I had been through this lecture from an MIT open course ware.
Basic theme of this lecture is to explain the big picture of calculus, which is nothing but, relating functions.
In specific, below function f1(distance vs time) is being depicted(drawn) from the given behaviour of f2(speed vs time).
My question:
As per function f2, I understand that the vehicle is moving with an accelerated speed. In function f1, I would like to understand, How is the initial value of distance close to zero(as highlighted above)?
The graph could have been displayed better. From the speed-vs-time graph we can see that there is a linear relationship, namely $$s=s_0+at,$$ where $s$ is the speed, $s_0$ is the initial speed (the speed at tinme $t=0$), and $a$ is the acceleration. If we integrate this with respect to time we get $$x=\int(s_0+at) \ dt=\left(s_0x+\frac{1}{2}at^2\right)+c.$$ This is the expression for distance (we know this because the derivative of distance with respect to time is the speed or velocity), and we can see that it is a quadratic equation, and hence the curve is a parabola.
From his graph for distance v time, it passes through $(0,0)$, and all this means is that at time $t=0$ the distance of the object from the origin is $x=0$.