Solved Problem on One-dimensional Motion

A car starts a motion from the rest in a straight line. The car runs 100 m and 120 m in two successive seconds. Determine the acceleration of motion.

Problem data:

Initial speed of the car: v₀ = 0;
Distance traveled between t and (t+1) seconds: S₂ − S₁ = 100 m;
Distance traveled between (t+1) and (t+2) seconds: S₃ − S₂ = 120 m.

Problem diagram:

We choose a reference frame pointing to the right. The car starts from the origin, S₀ = 0.
After t seconds the start, the car runs a distance from S meters. Then in the time interval of 1 second, between t and (t+1) seconds, it travels 100 meters reaching the position (S+100) meters.

Figure 1

In the next time interval of 1 second, between t+1 and t+2 seconds, it travels the space of 120 meters, reaching the position S+220 meters.

Solution

The equation of displacement as a function of time with constant acceleration is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\frac{a}{2}t^{2}} \tag{I} \end{gather} \]

Writing this equation for the movement of the car between origin S₀ = 0 and the point S

\[ \begin{gather} S=0+0.t+\frac{a}{2}t^{2}\\ S=\frac{a}{2}t^{2} \tag{II} \end{gather} \]

Writing the expression (I) for the movement of the car between the origin S₀ = 0 and the point S+100

\[ \begin{gather} S+100=0+0.t+\frac{a}{2}(t+1)^{2}\\ S+100=\frac{a}{2}(t+1)^{2} \tag{III} \end{gather} \]

Writing the expression (I) for the movement of the car between the origin S₀ = 0 and the point S+220

\[ \begin{gather} S+220=0+0.t+\frac{a}{2}(t+2)^{2}\\ S+220=\frac{a}{2}(t+2)^{2} \tag{IV} \end{gather} \]

Equation (II), (III), and (IV) can be written as a system of three equations to three unknowns S, a and t

\[ \begin{gather} \left\{ \begin{array}{l} S=\dfrac{a}{2}t^{2}\\ S+100=\dfrac{a}{2}(t+1)^{2}\\ S+220=\dfrac{a}{2}(t+2)^{2} \end{array} \right. \tag{V} \end{gather} \]

subtracting the first equation of the second in the system (V)

\[ \begin{gather} \quad \cancel{S}+100=\frac{a}{2}(t+1)^{2}\\ \frac{\text{(-)} \qquad\quad \cancel{S}=\dfrac{a}{2}t^{2} \qquad\quad}{100=\dfrac{a}{2}(t+1)^{2}-\dfrac{a}{2}t^{2}} \end{gather} \]

factoring the term \( \frac{a}{2} \) on the right-hand side of the equation

\[ 100=\frac{a}{2}\left[(t+1)^{2}-t^{2}\right] \]

From Special Binomial Products

\[ (a+b)^{2}=a^{2}+2ab+b^{2} \]

expanding the first term in brackets

\[ \begin{gather} 100=\frac{a}{2}\left[t^{2}+2t+1-t^{2}\right]\\ 100=\frac{a}{2}\left(2t+1\right) \tag{VI} \end{gather} \]

Subtracting the second equation of the third in the system (V)

\[ \begin{gather} \qquad\; \cancel{S}+220=\frac{a}{2}(t+2)^{2} \qquad \\ \frac{\text{(-)} \quad \cancel{S}+100=\dfrac{a}{2}(t+1)^{2} \qquad}{120=\dfrac{a}{2}(t+2)^{2}-\dfrac{a}{2}(t+1)^{2}} \end{gather} \]

factoring the term \( \frac{a}{2} \) on the right-hand side of the equation

\[ 120=\frac{a}{2}\left[(t+2)^{2}-(t+1)^{2}\right] \]

the two terms between brackets are Special Binomial Products as the same used above

\[ \begin{gather} 120=\frac{a}{2}\left[t^{2}+2.2t+2^{2}-\left(t^{2}+2t+1\right)\right]\\ 120=\frac{a}{2}\left[t^{2}+4t+4-t^{2}-2t-1\right]\\ 120=\frac{a}{2}\left(2t+3\right) \tag{VII} \end{gather} \]

Expressions (VI) and (VII) can be written as a system of two equations to two unknowns, a e t

\[ \left\{ \begin{matrix} 100=\dfrac{a}{2}\left(2t+1\right)\\ 120=\dfrac{a}{2}\left(2t+3\right) \end{matrix} \tag{VIII} \right. \]

subtracting the first equation of the second in the system (VIII)

\[ \frac{\left. \begin{matrix} \quad\quad 120=\dfrac{a}{2}\left(2t+3\right)\\ (-) \quad 100=\dfrac{a}{2}\left(2t+1\right) \end{matrix} \quad \right.}{20=\dfrac{a}{2}(2t+3)-\dfrac{a}{2}(2t+1)} \]

factoring the term \( \frac{a}{2} \) on the right-hand side of the equation

\[ \begin{gather} 20=\frac{a}{2}\left[(2t+3)-(2t+1)\right]\\ 20=\frac{a}{2}\left[2t+3-2t-1\right]\\ 20=\frac{a}{\cancel{2}}.\cancel{2} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {a=20\;\text{m/s}^{2}} \]

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