The problem is-
A car leaves San Fransisco heading towards Denver going at the constant speed of 50 miles per hour. Another car leaves Denver heading towards San Fransisco an hour later, but going at a constant speed of 60 miles per hour. When the two cars meet, which car is closer to Denver?
Its a trick question in real life, but if you wanted to solve it, could you? If so, how?
2007-06-30 11:03:32 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
And by the way, you have to assume that they drive in a straight line, and that the problem doesn't give you the distance between the two cities.
2007-06-30 11:04:26 · update #1
When the two cars meet, they are at the same location. Therefore, neither car is closer to Dever than the other.
2007-06-30 11:10:51 · answer #1 · answered by sahsjing 7 · 1⤊ 0⤋
Car A leaves SF at 50 mph
Car B leaves DEN 1 hr later at 60 mph
Don't you need to know the distance between SF and DEN?
Visualize the problem this way-
in 2 hrs A has traveled 100 mi and B has traveled 60 miles
When their combined distance traveled equals the distance between SF and DEN they meet.
Distance A traveled = t x 50
Distance B traveled = (t-1) x 60
t x 50 + (t-1) x 60 = distance between SF and DEN
50t+60t - 60 = distance between SF and DEN
Solve for t (time A traveled; B traveled 1 hr less)
Compute the distance from DEN of A (at t) and B (at t-1).
The trick part of the question is how you define "meet". If it is front bumper to front bumper, then the car from DEN will be closest to DEN. If "meet" means they are side by side, then it still is a trick question because they would both be the same distance from DEN.
2007-06-30 11:21:48 · answer #2 · answered by skipper 7 · 0⤊ 0⤋
If you wanted to solve this, even though the solution is fairly obvious, you would only need to use algebra, calculus is unnecessary.
Since they are meeting, they are at the same position, thus are the same distance from Denver. One car may have traveled further from its original location than the other, but they are in the same position when they meet.
2007-06-30 11:15:44 · answer #3 · answered by msi_cord 7 · 0⤊ 0⤋
You don't need calculus for this one. You don't even need any arithmetic, if you happen to realize that the cars will be in the same place when they meet, hence, equidistant from Denver.
If, on the other hand, the cars were travelling at known but nonconstant speed and you wanted to know when or where they would meet, you might need calculus for that.
2007-06-30 11:38:15 · answer #4 · answered by devilsadvocate1728 6 · 0⤊ 0⤋
The one that comes from Denver
(That car is a car's length closer to Denver)
2007-06-30 11:11:57 · answer #5 · answered by Ernst S 5 · 0⤊ 0⤋
You couldn't use calculus - and there wouldn't be any way to solve it without knowing the distance between the cities.
2007-06-30 11:19:33 · answer #6 · answered by opalraindrops 2 · 0⤊ 0⤋