Problem 113 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A paticle moves towards east from a point \(A\) to a point \(B\) at the rate of \(4 \mathrm{~km} / \mathrm{h}\) and then towards north from \(B\) to \(\mathrm{C}\) at the rate of \(5 \mathrm{~km} / \mathrm{hr}\). If \(A B=12 \mathrm{~km}\) and \(B C=5 \mathrm{~km}\), then its average speed for its journey from \(A\) to \(C\) and resultant average velocity direct from \(A\) to \(C\) are respectively [2004] (a) \(\frac{13}{9} \mathrm{~km} / \mathrm{h}\) and \(\frac{17}{9} \mathrm{~km} / \mathrm{h}\) (b) \(\frac{13}{4} \mathrm{~km} / \mathrm{h}\) and \(\frac{17}{4} \mathrm{~km} / \mathrm{h}\) (c) \(\frac{17}{9} \mathrm{~km} / \mathrm{h}\) and \(\frac{13}{9} \mathrm{~km} / \mathrm{h}\) (d) \(\frac{17}{4} \mathrm{~km} / \mathrm{h}\) and \(\frac{13}{4} \mathrm{~km} / \mathrm{h}\)

Step-by-Step Solution

Answer

The average speed is \(\frac{17}{4} \text{ km/h}\) and average velocity is \(\frac{13}{4} \text{ km/h}\). Option (d).

Step 1: Calculate Total Distance

The total distance traveled from A to C is the sum of the distances AB and BC. Thus, it is calculated as: \( AB + BC = 12\, \text{km} + 5\, \text{km} = 17\, \text{km} \).

Step 2: Calculate Total Time

The time taken to travel from A to B is given by \( \text{Time}_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{12 \text{ km}}{4\, \text{km/h}} = 3\, \text{hours} \). Similarly, the time taken to travel from B to C is \( \text{Time}_2 = \frac{5 \text{ km}}{5 \text{ km/h}} = 1\, \text{hour} \). Thus, the total time is \( 3 + 1 = 4\, \text{hours} \).

Step 3: Calculate Average Speed

The average speed is calculated using the formula \( \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{17\, \text{km}}{4\, \text{hours}} = \frac{17}{4} \text{ km/h} \).

Step 4: Calculate Resultant Displacement

The displacement from A to C can be calculated using the Pythagorean theorem because AB and BC form a right-angle triangle at B. Thus, \( AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\, \text{km} \).

Step 5: Calculate Average Velocity

The average velocity is given by \( \text{Average Velocity} = \frac{\text{Displacement}}{\text{Total Time}} = \frac{13\, \text{km}}{4\, \text{hours}} = \frac{13}{4} \text{ km/h} \).

Key Concepts

Kinematics

In kinematics, we study how objects move through space over time. It's all about describing the motion, without focusing on why the motion happens. Some key aspects include the concepts of speed and velocity.

• Speed is a scalar quantity that refers to "how fast an object is moving." It's the rate at which distance is covered. For example, in the exercise, the speed of the particle traveling towards east is given as 4 km/h.
• Velocity, on the other hand, is a vector quantity. This means it has both magnitude and direction. While speed refers only to "how fast," velocity tells us "how fast and in which direction." The exercise asks for the average velocity of the particle traveling from point A to point C. Here, direction is factored in when finding the average velocity.

Understanding these terms is essential for calculating the average speed and velocity of the object in question, as seen in the solution steps.

Displacement Calculation

Displacement is different from the total distance traveled. It refers to the object's change in position from the initial point to the final point, along the shortest path between them.
This exercise involves calculating displacement using information about the path segments and the directions of motion. From the provided solution:

• First, we calculate the direct distance from start to end point using the Pythagorean theorem. This applies as the path from A to B, and then B to C, forms a right angle triangle at B.
• By recognizing AB as 12 km and BC as 5 km, we calculate AC, the direct path or displacement, as 13 km. The Pythagorean theorem states \( AC = \sqrt{AB^2 + BC^2} \).

The calculated displacement is then used to determine the average velocity, showing how displacement plays a critical role in this process.

Right-Angle Triangle

Triangles are one of the most useful geometric shapes in kinematic problems, especially the right-angle triangle. In this exercise, using the concept of the right-angle triangle formed by the particle's path helps us simplify the calculation of the displacement.

• A right-angle triangle is a triangle in which one angle is exactly 90 degrees. Formation of such a triangle occurs when an object moves in two perpendicular directions (like east to north).
• The sides of a right-angle triangle follow the Pythagorean theorem, \( a^2 + b^2 = c^2 \), where \( c \) represents the hypotenuse. In this exercise, \( AB \) and \( BC \) form the two perpendicular sides, with \( AC \) representing the hypotenuse.
• By applying this theorem, we could easily calculate the shortest path or displacement the particle covered from A to C, crucial for determining the average velocity.

By understanding the special nature of right-angle triangles, we can solve complex problems efficiently, as demonstrated in this exercise.