Constant acceleration problem to find the speed of a rocket

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Homework Help Overview

The problem involves a rocket moving upward from the Earth's surface with a time-dependent acceleration of 2.60t. The goal is to determine the speed of the rocket when it reaches a height of 240 units above the surface. The complexity arises from the non-constant acceleration, which suggests that standard kinematic equations may not be applicable.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to integrate the acceleration to find velocity due to its time-dependent nature. Questions arise regarding the limits of integration and how to determine the time at which the rocket reaches the specified height.

Discussion Status

The discussion is ongoing, with participants exploring the implications of non-constant acceleration and considering integration as a necessary approach. Guidance has been offered regarding the need for expressions of velocity and distance as functions of time.

Contextual Notes

There is an acknowledgment that the usual kinematic equations for constant acceleration do not apply, indicating a need to rethink the problem setup and assumptions.

mahrap
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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?
 
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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?
 

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