how many "g's" (a unit of acceleration equal to the acceleration of gravity i.e 9.8 m/s^2) can a rocket produce if it accelerates horizontally from 0 to 1600 km/h in 1.8 seconds? what distance will it cover during that time?
2007-03-09 20:26:35 · 5 answers · asked by princess 1 in Science & Mathematics ➔ Physics
v = u + a.t
v = 1600 km/h = 1600X1000 / 3600 m/s = 444.444...m/s
a = v/t since u = 0
a = 444.444/1.8 = 246.9 m/sec^2
This is nearly 25 g (25.19..)
s = u.t + 1/2 a.t^2
s = 0 + 246.9/2 X 1.8 X 1.8 = 400 m approx.
2007-03-09 20:54:50 · answer #1 · answered by Swamy 7 · 0⤊ 0⤋
1600 km/hr * 1 hr / 3600 s * 1000 m / km = 4000/9 m/s
4000/9 m/s / 1.8 = 247 m/s/s
247 / 9.81 = 25.17 G
x(t) = 1/2 a t^2 + v0 t + x0
v0 = 0, x0 = 0
x(1.8) = 400 m
2007-03-10 04:39:14 · answer #2 · answered by feanor 7 · 0⤊ 0⤋
the initial velocity =0 m/s
final velocity=1600 kmph=444.444 m/s
difference in velocity=444.444m/s
time=1.8s
acceleraton=444.444/1.8=246.9135m/s^2=approx 247
resultant acceleration=247-9.8
=237.2 m/s^2
using the formula
s=1/2*a*t^2
we get s as 384.264 m
2007-03-10 05:31:22 · answer #3 · answered by satwik 2 · 0⤊ 0⤋
Convert the velocity to meters per second. You can then get the acceleration in meters per second per second, and divide by 9.8 to get gravities.
2007-03-10 04:36:09 · answer #4 · answered by Anonymous · 0⤊ 1⤋
speed (in m/sec) = (16,000,000 m) / (3600 sec) = 444.44... m/sec.
acceleration (a) = speed/time = (444.44 m/sec)/(1.8 sec) = 246.914 m/sec^2
No. g's = 249.914/9.8 = 25.195
distance = a t^2 / 2 = (246.914)(1.8)(1.8)/2 = 400 m.
2007-03-10 04:46:54 · answer #5 · answered by pollux 4 · 0⤊ 0⤋