A rocket is launched straight up with constant acceleration...?
The rocket's position as a function of time is given by: x(t) = 1/2 * a0 * t^2 + v0 * t + x0 Its velocity as a function of time is given by: v(t) = a0 * t + v0 We assume that the rocket starts from the ground (i.e. x0 = 0) and that the rocket has no initial velocity (i.e. v0 = 0). So we have: x(t) = 1/2 * a0 * t^2 v(t) = a0 * t where a0 is the rocket's acceleration. The bolt falls off after 4 seconds. We plug in t = 4 to find the bolt's position and velocity at this time: x(4) = 1/2 * a0 * (4)^2 = 8 * a0 v(4) = 4 * a0 Now we make a new "position" function for the bolt as a function of time: h(t) = 1/2 * g * t^2 + v0 * t + h0 where: g = -9.8 m/s^2 v0 = 4 * a0 h0 = 8 * a0 We are told that the bolt will hit the ground after 6 seconds: h(6) = 0 = 1/2 * -9.8 * 6^2 + (4 * a0) * 6 + 8 * a0 Now solve for a0: -9.8 * 18 + 32 * a0 = 0 32 * a0 = 176.4 a0 = 176.4 / 32 = 5.5 m/s^2 So the rocket's initial acceleration is 5.5 m/s^2. Hope that helps.
A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the?
Let a be the lift-off acceleration of the rocket. 4 seconds after lift-off, the altitude and the velocity of the rocket are h = (1/2)at² = (a/2)×4² = 8 a v = at = 4a Let's take the moment the bolt falls off the rocket as instant 0, ground level as zero-altitude, and upward direction as positive z-axis direction. The equation for the free-falling motion of the bolt is z = -(1/2)gt² + v₀t + z₀, in which g = 9.8, v₀ = 4a, z₀ = h = 8a z = -4.9 t² + 4at + 8a As t = 7.10 s, the bolt hits the ground: z(7.10) = 0. Hence, the equation determining the lift-off acceleration a is 0 = -4.9(7.1)² + 4a(7.1) + 8a 28.4a + 8a = 247 a = 247/(28.4 + 8) a = 6.79 m/s² Answer: a = 6.79 m/s²
A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bold falls off the?
The initial speed of the bolt is not 58.86 m/s. Let a be the acceleration of the rocket. During the 4 sec lift off, the rocket has reached a heigth of h = (1/2)*a*t^2 with t=4, h = (1/2)*a^16 h = 8*a Its velocity at 4 sec is v = t*a v = 4*a The initial velocity of the bolt is thus 4*a. During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion, h = (1/2)*g*t^2 + V0*t Subsituting h0=8*a, t=6 and V0=-4*a into it, 8*a = (1/2)*g*36 - 4*a*6 Solving fot a, a = 5.52 m/s^2
A rocket is launched straight up with constant acceleration...?
The rocket's place as a function of time is given by skill of: x(t) = a million/2 * a0 * t^2 + v0 * t + x0 Its speed as a function of time is given by skill of: v(t) = a0 * t + v0 We assume that the rocket starts from the floor (i.e. x0 = 0) and that the rocket has no preliminary speed (i.e. v0 = 0). So we've: x(t) = a million/2 * a0 * t^2 v(t) = a0 * t the place a0 is the rocket's acceleration. The bolt falls off after 4 seconds. We plug in t = 4 to discover the bolt's place and speed today: x(4) = a million/2 * a0 * (4)^2 = 8 * a0 v(4) = 4 * a0 Now we make a clean "place" function for the bolt as a function of time: h(t) = a million/2 * g * t^2 + v0 * t + h0 the place: g = -9.8 m/s^2 v0 = 4 * a0 h0 = 8 * a0 we are informed that the bolt will hit the floor after 6 seconds: h(6) = 0 = a million/2 * -9.8 * 6^2 + (4 * a0) * 6 + 8 * a0 Now resolve for a0: -9.8 * 18 + 32 * a0 = 0 32 * a0 = 176.4 a0 = 176.4 / 32 = 5.5 m/s^2 So the rocket's preliminary acceleration is 5.5 m/s^2. desire that facilitates.
What was the rocket's acceleration?
Let a be the acceleration of the rocket during take-off. 4 seconds after lift-off, the velocity and the height of the rocket (and the bolt) is v1 = at = 4a. h1 = (1/2)at^2 = (1/2)a.4^2 = 8a Take that moment as t = 0, the bolt starts its own motion with acceleration -g = -9.8 m/s^2, initial velocity v1 = 40, initial height h1 = 8a. The equation for its displacement is h = -(1/2)gt^2 + v1t + h1 h = (-1/2)*9.8 t^2 + 4at + 8a h = -4.9 t^2 + 4at + 8a When t = 7.10 s, h = 0: 0 = -4.9*(7.1)^2 + 4a*(7.1) + 8a (28.4 + 8)a = 247 a = 247/36.4 a = 6.79 m/s^2
A bolt falls off of a rocket...?
