Solved Problem on One-dimensional Motion
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A motorcyclist is moving in the opposite direction of a reference frame. The magnitude of its initial speed is 25 m/s, at the initial time its position is −150 m, and the magnitude of deceleration is 2 m/s2. Determine:
a) The equation of displacement as a function of time;
b) The equation of velocity as a function of time;
c) The instant in which it passes through the origin of the reference frame;
d) The instant that its speed is zero.


Problem data:
  • Initial speed of the rider:    |v0| = 25 m/s;
  • Acceleration of the motorcyclist:    |a| = 2 m/s2;
  • Position in the initial instant:    S0 = −150 m.
Problem diagram:

We chose a reference frame oriented to the right.

Figure 1

Solution

a) The motorcyclist has acceleration, he is in Uniformly Accelerated Rectilinear Motion, given by
\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_0+v_0t+\frac{a}{2}\;t^2} \end{gather} \]
the initial position is already given in the problem, S0 = −150 m as the motorcyclist moves in the opposite direction to the orientation of the reference frame, its speed is negative (v<0), so we have v = −25 m/s. The motorcyclist is decelerating, its acceleration is in the opposite direction of velocity (a>0), thus a= 2 m/s2. The equation of displacement as a function of time
\[ \begin{gather} S=-150\;\mathrm m-\left(25\;\mathrm{\small{\frac{m}{s}}}\right)t+\frac{2\;\mathrm{\frac{m}{s^2}}}{2}\;t^2 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {S=-150-25t+t^2} \end{gather} \]

b) For the equation of velocity as a function of time is of type
\[ \begin{gather} \bbox[#99CCFF,10px] {v=v_0+at} \end{gather} \]
using the problem data
\[ \begin{gather} \bbox[#FFCCCC,10px] {v=-25t+2t} \end{gather} \]

c) When the motorcyclist passes through the origin, we have S = 0, substituting this value in the expression found in item (a)
\[ \begin{gather} 0=-150-25\;t+t^2 \end{gather} \]
This is a Quadratic Equation where the unknown is the value of time.

Solution of the Quadratic Equation   \( t^2-25\;t-150=0 \)
\[ \begin{gather} \Delta =b^{2}-4ac=(-25)^{2}-4\times 1\times (-150)=625+600=1225\\[10pt] t=\frac{-b\pm \sqrt{\Delta\;}}{2a}=\frac{-(-25)\pm\sqrt{1225\;}}{2\times 1}=\frac{25\pm 35}{2} \end{gather} \]
the two roots of the equation will be
\[ t_1=30\;\mathrm s\qquad \mathrm{e}\qquad t_2=-5\;\mathrm s \]

Since there is no negative time, we neglected the second root, it will go through the origin at t = 30 s (it starts the motion to the left of the origin, its speed decreases due to deceleration until it becomes zero, then changes direction and begins to move in the same direction of the reference frame until it passes through the origin).

d) When the speed is zero (v = 0), substituting this value in the expression found in item (b)
\[ \begin{gather} 0=-25\;\mathrm{\small{\frac{m}{s}}}+\left(2\;\mathrm{\small{\frac{m}{s^2}}}\right)t\\[5pt] \left(2\;\mathrm{\small{\frac{m}{s^2}}}\right)t=25\;\mathrm{\small{\frac{m}{s}}}\\[5pt] t=\frac{25\;\mathrm{\frac{\cancel m}{\cancel s}}}{2\;\mathrm{\frac{\cancel m}{s^\cancel 2}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {t=12.5\;\mathrm s} \end{gather} \]
This is the instant that the motorcyclist changes direction, and begins to move in the same direction of the frame of reference until passing through the origin.
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