A swan on a lake gets airbourne by flapping its wings and running on top of water. (a) if the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an avg. rate of 0.350m/s^2, how far will it travel before coming airbourne? (b) How long does it take?
2006-09-20 10:13:20 · 4 answers · asked by cheertee1126 1 in Science & Mathematics ➔ Physics
Initial velocity is 0
final velcity 6
acceleration 0.350
distance ?
v=u+at
6=0+0.350t
t=6/0.350
=17.14s
s=ut+0.5at^2
s=0+0.5x0.350x17.14^2
(where s is distace) hav go rest
2006-09-20 10:22:03 · answer #1 · answered by Morfydd E 1 · 0⤊ 0⤋
The time is easy. After 1 second it is going .35, after two it is going .7 and so on. After 6/.35=17.14 seconds it is going 6. For distance, use the formula X=1/2AT^2, plug in .35 for A and 17.14 for T.
2006-09-20 17:17:47 · answer #2 · answered by campbelp2002 7 · 0⤊ 0⤋
Use these formulas:
v^2=u^2+2as
v=u+at
The numbers you want to plug in to the formulas are:
v=6
a=0.35
u=0 (because the swan starts from rest)
2006-09-20 17:23:11 · answer #3 · answered by ndavos 2 · 0⤊ 0⤋
before the lake run out,depends on how big the lake is
2006-09-20 17:26:23 · answer #4 · answered by Anonymous · 0⤊ 0⤋