Find Velocitiy from an Acceleration vs. Displacement Graph

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SUMMARY

The discussion focuses on calculating the velocity of a rocket at specific displacements using the relationship between acceleration and displacement. The key equation used is the integral form of ads = vdv, which relates the area under the acceleration curve to the change in velocity. The correct velocity at S2 = 70m is determined to be 40 m/s by integrating the area under the acceleration curve, which totals 800. This area directly correlates to the velocity squared divided by 2, confirming the solution.

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  • Knowledge of the relationship between acceleration, velocity, and displacement.
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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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