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:
Solutions · 1
Official solution
Total time T=412+55=3+1=4 h.
\[\text{Average speed} = \frac{12+5}4=\frac{17}4\text{ km/h},\qquad \text{Average velocity}=\frac{\sqrt{12^2+5^2}}4=\frac{13}4\text{ km/h}\]