The simplest case in kinematics is that of a particle moving with constant acceleration, either zero or nonzero. The goal of this section is to derive the equations giving the velocity and position of the particle as a function of time.

We consider a particle moving along the $x$-axis with an initial velocity $v_{0x}$ starting at location $x_0$ at $t=0$.

Case 1: acceleration is zero: $a_x=0\ m/s^2$

The velocity of the particle is then constant and equal to its initial velocity $v_{0x}$ i.e.

v_x\left(t\right)=v_{0x}

The position $x\left(t\right)$ of the particle is then a linear function of time given by

v_x\left(t\right)=v_{0x}t+x_0

where $x_0$ denotes the initial position of the particle at time $t=0$.

In this case, the velocity $v_x\left(t\right)$ is constant and the position $x\left(t\right)$ varies linearly as shown in the graphs below.

Case 2: acceleration is a non-zero constant: $a_x\ne 0$

The velocity $v_x\left(t\right)$ of the particle is then a linear function of time given by

v_x\left(t\right)=a_xt+v_{0x}

where $a_x$ is the constant acceleration of the particle and $v_{0x}$ is the initial velocity of the particle at $t=0$.

The position $x\left(t\right)$ of the particle is then a quadratic function of time given by

x\left(t\right)=\frac{1}{2}a_xt^2+v_{0x}t+x_0

where $x_0$ denotes the initial position of the particle at time $t=0$.

In this case, the velocity $v_x\left(t\right)$ varies linearly and the position $x\left(t\right)$ varies quadratically as shown in the graphs below.

Case 3: a third useful formula derived from $x\left(t\right)$ and $v_x\left(t\right)$

This common substitution allows for a relationship between the speed of the particle and its displacement $\mathrm{\Delta }x=x\left(t\right)-x_0$ that is independent of time.

We consider a particle that starts at time $t_i$ at a position $x_0$ with an initial velocity $v_{0x}$ and reaches at time $t_f$ the position $x_f$ with a final velocity $v_{fx}$. Therefore, we can express $t_f$ as follows

v_{fx}=a_xt_f+v_{0x}\ \ \ \ \Rightarrow \ \ \ \ \ t_f=\frac{v_{fx}-v_{0x}}{a_x}

Substituting into $x\left(t\right)$ then yields

\begin{aligned} x\left(t_f\right)=\frac{1}{2}a_xt^2_f+v_{0x}t_f+x_0\ \ \ \ &\Rightarrow \ \ \ \ x_f-x_0=\frac{1}{2}a_xt^2_f+v_{0x}t_f \\ \\ &\Rightarrow \ \ \ \ \ x_f-x_0=\frac{1}{2}a_x{\left(\frac{v_{fx}-v_{0x}}{a_x}\right)}^2+v_{0x}\left(\frac{v_{fx}-v_{0x}}{a_x}\right) \\ \\ &\Rightarrow \ \ \ \ \ x_f-x_0=\frac{v^2_{fx}+v^2_{0x}-2v_{0x}v_{fx}}{2a_x}+\frac{v_{0x}v_{fx}-v^2_{0x}}{a_x} \\ \\ &\Rightarrow \ \ \ \ \ x_f-x_0=\frac{v^2_{fx}-v^2_{0x}}{2a_x} \\ \\ &\Rightarrow \ \ \ \ \ v^2_{fx}=v^2_{0x}+2a_x\left(x_f-x_0\right) \\ \\ &\Rightarrow \ \ \ \ \ v^2_{fx}=v^2_{0x}+2a_x\mathrm{\Delta }x \end{aligned}

For an object moving in a direction $x$ over a distance $\mathrm{\Delta }x$ with an acceleration $a_x$, you may write

v^2_{fx}=v^2_{0x}+2a_x\mathrm{\Delta }x

Summary of formulas:

The formulas derived above are summarized below as they are fundamental formulas for one-dimensional kinematics with constant acceleration.

\boxed{ \begin{aligned} x\left(t\right)&=\frac{1}{2}a_xt^2+v_{0x}t+x_0 \\ \\ v_x\left(t\right)&=a_xt+v_{0x} \\ \\ v^2_{fx}&=v^2_{0x}+2a_x\mathrm{\Delta }x \end{aligned} }

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