Motion with Constant Acceleration

Definition: Motion with Constant Acceleration

In motion with constant or uniform acceleration the acceleration remains constant in time.

$$\mathbf{a}=\text{const}$$

Satz: Laws of Motion with Constant Acceleration

A point pass is initially ($t=0$) located at ${\mathbf{r}}_0$ and begins to move with initial velocity ${\mathbf{v}}_0$ and constant acceleration $\mathbf{a}$.

The position $\mathbf{r}$ and velocity $\mathbf{v}$ of the point mass at any subsequent moment $t$ are given by:

$$\mathbf{v}(t)={\mathbf{v}}_0+\mathbf{a}t$$ $$\mathbf{r}(t)={\mathbf{r}}_0+{\mathbf{v}}_{0}t+\frac{1}{2}\mathbf{a}{t}^2$$

Proof

Since acceleration is the derivative of velocity, we need to solve the following initial value problem in order to determine a formula for $\mathbf{v}(t)$:

$$\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=\mathbf{a}\mathbf{v}(0)=\mathbf{v}$$

Integrate both sides with respect to $t$.

$$\int \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}\mathrm{dt}=\int \mathbf{a}\mathrm{dt}$$ $$\mathbf{v}(t)=\mathbf{a}t+\mathbf{c}$$

We must now determine $\mathbf{c}$ so that $v(t)$ satisfies the initial condition $\mathbf{v}(0)={\mathbf{v}}_0$.

$$\mathbf{v}(0)=\mathbf{a}\cdot 0+\mathbf{c}={\mathbf{v}}_0⟹\mathbf{c}={\mathbf{v}}_0$$

We just proved the formula for the velocity.

$$\mathbf{v}(t)={\mathbf{v}}_0+\mathbf{a}t$$

To find the formula for the position, we need to solve another initial value problem.

$$\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}=\mathbf{v}(t)\mathbf{r}(0)={\mathbf{r}}_0$$ $$\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}={\mathbf{v}}_0+\mathbf{a}t$$

Integrate both sides with respect to $t$.

$$\int \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\mathrm{dt}=\int {\mathbf{v}}_0+\mathbf{a}t\mathrm{dt}$$ $$\mathbf{r}(t)=\frac{1}{2}\mathbf{a}{t}^2+{\mathbf{v}}_{0}t+\mathbf{c}$$

We must now find the constant $\mathbf{c}$ so that it satisfies the initial condition $\mathbf{r}(0)=\mathbf{r}$.

$$\mathbf{r}(0)=\frac{1}{2}\mathbf{a}\cdot 0^2+{\mathbf{v}}_0\cdot 0+\mathbf{c}={\mathbf{r}}_0⟹\mathbf{c}={\mathbf{r}}_0$$

Thus

$$\mathbf{r}(t)={\mathbf{r}}_0+{\mathbf{v}}_{0}t+\frac{1}{2}\mathbf{a}{t}^2$$