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Rocket Acceleration? 
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#1
Jan1908, 05:31 AM

P: 79

1. The problem statement, all variables and given/known data
A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6 seconds later. What was the rocket's acceleration? 2. Relevant equations Vf=Vi + at y=Vi(t)+.5a(t^2) Vf^2=Vi^2+2(a)(y) 3. The attempt at a solution First problem of this type that I've seen in our homework questions. I'm wandering with what information I should begin with first. I know there is constant acceleration, the bolt falls off the rocket at t=4s after launch, and that t=6s the bolt hits the ground. What should I already notice about this problem? I know that there is something that I'm overlooking. 


#2
Jan1908, 07:26 AM

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Try working backwards. You know how long it takes the bolt to fall. Then you can calculate its initial altitude. Obviously, that is the "final" altitude of the rocket, whose initial altitude is zero. Then try to get the acceleration of the rocket.



#3
Jan1908, 08:26 AM

P: 79

I'm sorry, I'm still lost. Wouldn't I have to know the velocity or acceleration to find the altitude?



#4
Jan1908, 09:05 AM

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Rocket Acceleration?
You don't know the rocket's acceleration, so use a symbol for that  call it [itex]a[/itex]. Express the position and velocity of the rocket as a function of time [itex]t[/itex] and this unknown acceleration [itex]a[/itex]. You know the exact time at which the bolt shakes loose. What are the position and velocity of the bolt at this time?
The bolt accelerates downward due to gravity after shaking loose. What is the general expression for the height of a falling object given an initial height and velocity? You know the initial height and velocity as a function of [itex]a[/itex] and you know the time at which the height becomes zero. Solve for [itex]a[/itex]. 


#5
Jan1908, 09:17 AM

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1.) Write down a function for the ycoordinate [itex]y(t)[/itex] of the bolt as a function of time in the interval [itex]0\leq t\leq4[/itex]. Since it's traveling with the rocket, the function will be in terms of the acceleration [itex]a[/itex]. Using this function, determine the position and velocity of the bolt at [itex]t=4[/itex].
2.) Write down a function for the ycoordinate of the bolt as a function of time in the interval [itex]4\leq t\leq6[/itex]. Remember that now the bolt is in freefall, so what is its acceleration here? Use the fact that the position and velocity functions are continuous. 3.) You know that [itex]y(6)=0[/itex]. Use that fact to obtain the acceleration [itex]a[/itex] of the rocket. 


#6
Jan1908, 09:41 AM

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#7
Jan1908, 02:44 PM

P: 79

Still confused about putting it into time, we haven't worked on anything like thisputting into terms of time with accelerationthat I know of anyways.
I understand that the bolt is in free fall, a=9.8 and a=9.8 for the positive velocity. But the answer the book gives is a=5.5 m/s^2. I'm getting stuck thinking about the acceleration is constant, so I'm thinking that the upwards motion is also going to be, a=9.8 since the downward motion is a=9.8 m/s^2. So that is why I'm getting confused about the accelerationI'm trying to figure the distance from liftoff that the bolt fell off and then apply that to figure the acceleration and velocity of the shuttle to that same distance? Physics isn't my best subject I suppose??? 


#8
Jan1908, 03:06 PM

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P: 15,067

The upward acceleration is not 9.8 m/s^{2}. It is an unknown. You need to use different symbols for the upward acceleration experienced while the bolt is attached to the rocket and the downward acceleration experienced after the bolt shakes free from the rocket.
Dick gave a very good outline for attacking the problem. Please follow his steps. 


#9
Jan2008, 11:07 AM

P: 79

I have yet to solve this problem. For the upward acceleration the Vi will be 0m/s and vf will be some value and vice versa for the downward acceleration? So since there is constant acceleration on the upward and downward accelerations the velocity on the left side and right side will be equal, but with different accelerations.
>1.) Write down a function for the ycoordinate of the bolt as a function of time in the >interval . Since it's traveling with the rocket, the function will be in terms of the >acceleration . Using this function, determine the position and velocity of the bolt at . I'm not sure how to follow your steps Tom. I found Y by S=.5at. Velecity=29.4 and Y=176.4? 


#10
Jan2008, 11:19 AM

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Calling the acceleration of the rocket "a", the height after 4 seconds (when the bolt fell off) "h", and the velocity at that time "Vf", you know that Vf= 4a and you know that h= a(8). That gives you two equations for a, h, and Vf.
When the bolt falls, that Vf will be the "initial" velocity of the bolt: x= 4.9t^{2}+ Vf t+ h. You know that x= 0 when t= 6 so 4.9(36)+ 6Vf+ h= 0. You now have one more equation for a total of three equations for a, Vf, and h. 


#11
Jan2008, 11:44 AM

P: 79

Thanks to all! With everyone's help, yes I needed a lot of it, I finally have the correct answer and now I have instructions to go back to when the need arises. Thanks to everyone.



#12
Jan2008, 05:29 PM

P: 79

Is there a rule of thumb that says I should plug in different equations or using substitution like above or just use the basic kinematic equations? I'm sure if, maybe if 2 or more variables are unknown and you need one or the other to find the other missing variable of the two?



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