• Support PF! Buy your school textbooks, materials and every day products Here!

Kinematics problem, is it possible to solve with given information?

  • Thread starter pyx
  • Start date
  • #1
pyx
6
0

Homework Statement


A missile is launched from a silo 50m deep. After 20 seconds the engine stops, there is a 10 second period where the rocket is coasting straight up. At the end of this period, the rocket is 8000m above the ground.

a) What is initial acceleration of rocket?

b) What is rocket's velocity as it passes 8000m?

v0=0m/s
v1=?
v2=?
t0=0s
t1=20s
t2=t1+10s=30s
a0=?
a1=-9.8m/s^2
y0=-50m
y1=?
y2=8000m


Homework Equations


Kinematics equations


The Attempt at a Solution



Tried solving for the y1 value, but need v1, can't find v1 without a0, can't find a0 without y1 or v1. I am very lost. I feel like there isn't enough information to solve, but there is probably some trick to this and I am missing it. Help please! Thanks
 

Answers and Replies

  • #2
326
3

Homework Statement


A missile is launched from a silo 50m deep. After 20 seconds the engine stops, there is a 10 second period where the rocket is coasting straight up. At the end of this period, the rocket is 8000m above the ground.

a) What is initial acceleration of rocket?

b) What is rocket's velocity as it passes 8000m?

v0=0m/s
v1=?
v2=?
t0=0s
t1=20s
t2=t1+10s=30s
a0=?
a1=-9.8m/s^2
y0=-50m
y1=?
y2=8000m


Homework Equations


Kinematics equations


The Attempt at a Solution



Tried solving for the y1 value, but need v1, can't find v1 without a0, can't find a0 without y1 or v1. I am very lost. I feel like there isn't enough information to solve, but there is probably some trick to this and I am missing it. Help please! Thanks
Well, the first part of the question asks you to find the acceleration before the rocket has the constant acceleration, therefore having the constant velocity. To answer the question for the first part, we need to break into parts.. We know that:

→v_0 = 0 m/s
→t = 20 s
→d = 50 m

Then, this is enough to answer the question. You need to use this formula:

d = v_0 * t + ½ * a * t²

Given the values, solve for a.
 
  • #3
pyx
6
0
So

50m=0m/s(20s)+.5(a)(400s^2)
50m=0m+200s^2(a)
0.25m/s^2=a
so a = 0.25m/s^2?

That is a pretty slowly accelerating rocket...

I didn't think that d=50m. The rocket starts 50 meters underground, I don't know the height after 20 seconds. How can d=50m?
 
  • #4
326
3
So

50m=0m/s(20s)+.5(a)(400s^2)
50m=0m+200s^2(a)
0.25m/s^2=a
so a = 0.25m/s^2?

That is a pretty slowly accelerating rocket...

I didn't think that d=50m. The rocket starts 50 meters underground, I don't know the height after 20 seconds. How can d=50m?
Oops. My bad. Then, we have...

s = 50 + v_0t + ½at²

Good thing you catch that mistakes! Without you, I would not notice the mistakes I make!
 
  • #5
pyx
6
0
OK, with that I don't know s or a...so how do I solve that?
 
  • #6
326
3
OK, with that I don't know s or a...so how do I solve that?
s is the displacement
a is acceleration
 
  • #7
pyx
6
0
I am still completely lost. I need to know the height after 20 seconds, as well as the velocity at 20 seconds and 30 seconds and the acceleration of the first part. No idea. Every equation I use has 2 unknowns.
 
  • #8
326
3
I am still completely lost. I need to know the height after 20 seconds, as well as the velocity at 20 seconds and 30 seconds and the acceleration of the first part. No idea. Every equation I use has 2 unknowns.
Oops. Sorry about that.

Hm... Before 20 seconds, we know that:

→the rocket is 50 m above the ground.
→v_0 = 0
→v_f = ?
→a > 0
→t = 20

So you use this form d_1 = h_0 + v_0 * t + ½ * a * t² = ½ * a * t². Clearly, d_1 =50 + ½ * a * t².

Between 20 and 30 seconds, since the rocket coasts up, we can say that a = 0, and that v_0_n = some velocity. Also, we need to determine the time interval between 30 seconds and 20 seconds; otherwise, if you just let t = 30, then you are answering the question wrong. It's after 20 seconds, we have the rocket having the zero acceleration and constant velocity (so that is the 10 second difference.). We should get:

d_2 = v_0_2 * Δt [where Δt is the difference between the time periods.]

Add up the distance and set that equal to 8000. We obtain:

d_1 + d_2 = 8000
50 + ½ * a * 20² + v_0_n * 10 = 8000
½ * a * 20² + v_0_n * 10 = 7950

Actually, you are correct. This problem seems to be impossible to solve!
 
  • #9
pyx
6
0
The rocket starts at 50 meters BELOW ground. Not above.
 
  • #10
326
3
The rocket starts at 50 meters BELOW ground. Not above.
Then, it's just this:

½ * a * 20² + v_0_n * 10 = 8000

All we have is just a/30 = (sum of all accelerations = a + 0)/(10 + 20).

At the first 20 seconds, a > 0. Then, after 20 seconds, a = 0.
 
  • #11
pyx
6
0
There has got to be a way to solve this. What about substitutions? Would that help any?
 
  • #12
326
3
There has got to be a way to solve this. What about substitutions? Would that help any?
Oh! I forgot to mention. Between t = 20 and t = 30, we have the same velocity as usual (rocket coasting up). This goes the same after the acceleration at t = 20. Use the acceleration formula: a = v_f/t, so we have...

½ * v_f/20 * 20² + v_f * 10 = 8000

Solve for v_f and find a using a = v_f/(overall time elapsed) = v_f / 30.
 

Related Threads for: Kinematics problem, is it possible to solve with given information?

Replies
8
Views
1K
Replies
3
Views
1K
Replies
18
Views
3K
Replies
16
Views
814
Replies
4
Views
2K
Replies
1
Views
6K
Replies
1
Views
3K
Replies
48
Views
14K
Top