Menu
FISICAEXE
Solved Problem on One-dimensional Motion
advertisement

A worker leaves his house and walks for 600 meters in 5 minutes to the bus stop, as soon as he reaches the bus stop he takes the bus and travels for 40 minutes at a constant speed of 18 kilometers per hour to the factory where he works. If he took the entire route by bike, at a constant speed of 6 meters per second. How long would it take from home to the factory? Answer in minutes.

Problem data:

  • Distance from the house to the bus stop: ΔS1 = 600 m;
  • Time interval from house to bus stop: Δt1} = 5 min;
  • Time interval from the bus stop to the factory: Δt2 = 40 min;
  • Speed of the bus: vo = 18 km/h;
  • Speed of the bicycle: v3 = 6 m/s.

Problem diagram:

We choose a frame of reference oriented to the right (Figure 1)

Figure 1

Solution:

Converting the given time intervals in minutes to seconds and the bus speed given in kilometers per hour to meters per second used in the International System of Units (SI)

\[ \begin{gather} \Delta t_1=5\;\cancel{\mathrm{min}}\times\frac{60\;\mathrm s}{1\;\cancel{\mathrm{min}}}=300\;\mathrm s \\[10pt] \Delta t_2=40\;\cancel{\mathrm{min}}\times\frac{60\;\mathrm s}{1\;\cancel{\mathrm{min}}}=2400\;\mathrm s \\[10pt] v_2=18\;\frac{\cancel{\mathrm{km}}}{\cancel{\mathrm h}}\times\frac{1000\;\mathrm m}{1\;\cancel{\mathrm{km}}}\times\frac{1\;\cancel{\mathrm h}}{3600\;\mathrm s}=\frac{18}{3,6}\;\frac{\mathrm m}{\mathrm s}=5\;\mathrm{m/s} \end{gather} \]

The total displacement of the worker by bicycle, ΔS, will be the sum of the displacements by foot, ΔS1, and bus, ΔS2

\[ \begin{gather} \Delta S=\Delta S_1+\Delta S_2 \tag{I} \end{gather} \]

the foot displacement is already given in the problem ΔS1 = 500 m, as the bus moves with onstant speed, its speed coincides with the average speed, using the expression the average speed

\[ \begin{gather} \bbox[background:#99CCFF; padding:10px; border-radius:10px;] {\bar v=\frac{\Delta S}{\Delta t}} \end{gather} \]
\[ \begin{gather} v_2=\frac{\Delta S_2}{\Delta t_2} \\[5pt] \Delta S_2=v_2\;\Delta t_2 \\[5pt] \Delta S_2=\left(5\;\mathrm{\small{\frac{m}{\cancel s}}}\right)\left(2400\;\mathrm{\cancel s}\right) \\[5pt] \Delta S_2=12000\;\mathrm m \end{gather} \]

From expression (I), the total displacement will be

\[ \begin{gather} \Delta S=600\;\mathrm m+12000\;\mathrm m \\[5pt] \Delta S=12600\;\;\mathrm m \end{gather} \]

as the speed of the bicycle is constant, we use the expression of the average speed to know the total time of the route Δt

\[ \begin{gather} v_3=\frac{\Delta S}{\Delta t} \\[5pt] \Delta t=\frac{\Delta S}{v_3} \\[5pt] \Delta t=\frac{12600\;\mathrm{\cancel m}}{6\;\mathrm{\frac{\cancel m}{s}}} \\[5pt] \Delta t=2100\;\text{s} \end{gather} \]

converting this value to minutes

\[ \begin{gather} \Delta t=2100\;\cancel{\mathrm s}\times\frac{1\;\mathrm{min}}{60\;\cancel{\mathrm s}}=35\;\mathrm{min} \end{gather} \]
\[ \begin{gather} \bbox[background:#FFCCCC; padding:10px; border-radius:10px;] {\Delta t=35\;\mathrm{min}} \end{gather} \]
advertisement