## anonymous 5 years ago Canoeing. during the first part of a canoe trip Ken covered 60km at a certain speed.He then traveled 24 km at a speed that was 4km/h slower.If the total time for the trip was 8 hrs.,what was the speed on each part of the trip?

1. anonymous

Set up your equations. Let v be the speed of the first trip, t_1 the time taken in the first trip and t_2 the second time. Using the definition,$speed=\frac{distance}{time}$then$distance=(speed)(time)$For the first leg$60=vt_1$and the second$24=(v-4)t_2$We know also that $t_1+t_2=8$You have three equations in three unknowns which you can now solve.

2. anonymous

$t_2=8-t_1$which you sub. into the equation 24=(v-4)... to get$24=(v-4)(8-t)$where I've dropped the subscript on t_1. So$t=\frac{56-8v}{4-v}$

3. anonymous

Substitute this into the first equation,$60=v(\frac{56-8v}{4-v}) \rightarrow 240-60v=56v-8v^2$

4. anonymous

This is quadratic in v$8v^2-116v+240=0$

5. anonymous

which can be factored as$4(v-12)(2v-5)=0\rightarrow v=12,\frac{5}{2}$

6. anonymous

If v=5/2, t=24 hours, which is outside your 8 hour limit. The other solution, v=12km/h yields t=5 hrs. So the first part of the trip took 5 hours, the second part therefore took the remaining 3 hrs.

7. anonymous

Damn, I answered the wrong question.

Hey it looks good to me, I liked the step by step explanation. I am sure Anointed will appreciate it.

9. myininaya

I like too

Agree, I learned something.

11. anonymous

Cheers

12. anonymous

Looks good to me.These kind of equations are a little complicated but you made it look so easy! Thanks!

13. anonymous

You're welcome :)