I need help with a math problem!!!!!!

[Zej]

Okay, i need help with a math problem! FAST.

A car left a garage travelling at 80km/h. 15 minutes later, another car left the same garage, travelling in the same direction at 100km/h. How long will it take for the first car to catch up with the second car?

I got part of the solution done... D1 = the distance the first car travelled and D2 is the distance the 2nd car travelled and r = the rate.

Let t be the time they meet.

D1 = rt
D1 = 80t


D2 = r(t - 15)
D2 = 100(t - 15)

D1 = D2
Therefor, 80t = 100(t - 15)


NOW I'M STUCK!!! SOMEONE PLEASE HELP!!!

[Edited by Zej on 05-23-2000 at 07:18 PM]

[Sam Finch]

the cars pass each other 75 minutes after the first car sets off.

if t is the time in hours after the first car sets off we get the distace the first car has traveled = 80t
and the distance the second has traveled is 100 (t - 0.25)

so when thy pass

80t = 100 (t - 0.25)

4t = 5 (t - 0.25)

= 5t - 1.25

1.25 = 5t - 4t
= t

ie t = 1.25 hours or 75 minutes

[Zej]

THANK YOU SO MUCH!!!!

I also need help on this one now if it's not a problem.

The reason i'm inclosing this in a code is because the
indentation is all messed up if I leave in normally.

      t - 6        5t + 2
 -  --------   -   -------    =  -11
        3             6

[Edited by Zej on 05-23-2000 at 08:06 PM]

[Edited by Zej on 05-23-2000 at 08:07 PM]

[mr. vb]

t= -27

Yes, Batman's answer is right. If you want to keep the answer simple and not have all of the decimals, you could just use

       76
 t =  ----
        7

[Chris]

First I think we should figure out the equation for both car have travel (km)

First Car:
Let assume
x is the total distancs for the first have travel in 15 minutes.
y is the total time that first car have travel (in minutes)
z is the travel speed of first car

So,
x = yz/60

Second Car:
Let assume
a is the total distancs for the first have trave (in km).
b is the total time that first car have travel (in minutes)
c is the travel speed of first car

So,

a = bc/60

Now, we can combine this 2 equation and calculate the time needed for the second car to catch up the first car.

i. x = yz/60
ii. a = bc/60

Since x = a,
Therefore,

[b]yz/60 = bc/60[b]
[b]yz = bc[b]
Our final answer is y,

y = bc/z

But because the first car have strat travel in 15 minutes early than the second car so, the final solution
should become:

Let assume t is the time that firt car have travel before the second car start travel.

y-t = bc/z

Now, we can fill in the variable with the respective value:
Let assume second car have travel 1 hour (60 minutes)
t = 15 minutes

y-15 = (60*100)/80
y = 75 minutes

As a result, I think the fastest and flexible solution is the following equation:

y = (bc/z) + t

Where:
b = Total time for second to catch up with first car.
y = Total time that first car have travel.
z = Travel speed for the first car.
c = Travel speed for the second car.
y = Total time for the first car have travel.

and you can create a GUI enable user to enter the respective value and let your program to calculate it.

Hope this can help you.

[da_silvy]

firstly we know that at 15 minutes car
a has travelled
Car A --------->20km
Car B 0 Km

after 1hr and 15 minutes
Car a ----------> (80 + 20)km
Car B ----------> (100km/hr for 1 hr)100km

ok??

does any one agree with me?
well does anyone?

[da_silvy]

whoops

it was already answered sorry
i forgot that 75 minutes, 1.25 hours and 1hr 15 minutes are all the same thing

Zej, what year are you in to be doing this stuff?

Now, looking at your question closely, the first car WON'T catch up to the second car, it is already ahead of the second one. The second car will pass the first car, and the first car will never catch up.

So if that is the exact wording of the question, then you answer never.

[cady]

i might have this wrong, but if the 2nd car is the one that is suppose to catch up to the first(if the wording of the orginal question is wrong) than wouldn't it take 1 hour to catch up? the first car left 15 minutes earlier than the 2nd car, how long will it take to catch up, right? the first car has gone 20 km when the second one starts going 100 km per hour so the 1st car in 1 hour(from when the 2nd started) will be 100 km away and the second car after 1 hour will be 100 hm away. so it only takes the second car 1 hour to overtake the 1st car from the time that IT starts, not the first car. Is this totally wrong, sometimes i'm not very good at these kind of questions but i like trying so if i'm wrong please explain.

[Sam Finch]

You're right, It's one hour after the second starts or one hour and 15 minutes after the first one starts. You just have to decide which car you're saying starts at t=0

[cady]

How come i couldn't do these questions in school where it really mattered? anyway, the answer would have to be 1hr because it's the 2nd car that is catching up to the first and has only been driving 1hour when it does. the catch up time would have to start with the car that started second and is doing the catching up not the first that is ahead until then. let's do another one

hmm, my garage is only about 15 feet long, I must find one of those cars that can reach 75 mph before it has passed the garage door, think of the fuel economy!

The answer is 1 hour.

[da_silvy]

yes, but you have to include the time in which the first car has been going. therefore making it 1 hr and 15 minutes['

[cady]

I don't think so. You don't include the time of the first car because it isn't the one catching up. Only one car is doing the "catching up" and that is the second car and it only takes 1 hour, the first car is just driving not catching up to anything so it can drive for 10 hours and it doesn't matter, it's how long it takes car 2 to catch up with car 1 and that is 1 hour.

[Sam Finch]

that's why I said 75 minutes after the first car has left.

both answers are right as long as you say when you started your clock. You could say 45minutes before the first car runs out of petrol if you want as long as you specify that the first car has 2 hours worth of petrol in it. (don't try that in an exam)