need help with a math question....

General discussion forum about the 240sx, and a great place to introduce yourself to the board!
ni*2*ss*4*an*0*
Posts: 19
Joined: Mon May 31, 2004 10:25 am
Contact:

Post

two cars are racing on a track that is 1.95 miles long....car A finishes at a time of 1.28.974 and car B comes in at 1.28.997 how far in feet is car A ahead of car B at the finish line?:help


User avatar
Ajax
Posts: 1643
Joined: Wed Mar 12, 2003 3:16 pm
Car: 1991 240sx SE
2010 Mazda 5

Post

2.661 ft approximately.Good God I am a nerd.

ni*2*ss*4*an*0*
Posts: 19
Joined: Mon May 31, 2004 10:25 am
Contact:

Post

thanks for the fast answer god i must be dumb cause ive been trying to figure that out for a whole day

User avatar
Ajax
Posts: 1643
Joined: Wed Mar 12, 2003 3:16 pm
Car: 1991 240sx SE
2010 Mazda 5

Post

I work for a company that deals with plats of land, so I do a lot of math work. Of course this problem was much more fun, yep nerd it is for me then.

youngturk
Posts: 159
Joined: Wed Apr 28, 2004 6:28 pm

Post

Hey ni*2*ss*4*an*0*,

You need to know that you don't have all the data you need to come up with an exact answer. Ajax's answer makes some assumptions. I made my own calculation and came up with 2.661 feet as well (edit: my previous calculation of 3.0597 feet was incorrect).

The biggest assumption I made is that the two cars accelerated at a similar rate all the way up to the finish line. Since the respective finish times are pretty close and the track is not too long, it is okay to make such an assumption. However, if the finish times were off by a few seconds and the track was much longer, you don't have enough data provided to find the correct answer.

IlIkEmYz
Posts: 662
Joined: Fri Dec 05, 2003 6:40 pm
Car: 06 Z

Post

could someone explain the problem and show some work, im kind of rusty in math right now, but i just want to know how you guys got the answer.

User avatar
Ajax
Posts: 1643
Joined: Wed Mar 12, 2003 3:16 pm
Car: 1991 240sx SE
2010 Mazda 5

Post

That's true, I made an assumption of averaging the speed, which is realy about the only thing you can do given these criteria. Considering the difference in time is 23/1000ths of a second, the difference in acceleration should be negligible.Here's what I did:transfer miles to feet (5280 per mile). transfer time to decimal minutes (28.974 sec = .4829 mins).solve for average speed of slower car and multiply by faster cars average speed.In any case, the difference between my answer and youngturk's is less than 5 inches- that's pretty close considering it's almost a two mile track.

User avatar
Shift_Oversteer
Posts: 990
Joined: Mon Nov 10, 2003 3:03 am
Car: s13 convertible
Contact:

Post

Haha Im impressed guys.... Math is like an alien language to me!

User avatar
hannibal
Posts: 9680
Joined: Wed Sep 11, 2002 2:38 am
Car: Red Line to Glenmont
Location: Washington DC

Post

Ummm, the correct solution to that problem cant be determined from the information given.Is this a real problem (in a book)? Theres gotta be some more details.

User avatar
Ajax
Posts: 1643
Joined: Wed Mar 12, 2003 3:16 pm
Car: 1991 240sx SE
2010 Mazda 5

Post

From a truth standpoint, no it cannot be solved precisely, but from a real world standpoint, saying 2 1/2 to 3 feet should be good enough over a two mile course (that's between .02584 and .02913% of the course!). Pretty negligible.Hey, it could be a perfect circle track after one warm-up lap in whcih both cars get up to their top speed, and they both got tracked the exact distance around at that same top speed. Then I would be precise.

youngturk
Posts: 159
Joined: Wed Apr 28, 2004 6:28 pm

Post

Car A reached the finish line in 1 minute and 28.974 seconds on a 1.95 mile track.28.974 seconds is equal to 28.974 / 60 minutes = 0.4829 minutes.So Car A’s total finish time was 1.4829 minutes.

Avg. speed of Car A was 60 mins/ 1 hr x 1.95 miles / 1.4829 minutes= 78.8995 mph

78.8995 mph = 78.8995 mile / 1 hour x 5,280 feet / 1 mile x 1 hour / 60 minutes x 1 minute / 60 seconds78.8995 mph = 115.72 feet per second

Time difference between Car A and Car B = 28.997 – 28.974 seconds = 0.023 seconds

Assuming both cars were going the same speed when Car A crossed the finish line AND Car B’s acceleration was not significantly more than Car A’s, then the distance between the two cars was:

115.72 feet / sec x 0.023 seconds = 2.661 feet

If Ajax’s scenarios were the case, he would be correct IMO, since they would’ve driven the whole timed lap at their respective average speeds.

User avatar
Ajax
Posts: 1643
Joined: Wed Mar 12, 2003 3:16 pm
Car: 1991 240sx SE
2010 Mazda 5

Post

Bingo, but who cares about circle track racing!

IlIkEmYz
Posts: 662
Joined: Fri Dec 05, 2003 6:40 pm
Car: 06 Z

Post

thanks youngturk i so get it now, this reminds me of physics problems

ni*2*ss*4*an*0*
Posts: 19
Joined: Mon May 31, 2004 10:25 am
Contact:

Post

thank you guys for all the answers and problem solving...no the question was not out of a book its from my real life situation..the track record at the race track that i race at for my class (blackhawkfarmsraceway.com) is 1.28.974 held by a turbo miata. my time last weekend was 1.28.997..ive been trying to beat this time for two years and im getting closer..well i was talking to guy at the track about how close i was, and i said it had to be few feet..he said it was probably about 20 feet...so i wanted to see who was closer

youngturk
Posts: 159
Joined: Wed Apr 28, 2004 6:28 pm

Post

ni*2*ss*4*an*0*: You are so close, it is just a matter of time until you beat his record. Good luck.


Return to “240sx General Discussion”