Find Velocitiy from an Acceleration vs. Displacement Graph

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Homework Help Overview

The problem involves determining the speed of a rocket at specific displacements using an acceleration versus displacement graph. The rocket starts from rest, and the acceleration values at certain displacements are provided.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and velocity through integration, specifically referencing the equation ads = vdv. There is exploration of interpreting the integral graphically and calculating areas under the curve.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and interpretations of the integral. Some have reached a numerical value for the area under the curve, while others are questioning the next steps and the relationship between the integrals.

Contextual Notes

There appears to be some uncertainty regarding the correct interpretation of the area under the acceleration graph and its implications for calculating velocity. The discussion reflects an ongoing exploration of these concepts without a definitive resolution yet.

CGI
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Homework Statement


The a-s graph for a rocket moving along a straight horizontal track has been experimentally determined. If the rocket starts from rest, determine its speed at the instants s = 20m, 50m, and 70m, respectively. On the graph, S1 = 50m, a1 = 10 m/s^2, S2 = 70m, and a2 = 20 m/s^2

Homework Equations


ads = vdv [/B]

The Attempt at a Solution


I know that the solution for the velocity at S2 = 70m is V = 40 m/s, but I'm not sure how to get there.
I know we talked about the relationship for ads = vdv so I tried taking the integral on both sides to calculate for the velocity.

I tried to solve for the velocity at S2

The integral from 0 to v of vdv = the integral from 0 to 20 of ads
So, V^2/2 = 20s which means V^2 = 40s where S = 70 meters.
So I get V = √(40)(70) = 52.9, which is not correct.

20160116_194713.jpg
Any help would be much appreciated!
 
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CGI said:
The integral from 0 to v of vdv = the integral from 0 to 20 of ads

What is the graphical way to interpret the integral of ads from s = 0 to s = 70 m?
 
My best guess is the area of that section from 0 to 70? Is that in the right direction?
 
CGI said:
My best guess is the area of that section from 0 to 70? Is that in the right direction?
Yes.
 
Okay, so I got 800 for the total area underneath, but I'm not sure what the next step is
 
CGI said:
Okay, so I got 800 for the total area underneath, but I'm not sure what the next step is

Good.

How is the integral of ads related to the integral of vdv?
 
They're equal to each other? So I could set v^2/2 = 800 which in the end makes v = 40.

Did I think about that correctly?
 
CGI said:
They're equal to each other? So I could set v^2/2 = 800 which in the end makes v = 40.

Did I think about that correctly?
Yes.
 

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