I Deriving kinematic equation for position

[PeroK]

I don't know, what you guys really are after here, but are you maybe just discussing the Taylor expansion of a function in a very complicated way, which leads to confusion:
$$x(t)=x(t_0) + \dot{x}(t_0) (t-t_0) + \frac{1}{2} \ddot{x}(t_0) (t-t_0)^2 + \cdots = \sum_{k=0}^{\infty} \frac{1}{k!} x^{(k)}(t_0) (t-t_0)^k.$$
Here $x^{(k)}$ means the $k$-th derivative of the function $x(t)$.

That's a good way to avoid doing the integration for the OP's acceleration $a = \alpha t$:
$$x(t)=x_0 + v_0(t-t_0) + \frac 1 2 \alpha t_0(t-t_0)^2 + \frac 1 6 \alpha(t - t_0)^3$$

 

[vanhees71]

Hm, I get
$$v(t)=v_0+\int_{t_0}^t \mathrm{d} t' \alpha t' = v_0 + \frac{\alpha}{2} (t^2-t_0^2)$$
and then, integrating again and after some algebra,
$$x(t)=x_0 + v_0 (t-t_0) +\frac{\alpha}{6} (t-t_0)^2 (t+2 t_0) + v_0 (t-t_0)+x_0.$$
[EDIT:] Well, it's the same. So never mind ;-).

 

[PeroK]

Hm, I get
$$v(t)=v_0+\int_{t_0}^t \mathrm{d} t' \alpha t' = v_0 + \frac{\alpha}{2} (t^2-t_0^2)$$
and then, integrating again and after some algebra,
$$x(t)=x_0 + v_0 (t-t_0) +\frac{\alpha}{6} (t-t_0)^2 (t+2 t_0) + v_0 (t-t_0)+x_0.$$
[EDIT:] Well, it's the same. So never mind ;-).

Both methods simplify to:
$$x(t)=x_0 + v_0(t-t_0) + \frac \alpha 6 \big [t^3 - 3t_0^2 t + 2t_0^3 \big ]$$

 

[annamal]

This may be an unnecessary post now, but since I’ve written it, here's an explanation of what (I think) went wrong with @annamal's logic/working.

$a = \alpha t$ gives $\frac {dv}{dt} = \alpha t$.
Integrating from $t=t_i$ to $t=t_f$ gives
$v_f - v_i = \frac{\alpha(t_f^2 – t_i^2)}{2}$

In the above equation, you can choose two quantities as (dependent and independent) variables - the other quantities are then treated as constants. You must then stick to these choices.

Assuming we are not interested in the effects of changing $\alpha$ our four options are:

a) $t_f$ and $v_f$ are variables; so we can rename them $t$ and $v(t)$:
$v(t) - v_i = \frac{\alpha(t^2 – t_i^2)}{2}$
$v(t) = v_i + \frac{\alpha(t^2 – t_i^2)}{2}$
(This is the conventional choice.)

b) $t_i$ and $v_i$ are variables; so we can rename them $t$ and $v(t)$:
$v_f - v(t) = \frac{\alpha(t_f^2 – t^2)}{2}$
$v(t) = v_f - \frac{\alpha(t_f^2 – t^2)}{2}$

c) $t_f$ and $v_i$ are variables; so we can rename them $t$ and $v(t)$:
$v_f - v(t) = \frac{\alpha(t^2 – t_i^2)}{2}$
$v(t) = v_f - \frac{\alpha(t^2 – t_i^2)}{2}$

d) $t_i$ and $v_f$ are variables; so we can rename them $t$ and $v(t)$:
$v(t) - v_i = \frac{\alpha(t_f^2 – t^2)}{2}$
$v(t) = v_i + \frac{\alpha(t_f^2 – t^2)}{2}$

Note that choice d) is @annamal's equation $v(t) = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$.

The meaning of v(t) in the equation is the final velocity as a function of of the start time. It is not the velocity as a function of time during the motion, So it makes no sense to integrate it to find the displacement.

Thanks. That was the solution. I derivated as d) instead of b)