 400= 500 x 5 = 2500 (Maryanne)

5400+2500 = 7900

So Roger's speed was 900 and Maryanne's was 500

Answered in from 900, 500. I did it correctly, right? (Or did i screw up?)
I used the 6v = 5 (v - 400) = 7,900 equation.