Velocity from force/time graph

1. Oct 25, 2004

FancyNut

^^ 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.

2. Oct 25, 2004

Tide

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

3. Oct 25, 2004

FancyNut

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

4. Oct 25, 2004

Spectre5

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

5. Oct 25, 2004

FancyNut

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

6. Oct 25, 2004

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!

7. Oct 25, 2004

Spectre5

very good! :)