rates question

[sudi760mba]

Nicky and Cristina are running a 1000 meter race. Since Cristina is faster than Nicky, she gives a 12 second head start. If Cristina runs at a pace of 5 meters per second and Nicky runs at a pace of only 3 meters per second, how many seconds will Nicky have run before Cristina catches up to him?

This is a chase problem where people are moving in the same direction.

When setting up the equations:

Christina: 5 m/s * t
Nicky: 3 m/s * (t+12)

If Nicky is slower and gets the headstart why would we list Nicky to have t + 12? wouldn't be the other way around where Nicky gets 3t and Christina would be 12 seconds after as 5(t+12)?

Thanks.

[scoobydooby]

sudi,
headstart means Nicky starts 12 secs before Christina. Christina being faster matches to catch up with Nicky.
so Christina actually takes 12 secs less than Nicky (starts 12 secs after Nicky starts)
so Christina:t, Nicky: t+12 (12 secs more)

you could also take Nickey:t, Christina: t-12 in the equation

[x2suresh]

you can use relative velocity.

(5-3)t = 3*12

t= 18 sec

[deepoe]

isn't it :

5 - 3 = 2m/s

So 1000 / 2 = 500 seconds

500 + 12 = 512 so after 512 seconds Cristina catches him up?

[Mr2Bits]

you can use relative velocity.

(5-3)t = 3*12

t= 18 sec

I get 18 seconds as well. Think about it logistically.

Nicky starts 36 meters ahead of Christina.

After 18 seconds, Christina has run (5*18) 90 meters and Nicky has run 3*18+36(head start) = 54 + 36 = 90 marking the overtake.

18 seconds marks the meeting and overtake

[deepoe]

ah ok:D thanks!

[deepak_free]

We are looking for seconds which Nicky runs before Cristina catches up.
The answer should be 18+12=30 Secs