Constant acceleration problem to find the speed of a rocket

Click For Summary
The rocket's vertical acceleration is given by 2.60t, indicating that it is not constant. To find the speed when the rocket is 240 meters above the surface, standard kinematic equations cannot be used due to the variable acceleration. Instead, integration is required to derive velocity as a function of time and position. The discussion emphasizes the need to determine appropriate limits for integration based on the variable being integrated. Solving the problem involves using these derived expressions to find the desired velocity.
mahrap
Messages
37
Reaction score
0
A rocket starts from rest and moves upward from the surface of the earth. For the first 10 of its motion, the vertical acceleration of the rocket is given by 2.60t, where the -direction is upward.

What is the speed of the rocket when it is 240 above the surface of the earth?

v = v0 + at
change in x = (v + v0)t/2

I solved for t in the second equation by plugging in 240 for change in x, 0 for v0, and left v as v: meaning i solved for t in terms of v. Then i took this and plugged it in the t of the first equation with t=480/v, a=2.6(480/v) and v0=0. However i keep getting a wrong answer. What should i do?
 
Physics news on Phys.org
mahrap said:
A rocket starts from rest and moves upward from the surface of the earth. For the first 10 of its motion, the vertical acceleration of the rocket is given by 2.60t, where the -direction is upward.

What is the speed of the rocket when it is 240 above the surface of the earth?

v = v0 + at
change in x = (v + v0)t/2

I solved for t in the second equation by plugging in 240 for change in x, 0 for v0, and left v as v: meaning i solved for t in terms of v. Then i took this and plugged it in the t of the first equation with t=480/v, a=2.6(480/v) and v0=0. However i keep getting a wrong answer. What should i do?

The acceleration is not constant, so the usual kinematic equations for constant acceleration will not apply. I see an integral or two in your future :smile:
 
What would be the limits of my integration. How would I find the time where the rocket is at the given height to integrate the acceleration to get velocity?
 
mahrap said:
What would be the limits of my integration. How would I find the time where the rocket is at the given height to integrate the acceleration to get velocity?

Integrate to find expressions for velocity as a function of time and distance as a function of time. Then use these expressions to solve the particular problem.

As for the limits, well what variable are you integrating over?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
11
Views
1K
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K