Kinematics of Constant acceleration word problem

[pistolpete333]

So the problem is:
a 1000 kg rocket is fired straight up, a 100 kg rocket is fired straight up with a constant acceleration for 16 seconds. It then stops accelerating and 20 seconds after launch reaches a height of 5100 m. Ignore air resistance.
Find the acceleration of the rocket for the first 16 seconds and the velocity of the rocket at 5100 m.


So in my physics class the only equations we have learned are the kinematics of constant acceleration in 1 dimension equations, and that is what the chapter this problem is from is based on.

I first tried using y = y_initial + (v_initial * t) + (1/2 * g * t^2) , but there are to many unknown values to make any use of it.
The values I were able to figure out from the problem are:
First 16 seconds
x = ?
x_initial = ?
v = ?
v_initial = ?
g = ?
t = 16

Free Fall
x = 5100
x_initial = ?
v = ?
v_initial = ?
g = -9.8
t = 4

basically I don't know where to go from here, and am just looking for some direction. Also I am unsure whether the 1000 kg plays any part in the problem, since we have not done anything with weight in our class, but if it does, feel free to provide me with some enlightenment. Thanks for any help and if any extra info is needed let me know.

 

[TSny]

So the problem is:
a 1000 kg rocket is fired straight up, a 100 kg rocket is fired straight up with a constant acceleration for 16 seconds. It then stops accelerating and 20 seconds after launch reaches a height of 5100 m. Ignore air resistance.
Find the acceleration of the rocket for the first 16 seconds and the velocity of the rocket at 5100 m.

I first tried using y = y_initial + (v_initial * t) + (1/2 * g * t^2) , but there are to many unknown values to make any use of it.
The values I were able to figure out from the problem are:
First 16 seconds
x = ?
x_initial = ?
v = ?
v_initial = ?
g = ?
t = 16

Free Fall
x = 5100
x_initial = ?
v = ?
v_initial = ?
g = -9.8
t = 4

You should be able to fill in values for x_initial and v_initial for the start of the first 16 seconds of the trip. Also, I wouldn't use g for the acceleration in the first part, because g is reserved for the acceleration due to gravity. Maybe just use "a" for the acceleration in the first part.

Can you use your equations for constant acceleration to write down an equation for x at the end of the first 16 seconds in terms of the unknown acceleration? Likewise for the velocity at the end of the first 16 seconds?

By the way, if you click on "go advanced" before entering a reply, you will see some formatting icons that will allow you to use subscripts, etc.

 

[pistolpete333]

Thanks for pointing out the initial values of x and v and using g instead of a, I guess I forgot to type out those values. I have been trying to do that but am having no luck. The equations i was given were
x = x_initial + V_initial * t + (1/2) * a * t^2
V = V_initial + a * t
V^2 = (V_initial)^2 + 2a(x-x_initial)

Also where is the go advanced option. Its probably pretty obvious and I am just missing it.

 

[TSny]

Thanks for pointing out the initial values of x and v and using g instead of a, I guess I forgot to type out those values. I have been trying to do that but am having no luck. The equations i was given were
x = x_initial + V_initial * t + (1/2) * a * t^2
V = V_initial + a * t

Also where is the go advanced option. Its probably pretty obvious and I am just missing it.

For the first part of the flight, plug in values for all of the symbols that you know for the two equations above.

The "go advanced" is just below the white box where you type your message. However, if you don't see it then the "go advanced" might only be available for those who have been members for a while. I'm not sure.

 

[Nickie]

Hello,

Thank you for providing the problem and the values you have already identified. To solve this problem, we will need to use the kinematics equations of constant acceleration in one dimension. These equations are:

1. x = x0 + v0t + 1/2at^2
2. v = v0 + at
3. v^2 = v0^2 + 2a(x-x0)

Let's start by looking at the first 16 seconds of the rocket's motion. We know that the rocket has a constant acceleration, and we are looking for the acceleration (a) and the initial velocity (v0). We also know that the initial position (x0) and initial time (t) are both zero. So, we can use equation 1 to solve for a:

x = x0 + v0t + 1/2at^2
5100 = 0 + 0 + 1/2a(16)^2
5100 = 128a
a = 5100/128 = 39.84 m/s^2

Now, we can use equation 2 to solve for the initial velocity:

v = v0 + at
0 = v0 + (39.84)(16)
v0 = -637.44 m/s

So, for the first 16 seconds, the rocket has an acceleration of 39.84 m/s^2 and an initial velocity of -637.44 m/s.

Next, we need to find the velocity of the rocket when it reaches a height of 5100 m. To do this, we will use equation 3. We already know the acceleration (a) and the initial velocity (v0) from our previous calculations. We also know the final position (x) and initial position (x0) from the problem statement. So, we can use equation 3 to solve for the final velocity (v):

v^2 = v0^2 + 2a(x-x0)
v^2 = (-637.44)^2 + 2(39.84)(5100-0)
v^2 = 406156.97
v = 637.44 m/s

So, when the rocket reaches a height of 5100 m, it has a velocity of 637.44 m/s.

As for the 1000 kg rocket, it does play a role in the problem.