Integral Calculus question I am attempting to finish this question.
Could anyone help me with the first part, i) as I am stuck on this.

ii) $$ v = u + at     
5 = 2 + a x 60
a = \frac{5-2}{60} = \frac{1}{2} m/s^2$$
$$v^2 = u^2 + 2 as    
5^2 = 2^2 + 2 x \frac{1}{20} x s$$
$$ s = \frac{20(25-4)}{2} = 210m$$
iii)
$$ v = u + at = 2 + \frac{1}{20} x 120 = 8m/s$$
$$v^2 = u^2 + 2as
8^2 = 2^2 + 2 x \frac{1}{20} x s$$
$$s = \frac{20(64-4)}{2} = 600m$$
 A: I'm not quite sure what you're doing in your answers for (ii) and (iii). It seems you're using that, for constant acceleration (which is not even true here), the velocity at time $t$ is the initial velocity plus the acceleration times $t$. However, you already have the velocity equation, but you don't directly have the acceleration.
Instead, as bounceback stated in the comment, note that $v(t) = \frac{ds(t)}{dt}$, where $v(t)$ is the velocity/speed, and $s(t)$ is the position, so $\int v(t)dt = s(t) + c$. Thus, since you're given that
$$v(t) = -0.2t^2 + 0.7t + 0.5 \tag{1}\label{eq1}$$
you get by integrating with respect to $t$ that
$$s(t) = -\frac{0.2}{3}t^3 + \frac{0.7}{2}t^2 + 0.5t + C \tag{2}\label{eq2}$$
for some constant $C$, with it being equal to $s(0)$. Thus, the distance moved from the position at the start time $t = 0$ is
$$D(t) = s(t) - s(0) = -\frac{0.2}{3}t^3 + 0.35t^2 + 0.5t \tag{3}\label{eq3}$$
Parts (ii) and (iii) can now easily be determined by using \eqref{eq3} to get $D(1) = 0.78\bar{3}$ metres and $D(2) = 1.8\bar{6}$ metres, respectively.
