# To find rocket's acceleration and velocity

• brunettegurl
Vcutoff but i still didnt get the answer...the gravity will be negative thoughout the calcs...There might be a typo in the original problem. Check to make sure that your Vcutoff is actually 9.80 m/s^2. If it is, that would change the equation to:Vf2 = (9.80*a)2 - 2*g*(2023 - 1/2*a*9.80)

## Homework Statement

A 1203.0 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 11.2 s, then the motor stops. The rocket altitude 24.1 s after launch is 2023.0 m. You can ignore any effects of air resistance. What was the rocket's acceleration during the first 11.2 s? What is the rocket's velocity as it passes through a cloud 2023.0 m above the ground?

## Homework Equations

(a)d=Vit + 1/2 at^2...
(b) Vf= Vi + at...
(c) d= di + Vf (t) - 1/2 at^2

## The Attempt at a Solution

So I used the first equation to findthe distance the rocket traveled in the first 11.2 seconds and subtract that with altitude given...and know that the starting speed is 0m/s i tried to determine the speed using vfinal {throughout the first 2 eqns. i used a= -9.80 m/s^2}and then plugged it into the last equation to determine the acceleration..im gettin the wrong answer ..pls help

Well, first thing to think about: if an object is accelerating at -9.8 m/s^2, and its initial velocity is 0, is it moving down or up? Does that make sense for your problem?

-9.8 m/s^2 is the acceleration of gravity. But that's not the only acceleration involved - the rocket engine adds some acceleration of its own.

They want you to find 2 things. Vfinal and the initial a.

I'd look to develop 2 equations that are made of the 2 unknowns, and then solve.

For instance

Vcutoff = 11.2*a

and

Xcutoff = 1/2*a*(11.2)2

That yields ...

Vfinal2 = Vcutoff2 - 2*g*(2023 - Xcutoff)

Substituting into that then should yield your first equation in the 2 unknowns you need to find.

You just need another equation in the 2 unknowns then and you can solve.

so if i use the equation:: Vfinal2 = Vcutoff2 - 2*g*(2023 - Xcutoff) and sub into it the vcutoff and xcutoff i will still have the 2 unknowns in the same questions..this is a really stupid question but 2 eliminate one of the unknowns i wld have to use the other unknown correct?? but there's no additional info. given in the question to do this...

brunettegurl said:
so if i use the equation:: Vfinal2 = Vcutoff2 - 2*g*(2023 - Xcutoff) and sub into it the vcutoff and xcutoff i will still have the 2 unknowns in the same questions..this is a really stupid question but 2 eliminate one of the unknowns i wld have to use the other unknown correct?? but there's no additional info. given in the question to do this...

Sure there is.

Consider how the cutoff velocity is affected by gravity. How does that relate to Vfinal?

in the cutoff velocity the acceleration inclds gravity right?? and I am not really sure how it relates to vfinal...

brunettegurl said:
in the cutoff velocity the acceleration inclds gravity right?? and I am not really sure how it relates to vfinal...

The statement of the problem - constant acceleration to cutoff - means that gravity was taken into account in determining the acceleration to that point.

But after cutoff it is being slowed by gravity - there is no more propellant. So ...

Vfinal = Vcutoff - g*t

You can figure the time. You know g, so ...

and the time wld b the differnce between 11.2 s and 24.1 s where the altitude was measured ..correct??

brunettegurl said:
and the time wld b the differnce between 11.2 s and 24.1 s where the altitude was measured ..correct??

That looks right.

ok so i did that and my hwk app. is telling me that the answer I am getting is wrong any more hints u can give me wld b really appreciated

brunettegurl said:
ok so i did that and my hwk app. is telling me that the answer I am getting is wrong any more hints u can give me wld b really appreciated

It look's like a complicated calculation. Looks like you get a quadratic equation don't you? Are you sure you solved the quadratic correctly?

OK. I see the quadratic in a seems to cancel out the quadratic term., so it is a little simpler.

Maybe write out your equation with all the substitutions. There seems to be ample opportunity for arithmetic slip ups in this.

this is what i get as a quadratic equation:: 125.4a^2 + 1180.48a- 52800.002=0

and for my acceleration value i got 16.345m/s^2

brunettegurl said:
this is what i get as a quadratic equation:: 125.4a^2 + 1180.48a- 52800.002=0

and for my acceleration value i got 16.345m/s^2

As just mentioned when I worked it out, I lose the quadratic in a. It seems to cancel out.

Vf2 = (11.2*a)2 - 2*g*(2023 - 1/2*a*11.22)

Vf = 11.2*a - 12.9*g

(11.2*a - 12.9*g)2 = (11.2*a)2 - 2*g*(2023 - 1/2*a*11.22)

The (11.2*a)2 term on both sides cancels. So I don't see an a2 term.

i went over my calcs. and i realized i 4got 2 square my Vcutoff but i still didnt get the answer...the gravity will be negative thoughout the calcs right?

brunettegurl said:
i went over my calcs. and i realized i 4got 2 square my Vcutoff but i still didnt get the answer...the gravity will be negative thoughout the calcs right?

g is negative, but I accounted for that directly by putting its direction in the equation.

thank u so much...