change in position = (35 m) − (30 m) = 5 m
change in time = (1 s) −  (0 s)  = 1 s
\frac{change in position}{change in time} = \frac{5 m}{1 s} = 5\frac{m}{s}

average velocity = \frac{change in position}{change in time} = 5\frac{m}{s}

x(t) = (5\frac{m}{s}) t ???

x(t) = (5\frac{m}{s}) t + 30 m

instantaneous velocity =  \frac{dx}{dt}

v(t)  = \frac{d}{dt} [ (5\frac{m}{s}) t + 30 m ]

x(t)  =  ∫ v dt

x(t)  =  ∫ (5\frac{m}{s}) dt

= (5\frac{m}{s}) ∫ dt

=  (5\frac{m}{s}) t  ???
x(t)  =  (5\frac{m}{s}) t + C

instantaneous velocity  v(t) =  \frac{dx}{dt}

change in position  (x_{2} − x_{1}) =  ∫_{t_{1}}^{t_{2}} v(t) dt

average acceleration = \frac{change in velocity}{change in time} = 0.6 \frac{m}{s^{2}}

v(t) = (0.6 \frac{m}{s^{2}}) t − (0.6 \frac{m}{s})

instantaneous acceleration =  \frac{dv}{dt}

a(t)  = \frac{d}{dt} [ (0.6 \frac{m}{s^{2}}) t − (0.6 \frac{m}{s}) ]
v(t)  =  ∫ a dt

a(t) = 9.8 \frac{m}{s^{2}} downwards

a(t) = −9.8 \frac{m}{s^{2}}

v_{blue}(t) = (−9.8 \frac{m}{s^{2}}) t

v_{green}(t) = (−9.8 \frac{m}{s^{2}}) t
x(t)  =  ∫ v(t) dt