A motorcyclist moves in the opposite direction of an oriented frame of reference, the magnitude of its speed
is 40 m/s, and initially, his position is −150 m. Find:
a) The displacement as a function of time;
b) In what time he passes the origin.
Problem data:
- Speed of the motorcyclist: |v| = 40 m/s;
- Position at initial time: S0 = −150 m.
Problem diagram:
We choose a reference frame with a positive direction oriented to the right.
Solution
a) We want to find a function of the type
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S=S_0+vt}
\end{gather}
\]
the initial position is given, in the problem
S0 = −150 m, as the motorcyclist moves
in opposite direction of the reference frame, its speed is negative (v<0), we have
v = −40 m/s. The function for displacement will be
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{S=-150-40t}
\end{gather}
\]
b) When the motorcyclist passes through the origin, we have
S = 0, substituting this value in the
expression found in item (a)
\[
\begin{gather}
0=-150\;\mathrm m-\left(40\;\mathrm{\small{\frac{m}{s}}}\right)t\\[5pt]
\left(40\;\mathrm{\small{\frac{m}{s}}}\right)t=-150\;\mathrm m\\[5pt]
t=-\frac{150\;\mathrm{\cancel m}}{40\;\mathrm{\frac{\cancel m}{s}}}\\[5pt]
t=-3.75\;\mathrm{s}
\end{gather}
\]
As there is no negative time, this means that the motorcyclist
does not pass through the origin of the reference frame
(he starts the motion to the left of the origin and moves to the left away from the origin).