David left Boston at 25 km/h. 45 minutes later Peter rode out of Boston at 25 km/h. how far down the road will David be when Peter catches up to him?

Question

This is a problem on my homework that I don't know how to do. 

Answer 1

Hello, thank you for taking the time to post your question!

Is it possible that you listed the incorrect rate for one of the riders? The way that it’s phrased now I think that it would be impossible for Peter to catch David because they are moving at the same speed. So if David had a head start and they are riding at the same rate it would not be possible for Peter to catch up.

I hope that helps get you moving in the right direction! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

d = r * t 

So for peter: d1 = r1 * t1

For David: d2 = r2 * t2

 

We know that david had a head start of 45 minutes so t2 = t1+45/60 and we want to know when they catch each other so when d1 = d2 = d

 

Plugging back 

 

d = r1 * t1

d = r2 * (t1+45/60)

 

Any two equations that are equal to the same thing are equal to each other(Transitive property and this is important to know and understand)

 

so

 

r1 * t1 = r2 * (t1+45/60)

r1 * t1 = r2* t1 + 45/60 *r2

r1 * t1 - r2 * t1 = 45/60 * r2

Factor

 

t1 * (r1 - r2) = 45/60*r2

Divide

 

t1 = (45/60)*r2/(r1-r2)

 

With r2 = 25 and r1 = 25

 

t1 approaches infinitely(division by zero).

 

The problem here is that both are moving at the same speed, did you put the wrong speed for Peter?

 

It makes sense that Peter never catches David because David left first and Peter is moving at the same speed. In order for Peter to catch David he needs to be moving at least a  little bit faster. However the equation still holds, just need the correct rates to plug in(The wording makes it sound like they should actually catch each other). Once you find the time just plug into d1 = r1 * t1 to find the total distance that was covered when they cross.

 

You can also solve this by graphing, going back to the original distance equations:

 

y = r1 * t

y = r2 * t + 45/60 * r2

 

With r1 = r2 you can see that these are two parallel lines that never cross. if r1 > r2, they cross once with t > 0 and if r2 > r1 they cross once with t < 0 (but negative times don't mean much in the physical world).