Find Velocitiy from an Acceleration vs. Displacement Graph

CGI

1. 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

2. Homework Equations

3. 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!

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TSny

Homework Helper
Gold Member
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

Homework Helper
Gold Member
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

Homework Helper
Gold Member
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

Homework Helper
Gold Member
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.

"Find Velocitiy from an Acceleration vs. Displacement Graph"

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