Finding the Final Velocity unsing a Force by Time Graph.

[OliTheNinja]

Homework Statement

The graph below shows the force applied to a 3.8 kg cart initially at rest but free to move on a horizontal track. Calculate the final velocity of the cart, after being subjected to the forces illustrated in the graph.

The graph is attached.

Homework Equations

Okay so I know that the area in a Force by Time graph is the impulse, but I'm not exactly sure how to get to that.

The Attempt at a Solution

So I first tried 10N*24s=240N*s, 240N*s=240kg*m/s, 240kg*m/s=3.8kg*Change in velocity, 240/3.8=63.158 m/s

WRONG!

I then tried doing 9N*6s=54N*s, 54/3.8=14.211 m/s

WRONG AGAIN!

I also tried 10N*17s=170N*s, 170/3.8=44.737 m/s

WRONG WRONG WRONG!

Help Please? If I can find the impulse, I know all I have to do is divide it by 3.8, since the impulse is equal to the change in momentum.

Thanks :)

 

[HallsofIvy]

Yes, the area under the curve is the impulse, mv. If you draw vertical lines at t= 6 and t= 17, you divide the region under the graph into three simple figures: from t= 0 to t= 6 you have a trapezoid with "height" (horizontally) of 6 and "bases" of length 9 and 4. From t= 6 to 17 you have a rectangle with height 4 and base 11. from t= 17 to t= 24, you have a trapezoid with height 7 and bases of length 4 and 10.

The area of a rectangle is "base times height" and the area of a trapezoid is "average of the two bases times height". The area under the graph is the sum of those three areas.