Analyzing Velocity vs time graph So I did an experiment on conservation of momentum, where two carts with different mass start at rest and get accelerated and we measure the speed. What I observed was that after the acceleration, they started to decrease in velocity. I thought this was due to presence of frictional force, and I used regression line to get the "a" of friction. And here's what I got, and from here, I cannot seem to go further because the graph seems like it's contradictory. Shouldn't the slope of both of them be the same as a = ug? (where u is the coefficient of kinetic friction)

 A: Couple of issues here.
First, assuming these “carts” have wheels we are dealing with rolling resistance (aka rolling friction) not kinetic friction (aka sliding friction). So the force opposing motion is 
$$F=C_{cc}N$$
Where $C_{cc}$ is the coefficient of rolling resistance and $N$ is the normal force =$mg$.
Second, unlike the coefficient of kinetic friction, the coefficient of rolling resistance may be a function of the normal force, depending on the nature of the wheel. For pneumatic tires the coefficient may depend on how much tire is in contact with the surface. For instance rolling resistance increases with under inflated tires. Equivalent to an under inflated tire is the same tire supporting more weight. 
I bring this up because I notice the magnitude of the coefficient $m$ (not to be confused with m for mass) for your red cart equation is greater than the blue and the red cart weighs twice the blue. Perhaps the coefficient of rolling resistance is higher for the red cart? 
Just a thought. Hope it helps.
A: The graphs make sense, albeit a bit misleading because of the different scales for red and blue.
Friction opposes motion, so if in $y=b + m t$ if $b$ is positive then $m$ must be negative (blue case) and if $b$ is negative then $m$ must be positive (red case).
The way I read it, we have
$$\begin{array}{l|l}
\text{before, t<2.5} & \text{after, t>2.5} \\
\hline
v_{\rm blue} = 0 & v_{\rm blue} = v_{\rm blue} = 0.333 - 0.008\,t \\
 v_{\rm red} = -0.15 + 0.0156 \,t  & v_{\rm red} = ?
\end{array}$$
So the $a_{\rm red} = 0.0156$ due to friction is attributed to $$a_{\rm red} = \mu_{\rm red}\,m_{\rm red}g$$ and the $a_{\rm blue} = 0.008$ attributed to $$a_{\rm blue} = \mu_{\rm blue}\,m_{\rm blue} g$$
