A particle moves towards east from a point A to a point B at the rate of 4 km/h and then towards north from B to C at the rate of 5 km/h. If AB = 12 km and BC = 5 km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are respectively
- A. 17/4 km/h and 13/4 km/h
- B. 13/4 km/h and 17/4 km/h
- C. 17/9 km/h and 13/9 km/h
- D. 13/9 km/h and 17/9 km/h
Answer: A. 17/4 km/h and 13/4 km/h
Explanation: Time for AB = 12/4 = 3 h; time for BC = 5/5 = 1 h; total time = 4 h. Total path length = 12 + 5 = 17 km ⇒ Average speed = 17/4 km/h. Net displacement AC = √(12² + 5²) = 13 km ⇒ Magnitude of average velocity = 13/4 km/h.