Problem 66 - Physics for Scientists and Engineers with Modern Physics

Question

Two cars approach a street corner at right angles to each other (see Fig. 3-35). Car 1 travels at \(35 \mathrm{~km} / \mathrm{h}\) and car 2 at \(45 \mathrm{~km} / \mathrm{h}\). What is the relative velocity of car 1 as seen by car 2 ? What is the velocity of car 2 relative to car \(1 ?\)

Step-by-Step Solution

Answer

The relative velocity of Car 1 seen by Car 2 is approximately 57.0 km/h at an angle. The velocity of Car 2 relative to Car 1 is the same magnitude but in the opposite direction.

Step 1: Understanding the Concept of Relative Velocity

The problem involves determining the velocity of one object relative to another. This requires calculating the relative velocity by considering velocities of the two objects involved. The formula for relative velocity of object A as seen by object B is given by \( \vec{v}_{A/B} = \vec{v}_A - \vec{v}_B \). Since both vehicles travel perpendicular to each other, the velocity subtraction needs to be done using vector calculation.

Step 2: Setting Up Relative Velocity Equations

Consider Car 1 moving along the x-axis at \(35 \text{ km/h} \) and Car 2 moving along the y-axis at \(45 \text{ km/h} \). The velocity of Car 1 is \( \vec{v}_1 = 35 \hat{i} \text{ km/h}\) and that of Car 2 is \( \vec{v}_2 = 45 \hat{j} \text{ km/h}\), where \( \hat{i} \) and \( \hat{j} \) are unit vectors along the x and y-axes respectively.

Step 3: Calculating Relative Velocity of Car 1 as Seen by Car 2

The relative velocity of Car 1 as seen by Car 2 is calculated using:\[ \vec{v}_{1/2} = \vec{v}_1 - \vec{v}_2 \]Substituting the velocity vectors, we have:\[ \vec{v}_{1/2} = 35 \hat{i} - 45 \hat{j} \text{ km/h} \]This indicates a vector with components along both axes.

Step 4: Calculating Magnitude of the Relative Velocity

To find the magnitude of the relative velocity of Car 1 as seen by Car 2, use the Pythagorean theorem:\[ |\vec{v}_{1/2}| = \sqrt{(35^2 + 45^2)} \text{ km/h} \]Calculating gives:\[ |\vec{v}_{1/2}| = \sqrt{1225 + 2025} \text{ km/h} = \sqrt{3250} \text{ km/h} \approx 57.0 \text{ km/h} \]

Step 5: Determining Velocity of Car 2 Relative to Car 1

The velocity of Car 2 relative to Car 1 is given by:\[ \vec{v}_{2/1} = \vec{v}_2 - \vec{v}_1 \]Substituting the velocity vectors gives:\[ \vec{v}_{2/1} = 45 \hat{j} - 35 \hat{i} \text{ km/h} \]Notice that this results in a vector of similar magnitude in the opposite direction.

Step 6: Magnitude of Relative Velocity of Car 2 Relative to Car 1

Since velocities are vector quantities, the magnitude of \( \vec{v}_{2/1} \) is the same as \( \vec{v}_{1/2} \). Hence, the magnitude is:\[ |\vec{v}_{2/1}| = \sqrt{3250} \text{ km/h} \approx 57.0 \text{ km/h} \]

Key Concepts

Vector Subtraction

Vector subtraction is an essential concept when dealing with relative velocities. It involves subtracting one vector from another to find the difference between them. In the context of this problem, determining the relative velocity of two cars requires the subtraction of their velocity vectors. This is expressed through the formula:

• For Car 1 relative to Car 2: \( \vec{v}_{1/2} = \vec{v}_1 - \vec{v}_2 \)
• For Car 2 relative to Car 1: \( \vec{v}_{2/1} = \vec{v}_2 - \vec{v}_1 \)

By using vector subtraction, you essentially take into account the direction and magnitude of both velocities. When Car 1, traveling at 35 km/h along the x-axis, and Car 2, at 45 km/h along the y-axis, are involved, you subtract component-wise. Therefore, \( \vec{v}_{1/2} = 35 \hat{i} - 45 \hat{j} \) km/h, resulting in a single vector that represents the relative velocity between the two cars.

Pythagorean Theorem

The Pythagorean theorem is a valuable tool when calculating vector magnitudes, particularly in right-angled scenarios. Given the right-angle paths of Car 1 and Car 2, this theorem helps us find the magnitude of the relative velocity vectors. Once the vectors \( 35 \hat{i} - 45 \hat{j} \) km/h and \( 45 \hat{j} - 35 \hat{i} \) km/h are established, we calculate the magnitude using:\[ |\vec{v}_{1/2}| = \sqrt{(35^2 + 45^2)} \text{ km/h} \]Applying the Pythagorean theorem \( \sqrt{a^2 + b^2} \), where \( a = 35 \) and \( b = 45 \), you determine the magnitude of Car 1's velocity relative to Car 2 as \( 57.0 \text{ km/h} \). This method not only provides the size of the resultant vector but also confirms the symmetry between the two relative velocity descriptions: │\vec{v}_{1/2}│ and │\vec{v}_{2/1}│ both have the same magnitude.

Velocity Vectors

Velocity vectors are fundamental in describing motion. They provide information about the direction and speed of moving objects. For Car 1 and Car 2, each is moving in a straight line at a constant speed, hence vector representation is used.

• Car 1's vector is \( \vec{v}_1 = 35 \hat{i} \text{ km/h} \)
• Car 2's vector is \( \vec{v}_2 = 45 \hat{j} \text{ km/h} \)

Here, \( \hat{i} \) and \( \hat{j} \) are unit vectors indicative of their respective axis directions. Understanding these vectors is crucial because the vector notation clearly indicates both speed and direction — Car 1 moves horizontally, while Car 2 moves vertically. When comparing their velocities, it's their vector difference that reveals how each car 'sees' the other as moving with respect to themselves. This concept is central to solving problems involving multiple moving vehicles, as in this exercise, providing a clear picture of their movement dynamics.