oh yes, Thank a lot. I just think crazy out of a box when velocity was too high than the graph expected to be drawn.
I attached the area for the graphs.
they would draw the wrong coordinate for velocity
I have got some ideas, can it is possible to assume that only during 5 seconds, a =+1, so velocity changes to +1. and from 5-10, it drops to 0 since a=-1 and from 10-15s, a remains -1 and velocity is now -1., if 15-20s, velocity -2...
Hi, since Fnet unchanges during interval time above, a=Fnet/m=+1, so what about velocity ? does it follow the equation v=vo+ a.t?
v(0)=0, v(5)= 0+1x5=5? I wonder why the answersheet covers only maximum of +2 of velocity ( in vertical line) for the graph to be drawed
After 5s, a(t>5)=Fnet(t>5)/m...
Thank for your guiding Simon, So I am still in trouble with the time t=0s, does F2 has magnitude of 2N and right ward direction?
if mass is at rest means it has V instantaneously =0, that means since F2 applied forced, There is unbalanced force act on mass, Fnet=F2-F1, a=1 and in positive...
1. problem
Two forces, F1 and F2(t), act on a mass m = 1.0 kg. Force 1 F1 is to the left in the negative x-direction and force F2(t) is to the right in the positive x-direction. F1 is constant with a magnitude of 1.0 N whereas F2(t) has a magnitude of 2.0 N for the time interval 0.0 s ≤ t ≤ 5.0...