In this display we take the acceleration to be constant at 2 meters per second per second (MPSPS). The initial velocity is taken to be zero. The resulting expression for v is

v=2MPSPS * tS .

This may be a good place to remind you of the importance of keeping your units straight. Disregarding the units, we would write

v=2*t ,
which is numerically correct. Physically though it makes no sense. Velocity can not be equal to time. Attaching the units to the equation we get
v=2 meter/1 second/1 second * t second .
In this expression we have the units of seconds in the denominator twice and the numerator once. We may in effect cancel out one of the denominator seconds with the numerator seconds so that the units on v are left as meter/second which is physically correct. I will try to always tag numerical values with units if there could be any confusion.

As you can tell from the form of

v=f(t)=2*tMPS ,
velocity will increase linearly with time. This is shown on the display. At any value of time, t, along the horizontal axis, the value of acceleration is 2 and the value of velocity is 2*t. What may be less obvious is that at any t, the value of velocity is equal to the area under the acceleration curve, here a flat line, between zero and the time t. This makes sense when you think about it. Acceleration is a rate of change of velocity, so based on the units, if I multiply the change in velocity with respect to time, by time, the result will have units of velocity. On a plot of velocity versus time, as we have here, the area under the acceleration curve is exactly that, the product of acceleration times time.