A small plane begins its descent. If the plane has an altitude of 4300 feet after 2 minutes and an altitude of 3100 feet after 5 minutes, what is the planes's constant rate of descent? what is the answer
2006-10-31 08:17:28 · 11 answers · asked by Felicia F 1 in Science & Mathematics ➔ Mathematics
Rate of descent = (4300 - 3100)/(5 - 2)
= 1200/3
= 400 ft/mn
2006-10-31 08:21:32 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
the difference in altitude = 4300 - 3100 feet = 1200 feet.
the elapsed time = 5 - 2 = 3 minutes.
therefore the rate of descent is:
1200 feet/3 minutes = 400 feet per minute.
2006-10-31 16:21:34 · answer #2 · answered by jonathon.shine@rogers.com 2 · 0⤊ 0⤋
4300 - 3100 = 1200
5 - 2 = 3
1200 / 3 = 240
(That's the difference between the 2 heights and that's divided by 3)
The constant rate of decent is...
400 Feet / Minute
2006-10-31 16:27:15 · answer #3 · answered by Anonymous · 0⤊ 0⤋
You have two points: (2, 4300) and (5, 3100)
The constant rate is (y2-y1)/(x2-x1) that is (3100-4300)/(5-2)=
-400... the negative signal means it is going down.
Answer the plane is going down in the rate of 400feet/min
2006-10-31 16:28:35 · answer #4 · answered by vahucel 6 · 0⤊ 0⤋
400' per minute as you said after 5 mins is that from beginning or since at 4300 ?..if from beginning it is 400' per minute. If from 4300 and it is at 5 minutes from there it would be 240' per minute
2006-10-31 16:23:53 · answer #5 · answered by colinhughes333 3 · 0⤊ 1⤋
400 Feet / Minute
2006-10-31 17:18:19 · answer #6 · answered by CJ 3 · 0⤊ 0⤋
400 feet per minute?
2006-10-31 16:19:37 · answer #7 · answered by lcritter55118 4 · 0⤊ 0⤋
u=(x1-x2)/(t1-t2)=(4300-3100)/(5-2) = 400 ft/min
2006-10-31 16:23:26 · answer #8 · answered by Dimitri C 1 · 0⤊ 0⤋
(4300-3100)ft/(5-2)min=1200ft/3min=400ft/min
2006-10-31 16:22:30 · answer #9 · answered by yupchagee 7 · 0⤊ 0⤋
1300/3 ft/min
2006-10-31 16:20:22 · answer #10 · answered by Anonymous · 0⤊ 1⤋