2 Cars travelling from towns- challenging problem

i have a problem really bugging me

2 cars from different towns town A and town B travel in opposite directions from 1 town to another. they meet at town C.the car from A completed thejourney from C to B in 45 minutes at a steady rate of 64/kms an hour. The other car completes the journey from C to A in 20 minutes. how fast was the car from B going?

i know that the car from B goes from B to C in the same time as the car from A goes from A to C.

please show how i can do this problem, and please don't say "guess and check"

thanks

Re: 2 Cars travelling from towns- challenging problem

Can you make a diagram of your problem first?

Re: 2 Cars travelling from towns- challenging problem

A-->c------45m---->B-64km/h

A<-------20m---C<--B-xkm/h

find x

m=minutes

Re: 2 Cars travelling from towns- challenging problem

You need to assume a few more things:

1) They leave at the same time.

2) Both their velocites are constant during the journey (in particular, the A car is moving at 64 km/h during its trip from A to C).

If you put all the information you know into equations, you'll then have enough to solve the problem. The way I did it, I had 3 equations with 3 unknowns that I had to solve for the velocity of the B car. A square root was involved. You might or might not get that depending on how you set things up.

Hint#1: "meeting at C" and "leaving at the same time" together say that the time for the A car to travel from A to C is the same time it takes the B car to travel from B to C. You'll need to use that. (Hmmm... it's a hint that you already included in your original post, so I guess it ain't such a big hint.)

Hint #2: Keep careful track of what's known and what's unknown. And label your variables.

Hint #3: Be consistent with your units. One of the first things that jumps out at me in this problem is that the speed is in km per HOUR, but the times are in MINUTES. Convert it all into the same time units from the very beginning. I'd suggest hours, so that 45 minutes = 3/4 hour, and 20 minutes = 1/3 hour.

Good luck!

Re: 2 Cars travelling from towns- challenging problem

Hello, bigidiotzz!

Quote:

Car-1 travels from town A to town B.

At the same time, Car-2 travels from town B to town A.

They meet at town C.

Car-1 completed the journey from C to B in 45 minutes at a steady rate of 64 kph.

Car-2 completed the journey from C to A in 20 minutes.

How fast was Car-2 going?

Code:

` Car-1`

: 64t : 48 :

A * - - - - > * - - - > * B

C

Car-1 travels at 64 kph.

In the first $\displaystyle t$ hours, it travels $\displaystyle 64t$ km to town C.

Then it takes 45 minutes $\displaystyle (\tfrac{3}{4}\text{ hour})$ to get to town A.

Hence: .$\displaystyle CB \:=\:64\left(\tfrac{3}{4}\right) \:=\: 48\text{ km.}$

Let $\displaystyle x$ = speed of Car-2.

Code:

` A * < - - - - * < - - - * B`

: x/3 C xt :

Car-2

In the same $\displaystyle t$ hours, Car-2 travels $\displaystyle xt$ km to town C.

Then it takes 20 minutes $\displaystyle (\tfrac{1}{3}\text{ hour})$ to get to town A.

Hence: .$\displaystyle CA \:=\:x\left(\tfrac{1}{3}\right) \:=\:\tfrac{x}{3}\text{ km.}$

We have: .$\displaystyle AC = CA \quad\Rightarrow\quad 64t \,=\,\frac{x}{3} \quad\Rightarrow\quad t \:=\:\frac{x}{192}\;\;[1]$

We have: .$\displaystyle BC = CB \quad\Rightarrow\quad xt \,=\,48 \quad\Rightarrow\quad t \:=\:\frac{48}{x} \;\;[2]$

Equate [1] and [2]: .$\displaystyle \frac{x}{192} \:=\:\frac{48}{x} \quad\Rightarrow\quad x^2 \:=\:9216$

$\displaystyle \text{Therefore: }\:x \;=\;96\text{ km/hr}$

Re: 2 Cars travelling from towns- challenging problem

Thank you so much! Finally I understand how this works.