# Distance Problem

1. Feb 22, 2013

### Equilibrium

1. The problem statement, all variables and given/known data
This problem has already been solved but i got a quite few clarifications:
http://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.493396.html

"a boy on his bicycle intends to arrive at a certain time to a town that is 30 km away from his home .after riding 10 km, he rested for half an hour and as a result he was obliged to ride the rest of the trip 2km/hr faster ."
2. Relevant equations
v = d/t
t = d/v

3. The attempt at a solution
Why did he disregard the 10km?
i think the formula should've of look like this:
$$\frac{30}{s}=\frac{10}{s}+0.5+\frac{20}{s+2}$$
(total time riding on original speed for the whole trip) = (time riding on original speed (for 10km))+(time rested)+(faster velocity on the rest (for 20km))

the formula he used:
$$\frac{30}{s}=\frac{30}{s+2}+0.5$$

Last edited: Feb 22, 2013
2. Feb 22, 2013

### Simon Bridge

He has to go 20km in the time remaining after travelling 10km at the slow speed, and waiting half an hour.

The time remaining is $20/s = 0.5 + 20/(s+2)$ ... agreeing with your formulation.

$40(s+2) = (s+2)s + 40s$

rearranging:
$s^2+2s-80=0$

$s \in \{ 10,-8 \}$

possible answers are 10hr and -8hr ... pick the positive one.

So your question is, "how did he know that his version would be correct?"
Try plotting the velocity-time graph.

Last edited: Feb 22, 2013
3. Feb 22, 2013

### nasu

The solutions to the quadratic equation are +8 and -10.
The solution for the problem is then 8km/h.

In the link given he is solving a different problem. I suppose he did not read the problem carefully.

4. Feb 22, 2013

### Simon Bridge

Did I set up the quadratic incorrectly ...

the discriminant is 324
so $s =\frac{1}{2}(-2\pm\sqrt{324}=1\pm 9$ ... Oh I see: I misplaced a minus sign!
<mumble mumble grzzl>
... time for bed!