Example 2. A point moves along a straight line with constant acceleration. Find the law of motion of the point.
Solution. Let's choose time as the independent variable, then the distance traveled and the speed of the point will be functions of time. Denote the distance traveled in time $t$ by $s$, the speed of the point by $v$, and the acceleration by $w$. According to the condition, $w=a$, where $a$ is a constant. Let the initial values of the speed and the distance be given, i.e., $v=v_{0}$ and $s=s_{0}$ at $t=0$, where $v_{0}$ and $s_{0}$ are some constants.

Since the acceleration of the point during rectilinear motion is the derivative of the speed with respect to time, according to the condition of the problem, we can write the following first-order differential equation with separable variables:

$$
\frac{d v}{d t}=a \text { or } d v=a d t
$$

Integrating the last equation, we get $v=a t+C_{1}$. The value of the constant $C_{1}$ will be determined by the initial conditions: $v=v_{0}$ at $t=0$. Substituting these values of $v$ and $t$ into the expression for $v$, we find:

$$
v_{0}=a \cdot 0+C_{1}, C_{1}=v_{0}
$$

Therefore, the speed of the point's motion is expressed by the formula

$$
v=a t+v_{0}
$$

Substituting this formula into the expression for the speed $v$ through the derivative of the distance with respect to time $v=\frac{d s}{d t}$, we obtain a new first-order differential equation with separable variables

$$
\frac{d s}{d t}=a t+v_{0} \text { or } d s=\left(a t+v_{0}\right) d t
$$

Integrating this equation, we find the general solution of the problem

$$
s=\frac{1}{2} a t^{2}+v_{0} t+C_{2}
$$

Since $s=s_{0}$ at $t=0$, then $C_{2}=s_{0}$, therefore,

$$
s=\frac{1}{2} a t^{2}+v_{0} t+s_{0} \text {. }
$$

In particular, assuming in the expressions for $v$ and $s$ that $a=g, v_{0}=0, s_{0}=0, s=h$, we get the law of motion for a body falling freely in a vacuum:

$$
v=g t, h=\frac{1}{2} g t^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-318.jpg?height=403&width=489&top_left_y=1199&top_left_x=42)

Fig. 9.2