velocity from force/time graph...

[FancyNut]

http://img.photobucket.com/albums/v345/chronoga/ff336e76.jpg


^^ So according to that graph, force (and so acceleration) increase from t =0 to t=4 then force drops to zero and there's probably a constant deacceleration for two seconds....

How do I get the velocity at t = 6?

acceleration is changing as forces changes (increases) so I can't pin it down and Im guessing in order to get velocity at t = 6 I'll need velocity at t = 4 and acceleration between t = 4 and t= 6...


Any hints/ideas? Thanks for any help. :smile:

[Tide]

HINT: If force is proportional to the derivative of the speed then the speed should be proportional to the integral of the force! :-)

[FancyNut]

I alternated between staring at that hint and the graph for 15 minutes...


:redface:

[Spectre5]

the velocity is the integral of acceleration....acceleration and force are proportional as you know....now you know what an integral represents, right...?

[FancyNut]

... so velocity is the area under the force? 1/2*b*h = 1/2*4*10 = 20 for initial velocity at t = 4?

[FancyNut]

hah never mind I got it! the hight of the triangle should've been in acceleration values and not force.. so I divided ten by the mass and THEN calculated the area and that's the initial velocity which is equal to the final one at t = 6 because acceleration is 0 from t =4 to t=6.

Thanks! :biggrin:

[Spectre5]

very good! :)