Solved Problem on Kinematics
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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 Problem diagram

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

Sketch of the man walking on foot and by bus compared to the same route with a bicycle.
figure 1


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 (S.I.), we have
\[ \Delta t_{1}=5\;\cancel{\text{min}} \times \frac{60\;\text{s}}{1\;\cancel{\text{min}}}=300\;\text{s}\\ {}\\ \Delta t_{2}=40\;\cancel{\text{min}} \times \frac{60\;\text{s}}{1\;\cancel{\text{min}}}=2400\;\text{s}\\ {}\\ v_{2}=18\;\frac{\cancel{\text{km}}}{\cancel{\text{h}}} \times \frac{1000\;\text{m}}{1\;\cancel{\text{km}}} \times \frac{1\;\cancel{\text{h}}}{3600\;\text{s}}=\frac{18}{3,6}\;\frac{\text{m}}{\text{s}}=5\;\text{m/s} \]
The total displacement of the worker by bicycle (Δ S) will be the sum of the displacements by foot (Δ S1) and bus (Δ S2)
\[ \Delta S=\Delta S_{1}+\Delta S_{2} \tag{I} \]
the foot displacement is already given in the problem Δ S1 = 500 m, as the bus moves with constant speed, its speed coincides with the average speed, using the expression the average speed, we have
\[ v_{2}=\frac{\Delta S_{2}}{\Delta t_{2}}\\ \Delta S_{2}=v_{2}\;\Delta t_{2}\\ \Delta S_{2}=5 \times 2400\\ \Delta S_{2}=12000\;\text{m} \]
From expression (I) the total displacement will be
\[ \Delta S=600+12000\\ \Delta S=12600\;\text{m} \]
as the speed of the bicycle is constant, we use the expression of the average speed to know the total time of the route Δ t
\[ v_{3}=\frac{\Delta S}{\Delta t}\\ \Delta t=\frac{\Delta S}{v_{3}}\\ \Delta t=\frac{12600}{6}\\ \Delta t=2100\;\text{s} \]
converting this value to minutes, we have
\[ \Delta t=2100\;\cancel{\text{s}} \times \frac{1\;\text{min}}{60\;\cancel{\text{s}}}=35\;\text{min} \]
\[ \bbox[#FFCCCC,10px] {\Delta t=35\;\text{min}} \]


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