Problem 8 - Basic Engineering Mathematics

Question

Two cars, \(P\) and \(Q\), are travelling towards the junction of two roads which are at right angles to one another. Car \(P\) has a velocity of \(45 \mathrm{~km} / \mathrm{h}\) due east and car \(Q\) a velocity of \(55 \mathrm{~km} / \mathrm{h}\) due south. Calculate (i) the velocity of car \(P\) relative to \(\operatorname{car} Q\), and (ii) the velocity of car \(Q\) relative to car \(P\).

Step-by-Step Solution

Answer

The relative velocity magnitudes are both approximately 71.06 km/h.

Step 1: Define Velocities

Let's define the velocity vectors for cars \(P\) and \(Q\). Since car \(P\) moves due east, we have \(\vec{v}_P = 45 \hat{i} \text{ km/h}\). For car \(Q\), moving due south, we have \(\vec{v}_Q = -55 \hat{j} \text{ km/h}\). Here, \(\hat{i}\) and \(\hat{j}\) are unit vectors in the east and south directions, respectively.

Step 2: Calculate Relative Velocity of P with respect to Q

The relative velocity of car \(P\) with respect to car \(Q\) is given by \(\vec{v}_{PQ} = \vec{v}_P - \vec{v}_Q\). Substituting the values, we find \(\vec{v}_{PQ} = 45 \hat{i} - (-55 \hat{j}) = 45 \hat{i} + 55 \hat{j}\).

Step 3: Determine Magnitude of Relative Velocity of P with respect to Q

Use the Pythagorean theorem to calculate the magnitude: \(\lVert \vec{v}_{PQ} \rVert = \sqrt{45^2 + 55^2}\). Calculating, \(\lVert \vec{v}_{PQ} \rVert = \sqrt{2025 + 3025} = \sqrt{5050}\), which is approximately 71.06 km/h.

Step 4: Calculate Relative Velocity of Q with respect to P

The relative velocity of car \(Q\) with respect to car \(P\) is \(\vec{v}_{QP} = \vec{v}_Q - \vec{v}_P\). Substituting the values, \(\vec{v}_{QP} = -55 \hat{j} - 45 \hat{i}\), simplifying to \(-45 \hat{i} - 55 \hat{j} \).

Step 5: Determine Magnitude of Relative Velocity of Q with respect to P

The magnitude is the same as for \(\vec{v}_{PQ}\) because it involves the same speed, even though directionally opposite: \(\lVert \vec{v}_{QP} \rVert = \sqrt{45^2 + 55^2} = \sqrt{5050}\), also approximately 71.06 km/h.

Key Concepts

Vector Calculation

In physics, vectors are essential for understanding how objects move. A vector includes both a magnitude and a direction. Vectors can represent physical quantities like velocity, where both speed and direction matter. For example, knowing a car is traveling at 45 km/h is useful, but direction is missing. Is it heading east? North? That’s where vectors help! For two cars, P and Q, their velocities are represented as vectors:

• Car P's velocity: \( \vec{v}_P = 45 \hat{i} \text{ km/h} \), moving eastwards (hence it’s in the direction of unit vector \( \hat{i} \)).
• Car Q's velocity: \( \vec{v}_Q = -55 \hat{j} \text{ km/h} \), moving southwards (using \( \hat{j} \) with a negative sign because it heads south).

Calculating the relative velocity involves subtraction of one vector from another. If finding the velocity of car P relative to car Q: 1. Subtract Q’s velocity vector from P’s. 2. This yields: \( \vec{v}_{PQ} = \vec{v}_P - \vec{v}_Q = 45 \hat{i} - (-55 \hat{j}) = 45 \hat{i} + 55 \hat{j} \) In sum, factoring vectors adds precision, helping you grasp not just how fast, but in which direction you're moving relative to something else.

Pythagorean Theorem

The Pythagorean theorem is a powerful tool in geometry, especially for calculating vector magnitudes. It’s designed for right triangles and relates the squares of a right triangle's sides. For a right triangle with sides \(a\), \(b\), and hypotenuse \(c\), the relationship is \( c^2 = a^2 + b^2 \). In vector calculation, the Pythagorean theorem is used to find the magnitude of a resultant vector. When you calculate the relative velocity of cars P and Q, you apply this theorem to find how fast they’re moving concerning each other. Given \( \vec{v}_{PQ} = 45 \hat{i} + 55 \hat{j} \), the magnitude \( \lVert \vec{v}_{PQ} \rVert \) becomes: 1. Use \( \sqrt{45^2 + 55^2} \) to solve, as if creating a triangle with sides 45 and 55. 2. This results in \( \sqrt{2025 + 3025} = \sqrt{5050} \), roughly 71.06 km/h. Thus, thanks to this theorem, you can confidently determine actual speeds in multidirectional contexts.

Unit Vectors

Unit vectors are vectors with a magnitude of one. They indicate direction, serving as building blocks in vector mathematics. Commonly noted as \( \hat{i} \) and \( \hat{j} \) or \( \hat{k} \), each symbolizes a key axis direction in space.

• \( \hat{i} \) denotes the standard direction on the x-axis (east-west horizontal direction).
• \( \hat{j} \) represents the standard y-axis direction (north-south vertical direction).
• \( \hat{k} \) would indicate the z-axis for three-dimensional movements upward or downward, though not used in this specific problem.

By incorporating unit vectors, you can easily describe any vector's direction and simplify manipulation within calculations. For instance, in our scenario with cars P and Q:

• Car P moves east, corresponding to \( 45 \hat{i} \).
• Car Q moves south, expressed as \( -55 \hat{j} \), where negative shows movement opposite to the usual positive direction of \( \hat{j} \).

Due to this clear representation, adjusting vectors to analyze relative movements becomes intuitive and systematic.