At noon a private plane left Austin for Los Angeles, 2100 km away, flying at 500 km/h. One hour later a jet left Los Angeles for Austin at 700 km/h. At what time did they pass each other. Be sure to show your work!
*The person who answers with the most accuracy and detail gets 10 points! Good Luck.
2007-11-11 02:56:55 · 3 answers · asked by noel 1 in Education & Reference ➔ Trivia
2.15pm at 1575km to LA & 625km to Austin
Distance from Austin to LA:
1st plane:
1 pm = 700km
2 = 1400
3 = 2100
2nd plane:
2 pm (leaves an hr later so after 1 hr of travelling) = 1700km
3 = 1200
4 = 700
so they meet inbetween 2 and 3 pm
1st plane
at 2.30pm = 1750km
2nd plane
at 2:30pm = 1450km
so inbewteen 2pm and 2:30 pm
1st plane
at 2:15pm= 1575km
2nd plane
at 2:15pm= 1575km
ta da!!!!
:)
whoops i have jus realosed i have worked it out with plane 1 at 700km/h
and pane 2 at 500km/h
oh well u get the idea...
but if u do it the ther way around they meet at 2:20pm! :)
2007-11-11 03:19:10 · answer #1 · answered by PokeTheMantie™ 3 · 0⤊ 1⤋
If Austin is 0 and LA is 2100
At 1pm - plane 1 is at 500, plane 2 just leaving 2100
At 2pm, plane 1 ay 1,000, plane 2 at 1,400
At 2.20pm, plane 1 at 1,000 + 500/3 = 1,166.67
plane 2 at 1,400 - 700/3 = 1,400 - 233.33 = 1,166.67
******************
Or by algebra - 500t + 700 (t-1) = 2100
1200t = 2800
t = 28/12 = 2 1/3 = 2 hours 20 mins
2007-11-11 03:27:35 · answer #2 · answered by Beardo 7 · 0⤊ 0⤋
Obviously your not that smart you asked your question twice
2007-11-14 16:47:38 · answer #3 · answered by Patrick B 2 · 0⤊ 0⤋