If it starts from rest, moves with constant acceleration, and becomes airborne in time 7.10 s, what is its speed when it takes off?
2007-09-14 12:07:10 · 5 answers · asked by bg 1 in Science & Mathematics ➔ Physics
d = ut + (0.5)at^2
d = 266m
u = 0
t = 7.1 s
a = unknown
266 = 0.5*a*7.1*7.1
a = (266)/(0.5*7.1*7.1)
= 10.55 m/s^2
v = u + at
v = unknown
u = 0
a = 10.55 m/s^2
t = 7.1
v = 10.55*7.1 = 74.905m/s
2007-09-14 12:10:37 · answer #1 · answered by Savvy 2 · 3⤊ 1⤋
v = 2 * d / t (formulas of linear movement)
v = 2 * 266 / 7.1 = 74.92m/s or 269.74km/h
Review and study your formulas!
d = v * t / 2 = a * t² / 2 = v² / (2 * a) = distance in m
v = sqr(2*a*d) = 2*d / t = a t = velocity in m/s
t = sqr(2d/a) = v/a = 2d/v = time is seconds
a = vt = 2d/t² = v²/2d = acceleration in m/s²
(or learn how to transform and manipulate equations!)
2007-09-14 20:22:49 · answer #2 · answered by just "JR" 7 · 0⤊ 0⤋
180 MPH.
Signed Andy the cheater.
2007-09-15 10:23:33 · answer #3 · answered by andyman 3 · 0⤊ 0⤋
180 mph
2007-09-14 12:10:18 · answer #4 · answered by Radiator 4 · 0⤊ 1⤋
around about 180mph to 200 mph depending on its load and depending on if its heading into the wind
2007-09-14 12:24:50 · answer #5 · answered by Anonymous · 0⤊ 1⤋