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 - Mathematics MCQ AIEEE 2004 2026 | ExamTest.live

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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 A B = 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

Solution & Step-by-step Explanation

Total distance traveled = A B + BC = 12 + 5 = 17 km . Total displacement (direct distance A C): Since East and North are perpendicular, A C = 1 2^{2} + 5^{2} = 13 km . Time from A to B (T_{1}) = \frac{distance}{speed} = 12/4 = 3 hours . Time from B to C (T_{2}) = 5/5 = 1 hour . Total time = T_{1} + T_{2} = 3 + 1 = 4 hours . Average speed = \frac{Total Distance}{Total Time} = 17/4 km/h . Average velocity = \frac{Total Displacement}{Total Time} = 13/4 km/h .

Try it yourself before checking the explanation above.

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 A B = 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

\frac{17}{4} km/h and \frac{13}{4} km/h

\frac{13}{4} km/h and \frac{17}{4} km/h

\frac{17}{9} km/h and \frac{13}{9} km/h

\frac{13}{9} km/h and \frac{17}{9} km/h

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