Advanced Algebra

[cheecette]

I can't remember how to do problems where two people are doing the same thing at different speeds and you have to figure out how long it would take them to do it together.
I.e. Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

What is the basic formula set up for these kinds of problems?

 

[Deleted member 4993]

I can't remember how to do problems where two people are doing the same thing at different speeds and you have to figure out how long it would take them to do it together.
I.e. Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

What is the basic formula set up for these kinds of problems?

There is no "standard" equation for these - you have to develope everytime you do these!

Start naming variables.

The question is how much time will pass.. ? let that time = t hours

First biker is travelling at 6 mph - he would have travelled for (t+3) hours.

Distance travelled by first cyclist = speed * time ..............................................(1)

Second biker is travelling at 10 mph - he would have travelled for (t) hours.

Distance travelled by first cyclist = speed * time ..............................................(2)

Equate the distances fro (1) and (2) - and solve for 't'.

 

[soroban]

Hello, cheecette!

There is a back-door approach to these two-vehicle problems.

Two cyclists start biking from a trail's start 3 hours apart.
The second cyclist travels at 10 mph and starts 3 hours after the first cyclist who travels at 6 mph.
How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

. . (A) 2 hours . . (B) 4½ hours . . (C) 5¾ hours . . (D) 6 hours . . (E)7 ½ hours

The first cyclist \(\displaystyle A\) travels at 6 mph and has a 3-hour headstart.
. . He is 18 miles ahead of \(\displaystyle B.\)


Now \(\displaystyle B\) must make up the 18-mile differece.

\(\displaystyle B\) travels: .\(\displaystyle 10 - 6 \,=\,4\) mph faster than \(\displaystyle A.\)

\(\displaystyle \text{It will take him: }\;\frac{18}{4} \:=\:4\tfrac{1}{2}\text{ hours . . . answer (B)}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Why does this work?

The difference of their speed is 4 mph.

It is as if \(\displaystyle A\) is 18 miles ahead and then stops,
. . then \(\displaystyle B\) approaches him at 4 mph.

Get it?