Yes, when the object is in freefall on Earth, gravity = a = 9.8m/s^2 down. Signs are optional but you could say -9.8n/s^s.
In 2D, the gravity is always 9.8m/s^2 down on the object.
It doesn't matter how the object is moving or thrown, gravity is always down.
The graph of the acceleration is the graph of the derivative of the v-graph. Find the slope at each second and graph it onto a new graph. The a-graph is not linear.
Wait, I think it was the d for the 4th second, from 3-4s. Because the first second is 0-1s, second second is for 1-2s, etc.
So
Vf=vo+at
V@3=5+9.8*3 and V@4=5+9.8*4
Vf^2=Vo^2+2ad
(V@4)^2=(V@3)^2+2*9.8*d
Find d
B=µoNI/2r, B is magnetic force, µo is magnetic constant, N is number of turns, I is current, r is radius
I think that's it but that R is unused in the problem, and it concerns me that I missed something.
V=IR
Both: 12=(R1+R2)I;
R2 removed: 12=R1(I+.2)
R1 removed: 12=R2(I+.1)
Solve for R's
R1=12/(I+.2)
R2=12/(I+.1)
Plug into original equation
12=(12/(I+.2)+12/(I+.1))I
Solve for I
I=.1414
Plug I back into R1 and R2
R1=12/(.1414+.2)=35.15ohm
R2=12/(.1414+.1)=49.7ohm
First off, deceleration is constant, if the board is uniform.
The equation you use is d=.5(vo+vf)t. d, vo, vf is given. Also, don't forget to convert cm to m.