Solved Problem on One-dimensional Motion

Two cars depart from distant points A and B, their motions are described by the following equations

\[ \begin{gather} S_{A}=2t^{2}\\ S_{B}=300-2t^{2} \end{gather} \]

units of the International System of Units (S.I.).
Determine the distance between cars, when the magnitudes of their velocities are the same.

Problem diagram:

Analyzing the equations given in the problem, we see that cars have acceleration. The equation of the displacement as a function of time with constant acceleration is given by

\[ \bbox[#99CCFF,10px] {S+S_{0}+v_{0}t+\frac{a}{2}t^{2}} \]

From the equations, we see that the car A starts from the origin, S_0A = 0, with initial speed v_0A = 0, and acceleration a_A = 4 m/s², the car B starts from the rest, v_0B = 0, from an initial position S_0B = 300 m and acceleration a_B = −4 m/s² (Figure 1).

Figure 1

Solution

The equation of velocity as a function of time is given by

\[ \bbox[#99CCFF,10px] {v=v_{0}+a t} \]

For car A

\[ \begin{gather} v_{A}=v_{0A}+a_{A}t\\ v_{A}=0+4t\\ v_{A}=4t \tag{I} \end{gather} \]

For car B

\[ \begin{gather} v_{B}=v_{0B}+a_{B}t\\ v_{B}=0-4t\\ v_{B}=-4t \tag{II} \end{gather} \]

Imposing the condition that the magnitude of velocities must be equal, we find the instant o time in which the speeds of the two cars equals, equating expressions (I) and (II)

\[ \begin{gather} |\;v_{A}\;|=|\;v_{B}\;|\\ |\;4t\;|=|-4t\;|\\ 4t=4t\\t=t \end{gather} \]

This result indicates that for any t the speeds of the cars will be the same, we can at all times calculate the distance setting \( \Delta S=|\;\Delta S_{A}-\Delta S_{B}\;|\).

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .