The motion of a body is described by the displacement as a function of time
\[
\begin{gather}
S=-10+4 t
\end{gather}
\]
where the position is measured in kilometers and the time in hours. Find:
a) The initial position;
b) The speed;
c) The instant in which the body passes through the origin;
d) The position of the body at time 4 h.
Solution:
The function describing the displacement as a function of time is given by
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{S=S_0+vt}
\end{gather}
\]
with the following associations
\[
\begin{array}{c}
S & = & S_0 & + & v & t \\
& & \downarrow & & \downarrow & \\
S & = & -10 & + & 4 & t
\end{array}
\]
a) The initial position of the body
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{S_0=-10\;\mathrm{km}}
\end{gather}
\]
b) Speed of the body
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{v=4\;\mathrm{km/h}}
\end{gather}
\]
c) When the body passes the origin of the coordinate system, we have S = 0, substituting this value
in the given function
\[
\begin{gather}
0=-10\;\mathrm{km}+\left(4\mathrm{\small{\frac{km}{h}}}\right)t \\[5pt]
\left(4\mathrm{\small{\frac{km}{h}}}\right)t=10\;\mathrm{km} \\[5pt]
t=\frac{10\;\mathrm{\cancel{km}}}{4\mathrm{\frac{\cancel{km}}{h}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{t=2.5\;\mathrm{h}}
\end{gather}
\]
d) For t = 4 h the position will be
\[
\begin{gather}
S=-10\;\mathrm{km}+\left(4\;\mathrm{\small{\frac{km}{\cancel h}}}\right)(4\;\mathrm{\cancel h}) \\[5pt]
S=-10\;\mathrm{km}+16\;\mathrm{km}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{S=6\;\mathrm{km}}
\end{gather}
\]
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