Ok, so I just started physics and am at a complete loss of the way to do anything in the course.
The HW is online, which is alright but our class is really short and don't have much time with the instructor, so I was wondering if any of you could help me wit this problem.
You are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 41.0 km/h for 41.0 km. What would be the least speed (in km/h) needed for the rest of the trip to arrive in time for the interview?
This really isn't Physics, it's more of an Algebra problem, or a min Calculus problem, depending on how you're looking at it, but I digress:
Total distance 300 km
Leave at 8 AM
First hour 100 km 8-9 AM
Second hour 41 km 9-10
At this point you've gone 141 km / 300 km, just under half
We could set up an equation, but a little "bounds" will give you a smaller area to work with.
If you traveled the last hour at 100 km you'd be at 241 km / 300 km at 11 AM. No good.
If you traveled 150 km the last hour you'd be at 291 at 11 AM. In the subsequent 15 minutes you;d go 37.5 km giving you time to get there and find a parking spot, but this rough estimate probably won't be accepted, so off to an equation we go:
300-141 = 159 km/h, but this would put us exactly 300 km later at 11 rather then 11:15
So, we take 159 (which is what we have left to drive in 1.25 hours):
159-x-(1/4)x = ??
-159 = (-5/4)x
x = 127.2 km/h
This puts you at the spot 300 km away exactly at 11:15 AM. Doesn't account for you needing to park, get inside, or any real life thing(s)
This post has been edited by KYA: 05 February 2010 - 08:01 PM