Vector Algebra
Addition of Vectors / Speed and Velocity
Grade None
Question:
<p>A particle moves towards east from a point <em>A</em> to a point <em>B</em> at the rate of 4 km/h and then towards north from <em>B</em> to <em>C</em> at the rate of 5 km/h. If \(AB = 12\) km and \(BC = 5\) km, then its average speed for its journey from <em>A</em> to <em>C</em> and resultant average velocity direct from <em>A</em> to <em>C</em> are, respectively,</p>
<p>\(\dfrac{17}{4}\) km/h and \(\dfrac{13}{4}\) km/h.</p>
<p>\(\dfrac{13}{4}\) km/h and \(\dfrac{17}{4}\) km/h.</p>
<p>\(\dfrac{17}{9}\) km/h and \(\dfrac{13}{9}\) km/h.</p>
<p>\(\dfrac{13}{9}\) km/h and \(\dfrac{17}{9}\) km/h.</p>
Step-by-Step Solution
Key Concept: Average speed is the total distance traveled divided by total time, while average velocity is the net displacement divided by total time. These are fundamentally different quantities—speed is scalar, velocity is vector.
Step 1: Calculate total distance and total time Distance AB = 12 km at 4 km/h → Time_1 = 12/4 = 3 hours Distance BC = 5 km at 5 km/h → Time_2 = 5/5 = 1 hour Total distance = 12 + 5 = 17 km Total time = 3 + 1 = 4 hours Step 2: Calculate average speed Average speed = Total distance / Total time = 17/4 = 4.25 km/h Step 3: Calculate net displacement (A to C) Since the particle moves east then north, these directions are perpendicular. The path forms a right triangle with: AC^2 = AB^2 + BC^2 = 12^2 + 5^2 = 144 + 25 = 169 AC = 13 km (net displacement) Step 4: Calculate average velocity Average velocity = Net displacement / Total time = 13/4 = 3.25 km/h ∴ Answer: Average speed = 4.25 km/h (or 17/4 km/h), Average velocity = 3.25 km/h (or 13/4 km/h)
Correct Answer: D