Multiple choice

A person walks along a straight road from his house to a market $2.5\ km$ away with a speed of $5\ km/hr$ and instantly turns back and reaches the house with a speed of $7.5\ km/hr$. The average speed of the person during the time interval $0$ to $50$ $minutes$ is (in $m/s$):

  1. $\displaystyle 4\frac{2}{3}$
  2. $\displaystyle\frac{5}{3}$
  3. $\displaystyle\frac{5}{6}$
  4. $\displaystyle\frac{1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total distance = 2.5 km out + 2.5 km back = 5 km. Time out = 2.5/5 = 0.5 hr = 30 min. Time back = 2.5/7.5 = 1/3 hr = 20 min. Total time = 50 min. Average speed = Total distance / Total time = 5 km / (50/60) hr = 6 km/hr. 6 km/hr = 6 * (1000/3600) m/s = 60/36 = 5/3 m/s.

AI explanation

The time taken to reach the market is 2.5 km divided by 5 km/hr, which equals 0.5 hours or 30 minutes. The return journey speed is 7.5 km/hr, and in the remaining 20 minutes (which is 1/3 of an hour), the person covers a distance of 7.5 multiplied by 1/3, equaling 2.5 km. Thus, the total distance traveled in the 50 minutes is 2.5 km to the market plus 2.5 km returning, totaling 5 km. The average speed is 5 km in 5/6 hours, which equals 6 km/hr; converting this to meters per second gives 6 multiplied by 5/18, resulting in 5/3 m/s.

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