Body under Constant Acceleration

Theorem

Let $B$ be a body under constant acceleration $a$.

Then the following equations apply:

• $\mathbf v = \mathbf u + \mathbf a t$

• $\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$

• $\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$

where:

• $\mathbf u$ is the velocity at time $t = 0$
• $\mathbf v$ is the velocity at time $t$
• $\mathbf s$ is the displacement of $B$ from its initial position at time $t$
• $\cdot$ denotes the scalar product.

Proof

$B$ has acceleration $\mathbf a$, so if we let $\mathbf x$ be the vector corresponding to the position of $B$ at time $t$, we have $\frac{d^2\mathbf x}{dt^2}=\mathbf a$.

Solving this differential equation, we obtain $\mathbf x = \mathbf c_0 + \mathbf c_1t + \frac{1}{2} \mathbf at^2$, with $\mathbf c_0$ and $\mathbf c_1$ constant vectors.

Evaluating $\mathbf x$ at $t=0$ shows that $\mathbf c_0$ is the value $\mathbf x_0$ of $\mathbf x$ at time $t=0$.

Taking the derivative of $\mathbf x$ at $t=0$ shows that $\mathbf c_1$ corresponds to $\mathbf u$.

Therefore, since $\mathbf s = \mathbf x - \mathbf x_0$, we have $\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$, and by taking the derivative of $\mathbf x$, we have $\mathbf v = \mathbf u + \mathbf a t$.

Next, we dot $\mathbf v$ into itself using the previous statement.

From the linearity and commutativity of the dot product, we have $\mathbf v\cdot \mathbf v =(\mathbf u + \mathbf a t)\cdot(\mathbf u + \mathbf a t) = \mathbf u \cdot\mathbf u + 2 \mathbf u\cdot \mathbf a t + t^2\mathbf a\cdot \mathbf a$.

Again using the linearity of the dot product, we can factor out $\mathbf a$ to obtain $\mathbf v\cdot \mathbf v =\mathbf u \cdot\mathbf u + \mathbf a \cdot (2 \mathbf u t + t^2\mathbf a)$.

The expression in parentheses is $2\mathbf s$, so we have $\mathbf v\cdot \mathbf v =\mathbf u \cdot\mathbf u +2 \mathbf a \cdot \mathbf s$ and our proof is complete.

$\blacksquare$