velocity/acceleratio in one dimension question...

[FancyNut]

Man, this is another embarrasing question.

A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance.

First this question has no use for the 1000 kg because it's in a chapter before forces are introduced... or I think.

a- What was the rocket's acceleration during the first 16 s?

Well I know delta time... but nothing about position, nothing about initial or final velocity...

What is the rocket's speed as it passes through a cloud 5100 m above the ground?

It's acceleration is -9.8 in that second stage and I know delta time is 4 seconds (20-16) but with no velocity I don't know how to get any info here...


I should be solving these easy questions with no problem. :cry:

[Tide]

Well I know delta time... but nothing about position, nothing about initial or final velocity...

Um ... I think the point of the problem is for you to figure those out! BTW - if the rocket is "launched" I think it's safe to conclude its initial speed is 0. :-)

[FancyNut]

Um ... I think the point of the problem is for you to figure those out!

Yeah I know.. :o

BTW - if the rocket is "launched" I think it's safe to conclude it's initial speed is 0. :-)

I thought there is supposed to be an initial velocity > 0. =\

There's another problem in my text but with a launched cannon-- it had an initial velocity of 100 m/s so I thought this is the same too. :frown:


And thanks for the reply!

[Spectre5]

but the lauched cannon you were looking at the ball after it exited the cannon....then traveled through the air, correct?

With this, the whole thing is beginning to move....so its initial velocity will be zero.

[FancyNut]

ok I still have at least two unknows in each stage...

for the first one when the rocket is accelerating, I only initial velocity/time/position are zero and that it takes 16 seconds for the motor to stop accelerating... what equation should I use? whatever one I choose I need to know the final velocity before getting acceleration and vice versa... but I can't get either without knowing the other... o_O

and for the second part it's the same thing... time is 4 seconds and acceleration is -9.8 and final position is 5100 meters but I dont know the initial position (the final position of the first part) or the starting velocity (the final velocity of the first part) and whatever equation I use I always end up with TWO unknowns.

[FancyNut]

.....

:cry:

[FancyNut]

the only thing I have is :

a_0 = y_1/128

that's the acceleration for the first part of the problem where the rocket is accelerating... y is the distance from start until the motor stops. *bashes head into desk*

[FancyNut]

Can anyone please help?

I have not done anything (chem, physics, math) since yesterday because of worrying about this one easy problem. I'd say my confidence is moving with a negative acceleration of 450.52 right now while anxiety is a postive 952 m/s^2.

I just downloaded a trial version of Physics 101 SE and just like I expected I can't calculate final velocity without acceleration nor the acceleration without final velocity... and the fact that I know acceleration for the second part is -9.8 doesn't help because I don't know the starting velocity (which is the final one for the first part).

[FancyNut]

can I just get a tiny hint. ;__;

[Tide]

Can anyone please help?

I have not done anything (chem, physics, math) since yesterday because of worrying about this one easy problem. I'd say my confidence is moving with a negative acceleration of 450.52 right now while anxiety is a postive 952 m/s^2.

I just downloaded a trial version of Physics 101 SE and just like I expected I can't calculate final velocity without acceleration nor the acceleration without final velocity... and the fact that I know acceleration for the second part is -9.8 doesn't help because I don't know the starting velocity (which is the final one for the first part).

You certainly don't believe that because Physics 101 SE can't solve the problem for you that it cannot be done! That program only models a few simple situations and can hardly be called the definitive work!

Just break the problem down into parts. If the rocket accelerates upward at a given rate for a given time, how fast it is travelling 16 seconds later? How high off the ground will the rocket be at that time?

If your answer to the first question is v1 and your answer to the second question is H then ask yourself the following: A projectile with an initial speed of v1 is launched from a height H. How high will the projectile rise in 4 seconds?

You should be able to piece things together and answer your original questions.

[relativitydude]

Physics 101 SE can still help you.

"A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100"

In the distance formula, plug into 5100 m for D, 0 m/s for Vi and 16 s for time and solve for A. Now plug that a into the Instantaneous formula along with the time to get velocity at 5100 m.

Im getting 39.84375 m/s^2 and Vf = 637.5 m/s

[CartoonKid]

I think for the first 16s the rocket was accelerating and for the 4s later, it was actually decelerating. We have split the case into 2 part. the total displacement is 5100m which consists of d1 and d2.

[relativitydude]

Oh, I misread the problem.

[AVNguyen]

[QUOTE=FancyNut]the only thing I have is :

a_0 = y_1/128

Your problem is actually not as hard as you think.I will give you a tip:1)At the first part of the journey, the accelaration of the spaceship is also under the deaccelaration of the 2nd part.2) the initial speed of part 2 is the result of the speed which you get from part 1
Hope that you can solve it

[FancyNut]

Thanks for all the help but I solved this problem a long time ago.... Thanks again though. :D