Time taken to travel x meters with constant acceleration

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Homework Help Overview

The discussion revolves around a physics problem involving a rocket's motion under constant acceleration. The original poster is trying to determine the time taken to reach a specific height given an acceleration of 20 m/s² and a displacement of 1379.8 m.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between acceleration and velocity, questioning whether a constant acceleration leads to a linear increase in velocity over time. They also seek clarification on the necessary equations to solve for time.
  • Some participants question the initial conditions of the problem, specifically whether the rocket starts from rest, and discuss the relevant kinematic equations that could be applied.
  • Another participant provides the kinematic equation and attempts to apply it to the problem, leading to a calculation that raises questions about the validity of the result.

Discussion Status

Participants are actively engaging with the problem, providing clarifications and discussing the application of kinematic equations. There is a recognition of a mistake in the calculations, but no consensus has been reached on the final answer. Guidance has been offered regarding the interpretation of the equations and the initial conditions.

Contextual Notes

The original poster's confusion about the acceleration and the resulting calculations indicates a need for clarity on the concepts of kinematics and the proper application of the equations involved. The discussion reflects the challenges of interpreting the problem setup and the mathematical relationships at play.

the_cloud
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Hi, I'm trying to solve something that I'm very sure is simple, I just can't seem to find the right equation.

A rocket is traveling straight up, with an acceleration of 20m/s^2. I need to know how long it takes to reach a height of 1379.8m (this is the second part of a question).

I am confused about the acceleration. Does the fact that it's traveling at 20m/s mean that after 2 seconds it's traveling at 400 m/s and at 3 seconds it's traveling at 800 m/s?

And to work out the time it takes to travel that far the equation would be t = 1379.8 / ?

Thanks
 
Last edited:
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It's not "travelling at 20 m/s". The acceleration of 20 m/s^2 means that's the velocity increases by an additional 20 m/s every second. But you need another "given" number to solve this problem, so was it also given that it began at rest, so that the initial velocity can be set to zero? If that's number is also given, you have enough information. Then you need the topic called "kinematics", which provides four formulas. One of those formulas mentions these variables: displacement, initial velocity, acceleration and time.
 
a=\frac{dv}{dt}
dv={a}*{dt}
v=at+v_0
v=\frac{dx}{dt}
dx={v}*{dt}
x=x_0+{v_o}{t}+\frac{1}{2}{a}{t}^2
 
Yep, the starting velocity was 0. I'll look into kinematics and see if I can make sense of those formulas.
 
OK, so I've had a go. Can someone please check my answer and see if it's correct? (Sorry for the crappy formatting, I'm not sure how to do sub\superscripts)

initial velocity = 0;
acceleration = 20m/s^2;
displacement = 1379.8;

d = vi * t + 0.5 * a * t^2
1379.8 = 0m/s * t + 0.5 * 20m/s^2 * t^2
1379.8 = (0m) * t + 10m/s^2 * t^2
1379.8 / -10 = t^2
SQRT(-137.98) = t //?
t = 11.746 seconds

I'm not really sure how I ended up with a negative in my square root, but using the positive I get a number that seems reasonable.
 
Your answer is fine.
the_cloud said:
I'm not really sure how I ended up with a negative in my square root,
When you divided both sides by 10, for some reason you added a negative sign. (Perhaps you were mixing that up with subtracting 10 from both sides?)
 
Ah, yep I did. Thanks for everyones help.
 

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