Rocket lifting off" v = u + at (u = 0; t = 4) v = 4a distance travelled by rocket in 4 seconds s = ut + 1/2 at^2 s = 8a Therefore: s = 2v the bolf falling off... s = ut + 1/2 at^2 (this time.. u = v of the rocket, t = 6.4s) -s = 6.4v - 4.9(6.4)^2 (s is negative coz it's falling) = 6.4v -200.704 so now you have 2 equations with 2 unknowns.. s = 2v -s = 6.4v - 200 Solve simultaneously and sub back in to find [a] the the rocket. I got a = 5.97m/s^2
How do i find the acceleration of the rocket?
The first thing you do, is draw a diagram. After that, you can see that as soon as the bolt falls off, it stops accelerating up as part of the rocket, and starts accelerating down do to gravity. But it already has some upward velocity that it picked up since the launch. So you start with the position formula for the bolt, p = -16 t * t + v t. (Think of throwing a ball off a roof! This is in feet per second) Then you note that v must be 4a, becasue the bolt accerated for 4 seconds when it was on the rocket. Substitute... 0 = -16 t * t + 4a t The trick is not to do too much. You don't care how high the rocket was when the bolt fell off. Good Luck!
If I were to get in a rocket and accelerate at a rate to maintain 1g - how fast would I be going in 1 hour?
I'll do this with two systems. g=9.807 meters per second per second. Assuming constant acceleration g. Velocity is acceleration times time. V=gt. Neglecting any other net forces on our body in question. 1 hour has 3600 seconds. 9.807 * 3600= 35,305.2 m/s or about 35,305.2 * 3600s/hr * 1km/1000m= 128,098.72 km/hr. Final speed. In US units. 32.17 feet per second per second is g. 32.17ft/(s*s) * 3600 s = 115,812 ft/s5,280 feet in a mile 115,812ft/s * 1 mi/5280ft * 3600s/hr = 10,849 mi/hr
What are the factors that determine a rocket's launch and escape velocity?
Escape velocityThe escape velocity is the velocity needed to fully escape the gravity of an object. That means that if you reach this speed, you will neither fall down on this object nor orbit it.It depends on the mass and radius of the main object (most of the time a planet or a moon, but can be anything). The formula to get it is:[math]v_e=\sqrt{\frac{2GM}{R}}[/math]Where [math]G[/math] is the gravitational constant ([math]6.67×10^{−11} m^3/kg/s[/math]), [math]M[/math] is the mass of the main object and [math]R[/math] the radius of this object.Speed of rocketsThe speed needed by a rocket depends on where you want to go. We’ll only talk about orbiting the Earth, because it’s what we do the most. To orbit an object at a given altitude, you need a velocity which depends on the mass of the mass object:[math]v_o=\sqrt{\frac{GM}{a}}[/math]Where [math]a[/math] is the semi-major axis (the maximum radius of the elliptical orbit).Delta-vThe delta-v is the real speed a rocket must “produce” to go somewhere. For an orbit, it depends on the orbital speed seen above and on the atmospheric drag. Indeed, the atmosphere produces drag which decelerate the rocket. The rocket has to compensate this deceleration.A map of the delta-v needed for the objects of the Solar SystemThrustTo reach the needed velocity, you’ll need thrust. It’s the force produced by the engine(s) of your rocket. Without air drag, the formula to know what acceleration your thrust will give you is:[math]\overrightarrow{a}=\frac{F}{m}[/math]Where [math]F[/math] is the thrust and [math]m[/math] the mass of the rocket.Now, you know how to go to space!
Is centrifugation at 15,000 g comparable to a rocket who at launch will pull 20g?
Oh, interesting. When I read your question, I assumed you meant the rocket was pulling 20 g's because it was following a curved path (the same way an aircraft "pulls g's" when the pilot pulls back on the stick to pull the nose up so it follows a curved path, as in looping the loop. In reading some of the other replies, I now suspect you meant the rocket was accelerating at 20 g's along a straight line. My problem is that I know that rockets don't accelerate that fast in a straight line. They only accelerate at a few g's - like 2 to 3 g's. Maybe 4 g's. Some missile may well accelerate much faster, but to me, a spacecraft mechanical engineer, a rocket is something we use to put our spacecraft into space, and so I naturally assumed that kind of rocket. So the only way I would expect it to pull 20 g's would be to follow a curved path. Silly me! When I assumed it was due to curving the path of the rocket, I was going to say that is the exact same thing as curving the path of the bacteria in the centrifuge except for the much larger acceleration level in the centrifuge. But now I'll treat the question as though it's linear acceleration for the rocket and rotational (centripetal or centrifugal) for the centrifuge. The change of velocity comes about differently in the two cases, but there is no difference in the kind of acceleration. Acceleration is just acceleration. They are both measured in the same units. When you define the acceleration vector to be the time derivative of the velocity vector, that derivative works equally well for both linear acceleration and acceleration due to curving the velocity vector around. So, except for the rotational motion on top of it, the acceleration itself is identical. You can't tell the difference. A bacterium in a rocket accelerating in a straight line at 10,000 g would feel the same as being in a centrifuge at 10,000 g's. Or slamming into a wall and having the container smush so that the "deceleration" was 10,000 g's. That really just acceleration too. It would feel the same to the bacterium.