Jan and Mary

[Airs]

Jan can run at 7.5m/s and Mary at 8m/s. On a race track Jan is given a 25m head start, and the race ends in a tie. How long is the track?

I have attempted to come up with an equation, but have failed. Horribly. I tried 8x = 7.5(x-25) but it doesn't work out. Can someone explain to me what the equation would be and why?

 

[galactus]

You can set this up different ways by manipulating the d=rt formula.

You know the rates are \(\displaystyle R_{j}=\frac{15}{2}\) and \(\displaystyle R_{m}=8\)

d/t=r

It takes Jan 25/(15/2)=10/3 seconds to run the 25 m.

\(\displaystyle \L\\\frac{d}{t+\frac{10}{3}}=\frac{15}{2}\)

\(\displaystyle \L\\\frac{d}{t}=8\)

You now have two equations with two unknowns. Solve. OK?.

 

[Airs]

Thank you so much.

 

[Denis]

Something simple like this helps:

(Mary)@8...................x.............................>
.....25.....(Jan)@15/2..............x-25............>

since time = distance / speed (and time is the same for both):
x / 8 = (x - 25) / (15/2)
x / 8 = (2x - 50) / 15
15x = 8(2x - 50)
Wrap it up!

 

[soroban]

Hello, Airs!

As Galactus pointed out, there are many ways to solve this one.
Here's a back-door approach . . .

Jan can run at 7.5 m/s and Mary at 8 m/s.
On a race track Jan is given a 25m head start, and the race ends in a tie.
How long is the track?

Jan is already 25m ahead of Mary. .Mary must play catch-up.

Note that Mary runs \(\displaystyle 8\,-\,7.5\:=\:0.5\) m/s faster than Jan.
It is as if Jan is standing still and Mary runs to her at 0.5 m/s.

At 0.5 m/s, how long will take Mary to run 25m ?