Problem 66 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Two cars approach a street corner at right angles to each other (sce Fig. \(35 ) .\) 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 scen by car 2\(?\) What is the velocity of car 2 relative to car 1\(?\)

Step-by-Step Solution

Answer

The relative velocity magnitude is approximately 57.00 km/h for both cars as seen by each other.

Step 1: Understand the Problem

We have two cars moving towards a corner from perpendicular directions. Car 1 moves at 35 km/h and Car 2 at 45 km/h. We need to determine the relative velocities between the two cars.

Step 2: Define Relative Velocity

Relative velocity is the velocity of one object as observed from another moving object. If car 1's velocity is \(\mathbf{v}_1\) and car 2's velocity is \(\mathbf{v}_2\), the relative velocity of car 1 with respect to car 2 is given by \(\mathbf{v}_{1,2} = \mathbf{v}_1 - \mathbf{v}_2\) and vice versa for car 2 with respect to car 1.

Step 3: Represent Velocities as Vectors

Assign a coordinate system: Let Car 1 move along the positive x-axis, and Car 2 along the positive y-axis. Thus, \(\mathbf{v}_1 = 35 \hat{i}\) km/h and \(\mathbf{v}_2 = 45 \hat{j}\) km/h.

Step 4: Calculate Relative Velocity of Car 1 as Seen by Car 2

The relative velocity of car 1 with respect to car 2 is given by \(\mathbf{v}_{1,2} = \mathbf{v}_1 - \mathbf{v}_2 = 35 \hat{i} - 45 \hat{j}\).

Step 5: Calculate Magnitude of Relative Velocity of Car 1 as Seen by Car 2

To find the magnitude, use the Pythagorean theorem: \[\|\mathbf{v}_{1,2}\| = \sqrt{(35)^2 + (-45)^2} = \sqrt{1225 + 2025} = \sqrt{3250} \approx 57.00\, \text{km/h}.\]

Step 6: Calculate Relative Velocity of Car 2 as Seen by Car 1

Similarly, the relative velocity of car 2 with respect to car 1 is \(\mathbf{v}_{2,1} = \mathbf{v}_2 - \mathbf{v}_1 = 45 \hat{j} - 35 \hat{i}\) km/h.

Step 7: Calculate Magnitude of Relative Velocity of Car 2 as Seen by Car 1

The magnitude is the same as before (because subtraction is commutative, just with opposite signs in components): \[\|\mathbf{v}_{2,1}\| = \sqrt{(-35)^2 + 45^2} = \sqrt{1225 + 2025} = \sqrt{3250} \approx 57.00\, \text{km/h}.\]

Key Concepts

Understanding Vector Representation

In physics, when we talk about vector representation, we are looking into how we can express quantities that have both magnitude and direction. Velocity is one such quantity. Cars moving towards a corner can be perfectly represented using vectors. This is helpful because vectors help us understand not just how fast something is moving, but also in which direction.
In the given exercise, each car's velocity is represented as a vector:

• Car 1's velocity vector is expressed in the form \( \mathbf{v}_1 = 35 \hat{i} \) km/h. Here \( \hat{i} \) indicates the direction along the x-axis (say eastward).
• Car 2's velocity vector is \( \mathbf{v}_2 = 45 \hat{j} \) km/h, where \( \hat{j} \) points towards the y-axis (say northward).

These vectors are not just lines with arrows, but mathematical representations that allow us to perform operations like addition and subtraction easily. In relative velocity problems, such operations become crucial as they let us find out velocities of objects as observed from one another.

Using a Coordinate System

A coordinate system is essentially an arrangement that helps you plot or define positions and directions. Imagine it like a map where you need to know which way is north or east to navigate. In physics, we assign directions to the axes commonly known as x (horizontal) and y (vertical). This gives us a clear reference point.
When solving the exercise, placing Car 1's movement along the x-axis and Car 2's along the y-axis simplifies calculations.

• Car 1's movement in positive x-direction corresponds to its velocity vector being placed along the x-axis.
• Similarly, Car 2's motion is strictly along the y-direction.

This perpendicular alignment is what helps apply vector math effectively, as we can respect both direction and magnitude when doing calculations. As a result, it's easier to find the relative velocities by just subtracting these vectors.

Applying the Pythagorean Theorem

The Pythagorean theorem is a principle from geometry that helps us relate the sides of right-angled triangles. It states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. It's usually noted as: \[ c^2 = a^2 + b^2 \].
In the context of our exercise, the relative velocity vector of one car as seen from the other forms a right-angled triangle with the individual velocities. To find the magnitude of this relative velocity:

• The formula applied is \[ \|\mathbf{v}_{1,2}\| = \sqrt{(35)^2 + (-45)^2} \], which provides \(57.00\) km/h as calculated.
• This same formula helps find the magnitude of the relative velocity regardless of the perspective since the main difference is the direction, not the magnitude.

Therefore, the Pythagorean theorem becomes a simple yet powerful tool for transforming vector components into tangible velocities in real-world problems.