Find Velocitiy from an Acceleration vs. Displacement Graph

[CGI]

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.

Any help would be much appreciated!

 

[TSny]

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?

 

[CGI]

My best guess is the area of that section from 0 to 70? Is that in the right direction?

 

[TSny]

My best guess is the area of that section from 0 to 70? Is that in the right direction?

Yes.

 

[CGI]

Okay, so I got 800 for the total area underneath, but I'm not sure what the next step is

 

[TSny]

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?

 

[CGI]

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?

 

[TSny]

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.