Rocket Acceleration and Altitude Calculation

[Cfem]

(I apologize in advance if this should have been merged with my other topic.

Homework Statement

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

A) What was the rocket's acceleration during the first 16s?
B) What is the rocket's speed as it passes through a cloud 5100m above the ground?

Homework Equations

d = vi(t) + (1/2)(a)(t²)

The Attempt at a Solution

5100 = d_total
d_first = 0 + (1/2)(a)(16²)
d_second = v₀ - (1/2)(9.8)(16)
v0 = 16a

So the combining the equations:

d_total = d_first + d_second

5100 = 0 + (1/2)(a)(16²) + v₀ - (1/2)(9.8)(16)
5100 = 128a + 16a - 78.4
5178.4 = 144a
a = 35.96

Now, our homework is checked automatically, so I'm not sure if I'm rounding incorrectly or what, but any help would be appreciated.

 

[PhanthomJay]

Your calc looks good. I would use a =36 m/s^2, but if you look up the often confusing rule for significant figures, and since the mass has only one, then a=40 m/s^2. I don't know what your software is looking for; take a gamble and use 40, and see if it takes. I hope you have a second chance, however, just in case.

 

[Cfem]

That didn't work either, but it's good to hear that my calculations were correct, at least (I couldn't have forgotten that much ._.)

I'll e-mail the instructor, but a second confirmation would be nice.

 

[xcvxcvvc]

(I apologize in advance if this should have been merged with my other topic.

Homework Statement

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

A) What was the rocket's acceleration during the first 16s?
B) What is the rocket's speed as it passes through a cloud 5100m above the ground?

Homework Equations

d = vi(t) + (1/2)(a)(t²)

The Attempt at a Solution

5100 = d_total
d_first = 0 + (1/2)(a)(16²)
d_second = v₀ - (1/2)(9.8)(16)
v0 = 16a

So the combining the equations:

d_total = d_first + d_second

5100 = 0 + (1/2)(a)(16²) + v₀ - (1/2)(9.8)(16)
5100 = 128a + 16a - 78.4
5178.4 = 144a
a = 35.96

Now, our homework is checked automatically, so I'm not sure if I'm rounding incorrectly or what, but any help would be appreciated.

You forgot to multiply velocity by time to get distance

I'd set the problem up like this:
$$5100 = \frac{1}{2}a_{rocket}16^2 + (16a)4 - \frac{1}{2}9.81 * 4^2$$

edited out incorrect stuff

 

[PhanthomJay]

Ooops, sorry, you are right, Cfem and I forgot the 't' after the 'Vo'.

But your second equation is incorrect. The rocket's acceleration is a function of the net force acting on it, which includes it's weight, the rocket thrust, air drag, etc. The 9.8 m/s^2 acceleration of gravity should not be accounted for twice.

 

[Cfem]

That's like the fourth time I've forgotten what one variable for no reason.

Thanks for the help.