Solved Problem on Two-dimensional Motion and Relative Motion

Solved Problem on Relative Motion

On a windless rainy day, the rain falls vertically to the ground at a speed of 10 m/s. A car moves horizontally with a constant speed of 72 km/h relative to the ground.
a) What is the direction of the rain relative to the car?
b) What is the speed of the rain relative to the car?

Problem data:

Car speed relative to the ground: v_c = 72 km/h;
Rain speed relative to the ground: v_r = 10 m/s.

Problem diagram:

Figure 1

Solution

First, we must convert the car speed given in kilometers per hour (km/h) to meters per second (m/s) used in the International System of Units (SI)

\[ \begin{gather} v_c=72\;\frac{\;\mathrm{\cancel{km}}}{1\;\mathrm{\cancel{h}}}\times\frac{1\;\mathrm{\cancel h}}{3600\;\mathrm s}\times\frac{1000\;\mathrm m}{1\;\mathrm{\cancel{km}}}=\frac{72}{3.6}\;\mathrm{\frac{m}{s}}=20\;\mathrm{m/s} \end{gather} \]

a) The problem gives us legs \( {\vec v}_r \) and \( {\vec v}_c \) (Figure 1-A). We want to find the direction given by the angle θ, the tangent of the angle θ will be (Figure 1-B)

\[ \begin{gather} \tan \theta =\frac{\text{opposite leg}}{\text{adjacent leg}}=\frac{v_a}{v_c}=\frac{10\;\mathrm{\cancel{m/s}}}{20\;\mathrm{\cancel{m/s}}}=\frac{1}{2} \end{gather} \]

From the Trigonometry the angle θ will be the arc whose tangent is \( \frac{1}{2} \)

\[ \begin{gather} \theta =\arctan \left(\frac{1}{2}\right)=26.5° \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\theta=26.5°} \end{gather} \]

b) The magnitude of the velocity of the rain relative to the car is represented by \( {\vec v}_{r/c} \), (Figure 1-A). The magnitude of the velocity of the rain relative to the car will be obtained by applying the Pythagorean Theorem to the triangle in Figure 1-B. The magnitude of \( {\vec v}_{r/c} \) represents the hypotenuse of the triangle, which we want to find, and the magnitudes of \( {\vec v}_r \) and \( {\vec v}_c \) are the given legs

\[ \begin{gather} \bbox[#99CCFF,10px] {{\vec v}_r={\vec{v}}_{r/c}+{\vec v}_c} \end{gather} \]

the magnitude will be

\[ \begin{gather} v_{r/c}^2=v_a^2+v_c^2\\[5pt] v_{r/c}^2=(10\;\mathrm{m/s})^2+(20\;\mathrm{m/s})^2\\[5pt] v_{r/c}^2=100\;\mathrm{m^2/s^2}+400\;\mathrm{m^2/s^2}\\[5pt] v_{r/c}=\sqrt{500\;\mathrm{m^2/s^2}\;} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{r/c}=22.4\;\mathrm{m/s}} \end{gather} \]

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