Velocity/Time/displancement Question

[Aicelle]

Homework Statement

Question:Vectorville and Scalatown are 20.0km apart. A cyclist leaves Vectorville and heads for Scalartown at 20.0km/h. A second cyclist leaves Scalartown for Vectorville at exactly the same time at a speed of 15.0km/h.
a) Where will the two cyclists meet between the two towns?
b) How much time passes before they meet (in minutes)?

Answers(From the back of the textbook): a)11.4km from vectorville; b) 0.571h or 34.2 min

Homework Equations

d= v/t --> t = d/v

The Attempt at a Solution

*sigh* I've tried everything, used like 3 peices of lined paper to work this equation. I first deicided to use the formula d=v/t rearranged to t=d/v afterwards, since there is no acceleration in this question. Then I concluded that both cyclist have the same time since it says they started at the same time, but we don't know at what time they meet each other nor the displacement of either of them during that time. So we've got two unknowns, therefore I tried merging the two equations of motions. Cyclist 1: t = d/ 20m/s ; Cyclist 2: t= d/ 15 m/s ---> d2 = 15km/h (d1/ 20 m/s) [subbed in cyclist 1's equation for time]. Then I kept trying to solve this but with no avail ;( Please help~

 

[jdougherty]

Hi Aicelle, welcome to PF!

You're right so far, d_{2} = \frac{15 d_{1}}{20}. But you have one equation now, and two unknowns (d_{1} \textrm{ and }d_{2}). Can you think of another relationship between them? How do they relate to the total distance?

 

[tiny-tim]

Welcome to PF!

Hi Aicelle! Welcome to PF!

Question:Vectorville and Scalatown are 20.0km apart. A cyclist leaves Vectorville and heads for Scalartown at 20.0km/h. A second cyclist leaves Scalartown for Vectorville at exactly the same time at a speed of 15.0km/h.
a) Where will the two cyclists meet between the two towns?
b) How much time passes before they meet (in minutes)?
…
So we've got two unknowns, therefore I tried merging the two equations of motions. Cyclist 1: t = d/ 20m/s ; Cyclist 2: t= d/ 15 m/s ---> d2 = 15km/h (d1/ 20 m/s) [subbed in cyclist 1's equation for time]. Then I kept trying to solve this but with no avail ;( Please help~

Say they're going along the x-axis …

the first one starts at x = 0 and the second starts at x = 20 …

what equations do you get?

 

[Aicelle]

Hmmm...total distance= d2+d1...so I can change d1 into 20-d2?

wait... we can simplfy d2=15d1/20 to d2=3d1/4...

 

[Aicelle]

Edit:Omigosh i got it =D

Sub d2=3d1/4 into the 20=d1+d2 equation =DDD Thanks everyone!