Solved Problem on One-dimensional Motion

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

\[ \begin{gather} S_{A}=10t+3t^{2}\\ S_{B}=300-2t^{2} \end{gather} \]

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

Problem diagram:

Analyzing the equations given in the problem, we see that trucks 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 truck A starts from the origin, S_0A = 0, with initial speed v_0A = 10 m/s, and acceleration a_A = 6 m/s², the truck 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 truck A

\[ \begin{gather} v_{A}=v_{0A}+a_{A}t\\ v_{A}=10+6t \tag{I} \end{gather} \]

For truck 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 trucks equals, equating expressions (I) and (II)

\[ \begin{gather} |\;v_{A}\;|=|\;v_{B}\;|\\ |\;10+6t\;|=|\;-4t\;|\\ 10+6t=4t\\ -6t+4t=10\\ -2t=10\\t=\frac{10}{-2}\\ t=-5 \end{gather} \]

As there is no negative time, this result indicates that there is no t that satisfies the condition of the problem.

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