Solved Problem on One-dimensional Motion

From two small towns, connected by a straight-line road 10 km long, run two carts, pulled each by a horse and walking at the speed of 5 km/h. At the moment of departure, a fly, which was on the forehead of the first horse, start to fly at a speed of 15 km/h and will land on the forehead of the second horse. After a negligible time depart again, with the same speed as before, toward the first horse, until landing on his forehead. It continues in this trip until the two horses meet and the fly dies crushed between the two foreheads. How many kilometers went through the fly?

Problem data:

Distance between the two cities: ΔS = 10 km;
Speed of the first horse: v₁ = 5 km/h;
Speed of the second horse: v₂ = 5 km/h;
Speed of the fly: v_f = 15 km/h.

Problem diagram:

We choose a frame of reference pointing to the right and origin in the position of horse 1 (Figure 1), the horse 2 is moving in the opposite direction of the reference frame.

Figure 1

Solution

We want to find the total distance traveled by the fly, S_f.
The first horse starts from the origin, initial position S₀₁ = 0, the positive sign in the speed indicates that it moves in the same direction of the reference frame. The second horse starts from a 10 km distant city, S₀₂ = 10 km, it moves in the opposite direction of the reference frame, its speed is negative, v₂ = −5 km/h. The fly is also departing from the origin, S_0f = 0, and it also moves in the same direction of the reference frame, and its speed is positive.
The fly will travel between the two horses until they meet. The first thing we have to determine is the time interval that the horses take until the meet. As the speeds of the horses are constant, we write the equation of the displacement as a function of time for each horse given by

\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+vt } \tag{I} \end{gather} \]

applying the expression (I) for each horse

\[ \begin{gather} S_{1}=S_{01}+v_{1}t\\ S_{1}=0+5t\\ S_{1}=5t \tag{II-a} \end{gather} \]

\[ \begin{gather} S_{2}=S_{02}+v_{2}t\\ S_{2}=10-5t \tag{II-b} \end{gather} \]

The two horses meet when they are in the same position, with the condition that expressions (II-a) and (II-B) are equal.

\[ \begin{gather} S_{1}=S_{2}\\ 5t=10-5t\\ 5t+5t=10\\ 10t=10\\ t=\frac{10}{10}\\ t=1\;\text{h} \tag{III} \end{gather} \]

Applying the expression (I) to the fly

\[ \begin{gather} S_{f}=S_{0f}+v_{f}t\\ S_{f}=0+15t\\ S_{f}=15t \tag{IV} \end{gather} \]

to know how many kilometers the fly traveled, we substitute the time interval that the horses take to meet, found in (IV), into the expression (V)

\[ \begin{gather} S_{f}=15.1\\ S_{f}=15\;\text{km} \end{gather} \]

The fly will run 15 km until it was crushed by the horses.

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