Math equation. How to?

[markus.dnd]

If car would drive 15 km/h faster it would get to point B one hour earlier. If 10 km/h slower, it would take one hour more. Must find the speed of the car and distance it was driving.

 

I am really embarrassed, because i knew how to do it about 3 years ago. But totally forgot it by now.

 

I can give answers, but i need to know how i can get the answer.

 

Thank you all.

[Klaynos]

You have two unknown quantities (distance and speed).

 

So, you need two equations.

 

Can you think of any equations you can write down including these quantities and the ones you're given?

[markus.dnd]

Isn't time unknown as well?

[Klaynos]

Isn't time unknown as well?

 

 

Yeah, but you should be able to form equations where the t is unimportant...

 

After writing my reply I did wonder about this, I'll try and work threw it later to see what I come up with.

[D H]

You have two unknown quantities (distance and speed). So, you need two equations.[/quote']Isn't time unknown as well?

Time, distance, and speed are related by [math]d=v*t[/math], so if you know any two of the three you can compute the third. There are indeed two unknown quantities. They don't necessarily have to be distance and speed; you could, for example, solve for time and speed, leaving distance as a derived quantity.

 

markus, please write some of the relevant equations so we can give you a hand. Our policy here is to help people do their homework rather than doing it for them.

[markus.dnd]

well as i said. my problem is not that i need answers. I am studying for exams and found that there are things that i can not do a thing to. i don't see what i have to do with such exercise. And due to holidays i do not have possibility to consult with my teacher. I was hoping that someone could show or explain me how such exercise should be solved.

[D H]

There is a bit of a trick to this problem. The problem statement explicitly describes two trips: One in which the speed is 15 km/h faster, reducing the trip time by one hour and the second in which the speed is 10 km/h slower, increasing the trip time by one hour. The trick: there is an implicit third trip. Those speeds and times refer to some baseline trip. Call the speed and time on this trip v and t respectively. The distance for this baseline trip is d=v*t. The two stated trips cover this exact same distance, but with speeds and times that differ from the baseline.

 

Can you take it from here or do you need more help?

[markus.dnd]

Now i get it . Thank you.

 

I got the right answer as well. But all ready i got new problem.

 

(dam i should have paid more attention three years ago.)

 

I'll try to get to the end of that one without help. And if i can not, then i'll be back