Thread: Distance Speed and time with 2 variables?

1. Distance Speed and time with 2 variables?

this one is sort of long...

Maurice drove 400 km from Edmonton to Battleford in one hour less time than it took Martin to drive the same route from Battleford to Edmonton. If Maurice drove 20km/h faster than Martin, at what speed did each of them drive? Show a complete algebraic solution.

So far, all I have is...
Martin
Distance - 400 km
Speed - s
Time - t

Maurice
Distance - 400 km
Speed - s + 20
Time - t - 1

Not getting anywhere with this, though.

2. Originally Posted by xxlvh
this one is sort of long...

Maurice drove 400 km from Edmonton to Battleford in one hour less time than it took Martin to drive the same route from Battleford to Edmonton. If Maurice drove 20km/h faster than Martin, at what speed did each of them drive? Show a complete algebraic solution.

So far, all I have is...
Martin
Distance - 400 km
Speed - s
Time - t

Maurice
Distance - 400 km
Speed - s + 20
Time - t - 1

Not getting anywhere with this, though.
the data you have so far is fine.

So, $s$ is Martin's speed. so we need to solve for it. then we can find Maurice's speed as well.

you should recall that: $\mbox{Speed} = \frac {\mbox{Distance}}{\mbox{Time}}$

solving for time, we get:

$\mbox{Time} = \frac {\mbox{Distance}}{\mbox{Speed}}$

So, the time Martin takes is given by: $t = \frac {400}s$

and the time Maurice takes is given by: $t - 1 = \frac {400}{s + 20} \implies t = \frac {400}{s + 20} + 1$

equate the $t$'s:

$\Rightarrow \frac {400}s = \frac {400}{s + 20} + 1$

now solve for $s$ and continue

3. So multiplying it all by s(s+20) leaves you with...
400s + 8000 = 400s + s2 + 20s
s2 + 20s - 7980 = 0
s = 79.9 and -99.9 so i'm assuming only take 79.9 as the final solution?

4. Originally Posted by xxlvh
So multiplying it all by s(s+20) leaves you with...
400s + 8000 = 400s + s2 + 20s
s2 + 20s - 7980 = 0
s = 79.9 and -99.9 so i'm assuming only take 79.9 as the final solution?
yes. but it's 80