Assuming that the human body can survive a maximum continuous acceleration of 6.36 m/s2, calculate the minimum time in which a space ship can accelerate from 0 m/s to 3.0 ×107 m/s, that is, to a velocity that is 10% the velocity of light.
I have no idea how to even start this problem. I know im required to use one of the four basic formulas but i am unsure of which and how to go about doing it.. Thanks
2007-08-31 06:54:00 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Physics
use simple formula
v=u+at
v=final velocity
a=acceleration
t=time
u=initial velocity=0
3*10^7=0+6.36*t
or t=3*10^7/6.36
it is 4716981.132 seconds
or 1310.27 hours
or 54.6 days
2007-09-01 08:15:16 · answer #1 · answered by rahul v 2 · 1⤊ 0⤋
You just divide the velocity by the acceleration. (m/s)/(m/s^2) = s. Note that at this velocity you can ignore relativistic effects with insignificant error.
The answer is 3E7/6.36 = 4716981 s.
EDIT: Alexander is right. The shuttle crew is subjected to a maximum of 3g (~29.4 m/s^2) but only for a few minutes. Probably 20-30% above 9.8 m/s2 would be OK for a long-duration flight. You can then reduce the flight time accordingly.
2007-08-31 06:59:34 · answer #2 · answered by kirchwey 7 · 0⤊ 0⤋
You would use V = Vo + at
V is 3.0 x 10^7
Vo is 0
and a is 6.36 m/s^2
Plug these in and solve for t and you get 4716981 seconds
Which is also 54 days 14 hours 16 minutes and 21 seconds
2007-08-31 07:02:19 · answer #3 · answered by Matt C 3 · 0⤊ 0⤋
v = u + at. 3 x 10^7 = 0 + 6.36t
t = 3 x 10^ 7 / 6.36
2007-08-31 07:01:54 · answer #4 · answered by Pandian p.c. 3 · 0⤊ 0⤋
OK, every second you can go 6.36 m/s faster. So, at the end of the first second, you're going 6.36 m/s, at the end of the second second, you're going 12.72 m/s, at the end of the third second, you're going 19.08 m/s, and so on.
You need to get to 3.0 x 10 ^7 m/s. So,
V = A * T
Or,
T = V/A
So, divide 3.0 x 10^7 by 6.36. To three significant figures, it's 4.72 x 10^6, or 4,720,000 seconds. That's about 54.6 days.
2007-08-31 07:03:27 · answer #5 · answered by El Jefe 7 · 0⤊ 0⤋
v(t) = v(0) + at
assume v(0) = 0
and v(t) = 3(10^7) m/s
and acceleration, a =6.36 m/s^2
then the time t is
t = (v(t) - v(0))/a
= 3(10^7)/6.36
= 4716981.132075472 sec
= 78616.35220125786 min
= 1310.272536687631 hours
= 54.594 days too long !!!
2007-08-31 07:05:41 · answer #6 · answered by vlee1225 6 · 0⤊ 0⤋
I'll only bother to offer an informal proof.... It's possible to show, that in terms of gravitational potential, a solid sphere behaves the same as a point mass, at least above the surface of said sphere. Obviously, a point mass will maximize the acceleration due to gravity of any given mass configuration. Furthermore, the argument of minimization of gravitational potential energy is a valid one here. It can also be shown that, if an object moves from a location of higher gravitational potential, to a location of lower potential, the acceleration due to gravity MUST INCREASE for that object. This can be shown to be valid for any arbitrary mass configuration. The shape that minimizes gravitation potential is, both from standpoint of simple observation, and mathematics, is a solid sphere of course. Therefore, if a sphere minimizes gravitational potential, and if acceleration due to gravity always increases as potential decreases, then a sphere must MAXIMIZE acceleration due to gravity. Q.et D. ~WOMBAT
2016-05-17 23:21:47 · answer #7 · answered by ? 3 · 0⤊ 0⤋
Simple.
Divide desired velocity by accleration.
The provided value of maximum continuous acceleration 6.36 m/s² is misleading, because all of us live in greater gravitatation field g =9.8 m/s² for our entire lives.
2007-08-31 07:02:38 · answer #8 · answered by Alexander 6 · 0⤊ 0⤋
Ummm..42?
2007-08-31 07:01:02 · answer #9 · answered by lazyjbob 5 · 0⤊ 1⤋
Pandian and Matt are correct.
2007-08-31 07:04:27 · answer #10 · answered by Anonymous · 0⤊ 0⤋