Motion with Constant Acceleration in 1D

Theorem: Motion with Constant Acceleration in 1D

In one-dimensional Motion with Constant Acceleration the Position, Velocity and Acceleration vectors become scalars $r(t),v(t),a$.

In addition to the general laws of Motion with Constant Acceleration, in one dimension we also have:

$$v_{\text{avg}}(t)=\frac{v(t)+v_0}{2}$$ $$r(t)=r_0+t\cdot v_{\text{avg}}(t)$$ $$v^2=v_0^2+2a(r-r_0)$$

The last law makes it possible to compute the Velocity $v=v(t)$ of a Point Mass as a function of its Position $r=r(t)$.

Proof: First Law

$$v_{\text{avg}}(t)=\frac{r(t)-r(0)}{t-0}=\frac{r(t)-r_0}{t}=\frac{r_0+v_{0}t+\frac{1}{2}a{t}^2-r_0}{t}=v_0+\frac{1}{2}at$$

We add $v_0$ to both sides of the following equation and then multiply by $2$.

$$v_{\text{avg}}(t)=v_0+\frac{1}{2}at$$ $$v_{\text{avg}}(t)+v_0=2v_0+\frac{1}{2}at$$ $$2v_{\text{avg}}(t)+2v_0=4v_0+at$$ $$2v_{\text{avg}}(t)+2v_0=3v_0+(v_0+at)=3v_0+v(t)$$ $$2v_{\text{avg}}(t)=v_0+v(t)$$ $$v_{\text{avg}}(t)=\frac{v(t)+v_0}{2}$$

Proof: Second Law

$$r(t)=r_0+v_{0}t+\frac{1}{2}a{t}^2=r_0+t\left(v_0+\frac{1}{2}at\right)=r_0+t\cdot v_{\text{avg}}(t)$$

For an explanation for the last step, see the proof of the first law.

Beweis: Third Law

Rearrange the law for the Velocity:

$$t=\frac{v(t)-v_0}{a}$$

Susbtitute for $t$ in the law for Position:

$$r(t)=r_0+v_0\frac{v(t)-v_0}{a}+\frac{1}{2}a{\left(\frac{v(t)-v_0}{a}\right)}^2$$ $$2ar(t)=2a{r}_0+2v_0(v(t)-v_0)+(v(t)-v_0{)}^2$$ $$2a(r(t)-r_0)=2v_{0}v(t)-2v_0^2+v(t{)}^2-2v_{0}v(t)+v_0^2$$ $$2a(r(t)-r_0)=v(t{)}^2-v_0^2$$ $$v(t)=2a(r(t)-r_0)+v_0^2$$