A person travels from his house to his office in 30 minutes but he takes 45 minutes to travel from his office to house via the same route. What is the ratio of the average velocity while going and the average velocity while returning ?<\/p>\n
[amp_mcq option1=”1″ option2=”2\/3″ option3=”2″ option4=”3\/2″ correct=”option4″]<\/p>\n
When going from house to office:
\nLet the house be the origin (displacement = 0) and the office be at a displacement D (assuming a straight line path for simplicity, though the problem just says “same route” implying the distance is the same).
\nTime taken ($t_{going}$) = 30 minutes = 0.5 hours.
\nDisplacement while going = D.
\nAverage velocity while going ($v_{going}$) = Displacement \/ Time = D \/ 0.5 = 2D. (Taking direction from house to office as positive).<\/p>\n
When returning from office to house:
\nThe person travels back to the origin.
\nTime taken ($t_{returning}$) = 45 minutes = 0.75 hours.
\nDisplacement while returning = -D (from office position D back to house position 0).
\nAverage velocity while returning ($v_{returning}$) = Displacement \/ Time = -D \/ 0.75 = -4D\/3. (Taking direction from office to house as negative).<\/p>\n
The question asks for the ratio of the average velocity while going and the average velocity while returning. The options are positive numbers, implying the ratio of the magnitudes of the average velocities (average speeds for each leg, since distance = |displacement|).
\nMagnitude of $v_{going}$ = $|2D| = 2D$.
\nMagnitude of $v_{returning}$ = $|-4D\/3| = 4D\/3$.<\/p>\n
Ratio of magnitudes of average velocities = $|v_{going}| \/ |v_{returning}|$ = $(2D) \/ (4D\/3)$
\n$= 2 \\times (3 \/ 4) = 6 \/ 4 = 3 \/ 2$.<\/p>\n